Extensions of semigroups by symmetric inverse semigroups of a bounded finite rank

TL;DR

This paper investigates the algebraic and topological properties of semigroup extensions formed by symmetric inverse semigroups of bounded finite rank, describing their structure, regularity, and topological extensions.

Contribution

It provides a detailed description of the structure, regularity, and Green's relations of these semigroup extensions, and introduces the concept of strongly tight ideal series.

Findings

Semigroup extensions are regular, orthodox, inverse, or stable iff the base semigroup has these properties.

Green's relations are characterized for these extensions.

Existence of unique compact topological extensions for compact Hausdorff monoids.

Abstract

We study the semigroup extension $\mathscr{I}_\lambda^n(S)$ of a semigroup $S$ by symmetric inverse semigroups of a bounded finite rank. We describe idempotents and regular elements of the semigroups $\mathscr{I}_\lambda^n(S)$ and $\overline{\mathscr{I}_\lambda^n}(S)$ show that the semigroup $\mathscr{I}_\lambda^n(S)$ ($\overline{\mathscr{I}_\lambda^n}(S)$) is regular, orthodox, inverse or stable if and only if so is $S$. Green's relations are described on the semigroup $\mathscr{I}_\lambda^n(S)$ for an arbitrary monoid $S$. We introduce the conception of a semigroup with strongly tight ideal series, and proved that for any infinite cardinal $\lambda$ and any positive integer $n$ the semigroup $\mathscr{I}_\lambda^n(S)$ has a strongly tight ideal series provides so has $S$. At the finish we show that for every compact Hausdorff semitopological monoid $(S,\tau_S)$ there exists a unique its compact topological extension $\left(\mathscr{I}_\lambda^n(S),\tau_{\mathscr{I}}^\mathbf{c}\right)$ in the class of Haudorff semitopological semigroups.

Equations

$$a{a}^{-1}a=a\text{mbox}and{a}^{-1}a{a}^{-1}=a^{-1}.$$

$$x(\alpha \beta )=(x\alpha )\beta \text{mbox}ifx\in \mathrm{dom}(\alpha \beta )=\{y\in \mathrm{dom}\alpha :y\alpha \in \mathrm{dom}\beta \},\text{mbox}for\alpha ,\beta \in {\mathcal{I}}_{\lambda }.$$

$${\mathcal{I}}_{\lambda }^{\infty }=\{\alpha \in {\mathcal{I}}_{\lambda }:\mathrm{rank}\alpha \text{mbox}isfinite\}\text{mbox}and{\mathcal{I}}_{\lambda }^n=\{\alpha \in {\mathcal{I}}_{\lambda }:\mathrm{rank}\alpha ⩽n\},$$

$$\left(\begin{array}[]{ccc}x_{1}&\cdots&x_{n}\\ y_{1}&\cdots&y_{n}\\ \end{array}\right)$$

$$(a,b)\cdot(c,d)=\left\{\begin{array}[]{cl}(a,d),&\hbox{ if~{} }b=c;\\ 0,&\hbox{ if~{} }b\neq c,\end{array}\right.$$

$$(\alpha,a,\beta)\cdot(\gamma,b,\delta)=\left\{\begin{array}[]{cl}(\alpha,ab,\delta),&\hbox{ if }\beta=\gamma;\\ 0,&\hbox{ if }\beta\neq\gamma,\end{array}\right.$$

$${\mathcal{I}}_{\lambda }^n(S)=\left\{{\alpha }_S:\mathrm{Gr}(\alpha )\to S:\alpha \in {\mathcal{I}}_{\lambda }^n\right\}$$

$${\small{\left(\begin{array}[]{ccc}x_{1}&\cdots&x_{k}\\ s_{1}&\cdots&s_{k}\\ y_{1}&\cdots&y_{k}\end{array}\right)}},$$

$$\alpha_{S}={\small{\left(\begin{array}[]{ccc}x_{1}&\cdots&x_{k}\\ s_{1}&\cdots&s_{k}\\ y_{1}&\cdots&y_{k}\end{array}\right)}}$$

$$\alpha\cdot\beta=\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)\cdot\left(\begin{array}[]{ccc}c_{1}&\cdots&c_{j}\\ d_{1}&\cdots&d_{j}\end{array}\right)=\left(\begin{array}[]{ccc}a_{i_{1}}&\cdots&a_{i_{m}}\\ d_{j_{1}}&\cdots&d_{j_{m}}\end{array}\right)\neq 0\quad\hbox{~{}in~{}}\mathscr{I}_{\lambda}^{n},$$

$${\mathcal{J}}_{0}=\{0\}\cup\left\{\alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ 0_{S}&\cdots&0_{S}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}\colon 0_{S}\hbox{ is the zero of }S\right\}$$

$$\alpha_{S}\cdot\alpha_{S}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ a_{1}&\cdots&a_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ a_{1}&\cdots&a_{i}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}s_{1}&\cdots&s_{i}s_{i}\\ a_{1}&\cdots&a_{i}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ a_{1}&\cdots&a_{i}\end{array}\!\!\right)}}=\alpha_{S},$$

$$\begin{split}\alpha_{S}=&\;\alpha_{S}\cdot\beta_{S}\cdot\alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}\cdot{\small{\left(\begin{array}[]{ccc}c_{1}&\cdots&c_{k}\\ t_{1}&\cdots&t_{k}\\ d_{1}&\cdots&d_{k}\end{array}\right)}}\cdot{\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}=\\ =&{\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}\cdot{\small{\left(\begin{array}[]{cccccc}b_{1}&\cdots&b_{i}&c_{i+1}&\cdots&c_{k}\\ t_{1}&\cdots&t_{i}&t_{i+1}&\cdots&t_{k}\\ a_{1}&\cdots&a_{i}&d_{i+1}&\cdots&d_{k}\end{array}\right)}}\cdot{\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}=\\ =&{\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}t_{1}s_{1}&\cdots&s_{i}t_{i}s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}.\end{split}$$

$$aba=aandbab=b.$$

$$[A_{1},\ldots,A_{i}]^{(a_{1},\ldots,a_{i})}_{(b_{1},\ldots,b_{i})}=\left\{{\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}\colon s_{1}\in A_{1},\ldots,s_{i}\in A_{i}\right\}$$

$$[A{]}_{(b_{(1)\sigma },\dots ,b_{(i)\sigma })}^{(a_{(1)\sigma },\dots ,a_{(i)\sigma })}=[A{]}_{(b_1,\dots ,b_i)}^{(a_1,\dots ,a_i)}.$$

$$\begin{split}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{=}&{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}&\cdots&e_{i}\\ u_{1}&\cdots&u_{i}\\ f_{1}&\cdots&f_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}d_{1}&\cdots&d_{i}\\ u_{(1)\rho}&\cdots&u_{(i)\rho}\\ f_{(1)\rho}&\cdots&f_{(i)\rho}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}u_{(1)\rho}&\cdots&t_{i}u_{(i)\rho}\\ f_{(1)\rho}&\cdots&f_{(i)\rho}\end{array}\!\!\right)}}=\\ =&{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ t_{(1)\sigma}u_{((1)\rho)\sigma}&\cdots&t_{(i)\sigma}u_{((i)\rho)\sigma}\\ f_{((1)\rho)\sigma}&\cdots&f_{((i)\rho)\sigma}\end{array}\!\!\right)}}\end{split}$$

$$\begin{split}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{=}&{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}g_{1}&\cdots&g_{i}\\ v_{1}&\cdots&v_{i}\\ h_{1}&\cdots&h_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ v_{(1)\zeta}&\cdots&v_{(i)\zeta}\\ h_{(1)\zeta}&\cdots&h_{(i)\zeta}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}v_{(1)\zeta}&\cdots&s_{i}v_{(i)\zeta}\\ h_{(1)\zeta}&\cdots&h_{(i)\zeta}\end{array}\!\!\right)}}=\\ =&{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ s_{(1)\sigma^{-1}}v_{((1)\zeta)\sigma^{-1}}&\cdots&s_{(i)\sigma^{-1}}v_{((i)\zeta)\sigma^{-1}}\\ h_{((1)\zeta)\sigma^{-1}}&\cdots&h_{((i)\zeta)\sigma^{-1}}\end{array}\!\!\right)}}.\end{split}$$

$$s_1=t_{(1)\sigma }{u}_{((1)\rho )\sigma },\dots ,s_i=t_{(i)\sigma }{u}_{((i)\rho )\sigma },and{t}_1=s_{(1){\sigma }^{-1}}{v}_{((1)\zeta ){\sigma }^{-1}},\dots t_i=s_{(i){\sigma }^{-1}}{v}_{((i)\zeta ){\sigma }^{-1}}.$$

$$s_1=t_{(1)\sigma }{u}_1,\dots ,s_i=t_{(i)\sigma }{u}_i,t_1=s_{(1){\sigma }^{-1}}{v}_1,\dots t_i=s_{(i){\sigma }^{-1}}{v}_i.$$

$${\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{(1)\sigma}&\cdots&c_{(i)\sigma}\\ t_{(1)\sigma}u_{1}&\cdots&t_{(i)\sigma}u_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}u_{(1)\sigma^{-1}}&\cdots&t_{i}u_{(i)\sigma^{-1}}\\ b_{(1)\sigma^{-1}}&\cdots&b_{(i)\sigma^{-1}}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}d_{1}&\cdots&d_{i}\\ u_{(1)\sigma^{-1}}&\cdots&u_{(i)\sigma^{-1}}\\ b_{(1)\sigma^{-1}}&\cdots&b_{(i)\sigma^{-1}}\end{array}\!\!\right)}}$$

$${\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{(1)\sigma^{-1}}&\cdots&a_{(i)\sigma^{-1}}\\ s_{(1)\sigma^{-1}}v_{1}&\cdots&s_{(i)\sigma^{-1}}v_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}v_{(1)\sigma}&\cdots&s_{i}v_{(i)\sigma}\\ d_{(1)\sigma}&\cdots&d_{(i)\sigma}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ v_{(1)\sigma}&\cdots&v_{(i)\sigma}\\ d_{(1)\sigma}&\cdots&d_{(i)\sigma}\end{array}\!\!\right)}},$$

$$\begin{split}&{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{l}&\cdots&e_{k_{l}}^{l}\\ u_{1}^{l}&\cdots&u_{k_{l}}^{l}\\ f_{1}^{l}&\cdots&f_{k_{l}}^{l}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{r}&\cdots&e_{k_{r}}^{r}\\ u_{1}^{r}&\cdots&u_{k_{r}}^{r}\\ f_{1}^{r}&\cdots&f_{k_{r}}^{r}\end{array}\!\!\right)}}{=}\\ &={\small{\left(\!\!\begin{array}[]{ccc}e_{(1)\rho}^{l}&\cdots&e_{(i)\rho}^{l}\\ u_{(1)\rho}^{l}&\cdots&u_{(i)\rho}^{l}\\ c_{1}&\cdots&c_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}d_{1}&\cdots&d_{i}\\ u_{(1)\zeta}^{r}&\cdots&u_{(i)\zeta}^{r}\\ f_{(1)\zeta}^{r}&\cdots&f_{(i)\zeta}^{r}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}e_{(1)\rho}^{l}&\cdots&e_{(i)\rho}^{l}\\ u_{(1)\rho}^{l}t_{1}u_{(1)\zeta}^{r}&\cdots&u_{(i)\rho}^{l}t_{i}u_{(i)\zeta}^{r}\\ f_{(1)\zeta}^{r}&\cdots&f_{(i)\zeta}^{r}\end{array}\!\!\right)}}=\\ &={\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{l}&\cdots&e_{i}^{l}\\ u_{1}^{l}t_{(1)\rho^{-1}}u_{((1)\zeta)\rho^{-1}}^{r}&\cdots&u_{1}^{l}t_{(i)\rho^{-1}}u_{((i)\zeta)\rho^{-1}}^{r}\\ f_{((1)\zeta)\rho^{-1}}^{r}&\cdots&f_{((i)\zeta)\rho^{-1}}^{r}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ u_{(1)\sigma}^{l}t_{((1)\rho^{-1})\sigma}u_{(((1)\zeta)\rho^{-1})\sigma}^{r}&\cdots&u_{(i)\sigma}^{l}t_{((i)\rho^{-1})\sigma}u_{(((i)\zeta)\rho^{-1})\sigma}^{r}\\ f_{(((1)\zeta)\rho^{-1})\sigma}^{r}&\cdots&f_{(((i)\zeta)\rho^{-1})\sigma}^{r}\end{array}\!\!\right)}}\end{split}$$

$$\begin{split}&{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}g_{1}^{l}&\cdots&g_{i}^{l}\\ v_{1}^{l}&\cdots&v_{i}^{l}\\ h_{1}^{l}&\cdots&h_{i}^{l}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}g_{1}^{r}&\cdots&g_{i}^{r}\\ v_{1}^{r}&\cdots&v_{i}^{r}\\ h_{1}^{r}&\cdots&h_{i}^{r}\end{array}\!\!\right)}}=\\ &={\small{\left(\!\!\begin{array}[]{ccc}g_{1}^{l}&\cdots&g_{i}^{l}\\ v_{1}^{l}&\cdots&v_{i}^{l}\\ h_{1}^{l}&\cdots&h_{i}^{l}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ v_{(1)\kappa}^{r}&\cdots&v_{(i)\kappa}^{r}\\ h_{(1)\kappa}^{r}&\cdots&h_{(i)\kappa}^{r}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}g_{(1)\nu}^{l}&\cdots&g_{(i)\nu}^{l}\\ v_{(1)\nu}^{l}s_{1}v_{(1)\kappa}^{r}&\cdots&v_{(i)\nu}^{l}s_{i}v_{(i)\kappa}^{r}\\ h_{(1)\kappa}^{r}&\cdots&h_{(i)\kappa}^{r}\end{array}\!\!\right)}}{=}\\ &={\small{\left(\!\!\!\!\begin{array}[]{ccc}g_{1}^{l}&\cdots&g_{i}^{l}\\ v_{1}^{l}s_{(1)\nu^{-1}}v_{((1)\kappa)\nu^{-1}}^{r}&\cdots&v_{i}^{l}s_{(i)\nu^{-1}}v_{((i)\kappa)\nu^{-1}}^{r}\\ h_{((1)\kappa)\nu^{-1}}^{r}&\cdots&h_{((i)\kappa)\nu^{-1}}^{r}\end{array}\!\!\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ v_{(1)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((1)\kappa)\nu^{-1})\varsigma}^{r}&\cdots&v_{(i)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((i)\kappa)\nu^{-1})\varsigma}^{r}\\ h_{(((1)\kappa)\nu^{-1})\varsigma}^{r}&\cdots&h_{(((i)\kappa)\nu^{-1})\varsigma}^{r}\end{array}\!\!\right)}}\end{split}$$

$$d_1=h_{(((1)\kappa ){\nu }^{-1})\varsigma }^r,\dots ,d_i=h_{(((i)\kappa ){\nu }^{-1})\varsigma }^r.$$

$$\begin{split}&{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}{=}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{l}&\cdots&e_{k_{l}}^{l}\\ u_{1}^{l}&\cdots&u_{k_{l}}^{l}\\ f_{1}^{l}&\cdots&f_{k_{l}}^{l}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{r}&\cdots&e_{k_{r}}^{r}\\ u_{1}^{r}&\cdots&u_{k_{r}}^{r}\\ f_{1}^{r}&\cdots&f_{k_{r}}^{r}\end{array}\!\!\right)}}{=}\\ &={\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{l}&\cdots&e_{k_{l}}^{l}\\ u_{1}^{l}&\cdots&u_{k_{l}}^{l}\\ f_{1}^{l}&\cdots&f_{k_{l}}^{l}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ v_{(1)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((1)\kappa)\nu^{-1})\varsigma}^{r}&\cdots&v_{(i)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((i)\kappa)\nu^{-1})\varsigma}^{r}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{r}&\cdots&e_{k_{r}}^{r}\\ u_{1}^{r}&\cdots&u_{k_{r}}^{r}\\ f_{1}^{r}&\cdots&f_{k_{r}}^{r}\end{array}\!\!\right)}}=\\ &={\small{\left(\!\!\begin{array}[]{ccc}e_{(1)\rho}^{l}&\cdots&e_{(i)\rho}^{l}\\ u_{(1)\rho}^{l}&\cdots&u_{(i)\rho}^{l}\\ c_{1}&\cdots&c_{i}\end{array}\!\!\right)}}{\cdot}{\small{\left(\!\!\!\!\begin{array}[]{ccc}c_{1}&{\cdots}&c_{i}\\ v_{(1)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((1)\kappa)\nu^{-1})\varsigma}^{r}&{\cdots}&v_{(i)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((i)\kappa)\nu^{-1})\varsigma}^{r}\\ d_{1}&{\cdots}&d_{i}\end{array}\!\!\!\!\right)}}{\cdot}{\small{\left(\!\!\begin{array}[]{ccc}d_{1}&\cdots&d_{i}\\ u_{(1)\zeta}^{r}&\cdots&u_{(i)\zeta}^{r}\\ f_{(1)\zeta}^{r}&\cdots&f_{(i)\zeta}^{r}\end{array}\!\!\right)}}{=}\\ &={\small{\left(\!\!\!\!\begin{array}[]{ccc}e_{(1)\rho}^{l}&{\cdots}&e_{(i)\rho}^{l}\\ u_{(1)\rho}^{l}v_{(1)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((1)\kappa)\nu^{-1})\varsigma}^{r}u_{(1)\zeta}^{r}&{\cdots}&u_{(i)\rho}^{l}v_{(i)\zeta}^{l}s_{((1)\nu^{-1})\varsigma}v_{(((i)\kappa)\nu^{-1})\varsigma}^{r}u_{(1)\zeta}^{r}\\ f_{(1)\zeta}^{r}&{\cdots}&f_{(i)\zeta}^{r}\end{array}\!\!\!\!\right)}}\end{split}$$

$$s_1=u_{(1)\sigma }^l{v}_{(((1)\zeta ){\rho }^{-1})\sigma }^l{s}_{((((1){\nu }^{-1})\varsigma ){\rho }^{-1})\sigma }{v}_{(((((1)\kappa ){\nu }^{-1})\varsigma ){\rho }^{-1})\sigma }^r{u}_{(((1)\zeta ){\rho }^{-1})\sigma }^r$$

$$\dots \dots \dots \dots \dots \dots \dots \dots$$

$$s_i=u_{(i)\sigma }^l{v}_{(((i)\zeta ){\rho }^{-1})\sigma }^l{s}_{((((i){\nu }^{-1})\varsigma ){\rho }^{-1})\sigma }{v}_{(((((i)\kappa ){\nu }^{-1})\varsigma ){\rho }^{-1})\sigma }^r{u}_{(((i)\zeta ){\rho }^{-1})\sigma }^r.$$

$$s_1=x_1{t}_{(1)\sigma }{u}_1,\dots ,s_i=x_i{t}_{(i)\sigma }{u}_i,t_1=y_1{s}_{(1){\sigma }^{-1}}{v}_1,\dots ,t_i=y_i{s}_{(i){\sigma }^{-1}}{v}_i.$$

$$\begin{split}{\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}&={\small{\left(\!\!\begin{array}[]{ccc}c_{(1)\sigma}&\cdots&c_{(i)\sigma}\\ x_{1}t_{(1)\sigma}u_{1}&\cdots&x_{i}t_{(i)\sigma}u_{i}\\ b_{(1)\sigma}&\cdots&b_{(i)\sigma}\end{array}\!\!\right)}}={\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ x_{(1)\sigma^{-1}}t_{1}u_{(1)\sigma^{-1}}&\cdots&x_{(i)\sigma^{-1}}t_{i}u_{(i)\sigma^{-1}}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}=\\ =&{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ x_{(1)\sigma^{-1}}&\cdots&x_{(i)\sigma^{-1}}\\ c_{1}&\cdots&c_{i}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}\cdot{\small{\left(\!\!\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ u_{(1)\sigma^{-1}}&\cdots&u_{(i)\sigma^{-1}}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}}\end{split}$$

Full text

1. Introduction, motivation and main definitions

In this paper we shall follow the terminology of [11, 31].

If $S$ is a semigroup, then by $E(S)$ we denote the subset of all idempotents of $S$. On the set of idempotents $E(S)$ there exists the natural partial order: $e\leqslant f$ if and only if $ef=fe=e$.

A semigroup $S$ is called:

• •

regular, if for every $a\in S$ there exists an element $b$ in $S$ such that $a=aba$;

• •

orthodox, if $S$ is regular and $E(S)$ is a subsemigroup of $S$;

• •

inverse if every $a$ in $S$ possesses an unique inverse, i.e. if there exists an unique element $a^{-1}$ in $S$ such that

[TABLE]

It is obvious that every inverse semigroup is orthodox and every orthodox semigroup is regular. A map which associates to any element of an inverse semigroup its inverse is called the inversion.

Let $\lambda$ be an arbitrary non-zero cardinal. A map $\alpha$ from a subset $D$ of $\lambda$ into $\lambda$ is called a partial transformation of $X$. In this case the set $D$ is called the domain of $\alpha$ and is denoted by $\operatorname{dom}\alpha$. Also, the set $\{x\in X\colon y\alpha=x\mbox{ for some }y\in Y\}$ is called the range of $\alpha$ and is denoted by $\operatorname{ran}\alpha$. The cardinality of $\operatorname{ran}\alpha$ is called the rank of $\alpha$ and denoted by $\operatorname{rank}\alpha$. For convenience we denote by $\varnothing$ the empty transformation, that is a partial mapping with $\operatorname{dom}\varnothing=\operatorname{ran}\varnothing=\varnothing$.

Let $\mathscr{I}_{\lambda}$ denote the set of all partial one-to-one transformations of $\lambda$ together with the following semigroup operation:

[TABLE]

The semigroup $\mathscr{I}_{\lambda}$ is called the symmetric inverse semigroup over the cardinal $\lambda$ (see [11]). The symmetric inverse semigroup was introduced by V. V. Wagner [33] and it plays a major role in the theory of semigroups.

Put

[TABLE]

for $n=1,2,3,\ldots$. Obviously, $\mathscr{I}_{\lambda}^{\infty}$ and $\mathscr{I}_{\lambda}^{n}$ ($n=1,2,3,\ldots$) are inverse semigroups, $\mathscr{I}_{\lambda}^{\infty}$ is an ideal of $\mathscr{I}_{\lambda}$, and $\mathscr{I}_{\lambda}^{n}$ is an ideal of $\mathscr{I}_{\lambda}^{\infty}$, for each $n=1,2,3,\ldots$. Further, we shall call the semigroup $\mathscr{I}_{\lambda}^{\infty}$ the symmetric inverse semigroup of finite transformations and $\mathscr{I}_{\lambda}^{n}$ the symmetric inverse semigroup of finite transformations of the rank $\leqslant n$. The elements of semigroups $\mathscr{I}_{\lambda}^{\infty}$ and $\mathscr{I}_{\lambda}^{n}$ are called finite one-to-one transformations (partial bijections) of the cardinal $\lambda$. By

[TABLE]

we denote a partial one-to-one transformation which maps $x_{1}$ onto $y_{1}$, $\ldots$, $x_{n}$ onto $y_{n}$, and by [math] the empty transformation. Obviously, in such case we have $x_{i}\neq x_{j}$ and $y_{i}\neq y_{j}$ for $i\neq j$ ($i,j=1,\ldots,n$). The empty partial map $\varnothing\colon\lambda\rightharpoonup\lambda$ we denote by [math]. It is obvious that [math] is zero of the semigroup $\mathscr{I}_{\lambda}^{n}$.

Let $\lambda$ be a non-zero cardinal. On the set $B_{\lambda}=(\lambda\times\lambda)\cup\{0\}$, where $0\notin\lambda\times\lambda$, we define the semigroup operation “$\,\cdot\,$” as follows

[TABLE]

and $(a,b)\cdot 0=0\cdot(a,b)=0\cdot 0=0$ for $a,b,c,d\in\lambda$. The semigroup $B_{\lambda}$ is called the semigroup of $\lambda\times\lambda$-matrix units (see [11]). Obviously, for any cardinal $\lambda>0$, the semigroup of $\lambda\times\lambda$-matrix units $B_{\lambda}$ is isomorphic to $\mathscr{I}_{\lambda}^{1}$.

Let $S$ be a semigroup with zero and $\lambda$ be a non-zero cardinal. We define the semigroup operation on the set $B_{\lambda}(S)=(\lambda\times S\times\lambda)\cup\{0\}$ as follows:

[TABLE]

and $(\alpha,a,\beta)\cdot 0=0\cdot(\alpha,a,\beta)=0\cdot 0=0$, for all $\alpha,\beta,\gamma,\delta\in\lambda$ and $a,b\in S$. If $S=S^{1}$ then the semigroup $B_{\lambda}(S)$ is called the Brandt $\lambda$-extension of the semigroup $S$ [15, 19]. Obviously, if $S$ has zero then ${\mathcal{J}}=\{0\}\cup\{(\alpha,0_{S},\beta)\colon 0_{S}$ is the zero of $S\}$ is an ideal of $B_{\lambda}(S)$. We put $B^{0}_{\lambda}(S)=B_{\lambda}(S)/{\mathcal{J}}$ and the semigroup $B^{0}_{\lambda}(S)$ is called the Brandt $\lambda^{0}$-extension of the semigroup $S$ with zero [22].

A topological (inverse) semigroup is a Hausdorff topological space together with a continuous semigroup operation (and an inversion, respectively). A semitopological semigroup is a Hausdorff topological space together with a separately continuous semigroup operation.

The Brandt $\lambda$-extension $B_{\lambda}(S)$ (or the Brandt $\lambda^{0}$-extension $B^{0}_{\lambda}(S)$) of a semigroup $S$ we may to consider as a some semigroup extension of the semigroup $S$ by the semigroup $\lambda\times\lambda$-matrix units $B_{\lambda}$. An analogue of so extension gives the following definition.

2. The construction of of the semigroup extension $\mathscr{I}_{\lambda}^{n}(S)$

In this paper using the semigroup $\mathscr{I}_{\lambda}^{n}$ we propose the following semigroup extension.

Construction 2.1**.**

Let $S$ be a semigroup, $\lambda$ be a non-zero cardinal, $n$ and $k$ be a positive integers such that $k\leqslant n\leqslant\lambda$. We identify every element $\alpha\in\mathscr{I}_{\lambda}^{n}$ with its graph $\operatorname{\textsf{Gr}}(\alpha)\subset\lambda\times\lambda$ and put

[TABLE]

and every map from the empty map [math] into $S$ we identify with the empty map $\varnothing\colon\lambda\times\lambda\rightharpoonup S$ and denote its by [math]. An arbitrary element $0\neq\operatorname{rank}\alpha\leqslant n$ we denote by

[TABLE]

where \alpha=\left(\begin{array}[]{ccc}x_{1}&\cdots&x_{k}\\ y_{1}&\cdots&y_{k}\end{array}\right), and $\left((x_{1},y_{1})\right)\alpha=s_{1}$, $\ldots,$ $\left((x_{k},y_{k})\right)\alpha=s_{k}$. Also for $\alpha_{S}\in\mathscr{I}_{\lambda}^{n}(S)$ such that

[TABLE]

we denote $\mathop{\textsf{{d}}}(\alpha_{S})=\{x_{1},\ldots,x_{k}\}$ and $\mathop{\textsf{{r}}}(\alpha_{S})=\{y_{1},\ldots,y_{k}\}$.

Now, we define a binary operation “$\cdot$” on the set $\mathscr{I}_{\lambda}^{n}(S)$ in the following way:

• $(i)$

$\alpha_{S}\cdot 0=0\cdot\alpha_{S}=0\cdot 0=0$ for every $\alpha_{S}\in\mathscr{I}_{\lambda}^{n}(S)$;

• $(ii)$

if $\alpha\cdot\beta=0$ in $\mathscr{I}_{\lambda}^{n}$ then $\alpha_{S}\cdot\beta_{S}=0$ for any $\alpha_{S},\beta_{S}\in\mathscr{I}_{\lambda}^{n}(S)$;

• $(iii)$

if \alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}}, \beta_{S}={\small{\left(\begin{array}[]{ccc}c_{1}&\cdots&c_{j}\\ t_{1}&\cdots&t_{j}\\ d_{1}&\cdots&d_{j}\end{array}\right)}} and

[TABLE]

then \alpha_{S}\cdot\beta_{S}={\small{\left(\begin{array}[]{cccc}a_{i_{1}}&\cdots&a_{i_{m}}\\ s_{i_{1}}t_{j_{1}}&\cdots&s_{i_{m}}t_{j_{m}}\\ d_{j_{1}}&\cdots&d_{j_{m}}\end{array}\right)}}.

Simple verifications show that so defines binary operation on $\mathscr{I}_{\lambda}^{n}(S)$ is associative and hence $\mathscr{I}_{\lambda}^{n}(S)$ is a semigroup. It is obvious that $\mathscr{I}_{\lambda}^{1}(S)$ is isomorphic to the Brandt $\lambda$-extension $B_{\lambda}(S)$ of the semigroup $S$.

We remark that if the semigroup $S$ contains a zero $0_{S}$ then

[TABLE]

is an ideal of $\mathscr{I}_{\lambda}^{n}(S)$.

Also, we define a binary relation $\equiv_{0}$ on the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ in the following way. For $\alpha_{S},\beta_{S}\in\mathscr{I}_{\lambda}^{n}(S)$ we put $\alpha_{S}\equiv_{0}\beta_{S}$ if and only if at least one of the following conditions holds:

• $(1)$

$\alpha_{S}=\beta_{S}$;

• $(2)$

$\alpha_{S},\beta_{S}\in{\mathcal{J}}_{0}$;

• $(3)$

$\alpha_{S},\beta_{S}\notin{\mathcal{J}}_{0}$ and each of conditions

• $(i)$

$(x,y)\alpha_{S}$ is determined and $(x,y)\alpha_{S}\neq 0_{S}$;   and

• $(ii)$

$(x,y)\beta_{S}$ is determined and $(x,y)\beta_{S}\neq 0_{S}$

implies the equality $(x,y)\alpha_{S}=(x,y)\beta_{S}$.

It is obvious that $\equiv_{0}$ is an equivalence relation on the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

The following proposition prove by usual verifications.

Proposition 2.2**.**

The relation $\equiv_{0}$ is a congruence on the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

We define $\overline{\mathscr{I}_{\lambda}^{n}}(S)=\mathscr{I}_{\lambda}^{n}(S)/_{\equiv_{0}}$.

In this paper we study algebraic properties of the semigroups $\mathscr{I}_{\lambda}^{n}(S)$ and $\overline{\mathscr{I}_{\lambda}^{n}}(S)$. We describe idempotents and regular elements of the semigroups $\mathscr{I}_{\lambda}^{n}(S)$ and $\overline{\mathscr{I}_{\lambda}^{n}}(S)$ show that the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ ($\overline{\mathscr{I}_{\lambda}^{n}}(S)$) is regular, orthodox, inverse or stable if and only if so is $S$. Green’s relations are described in the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ for an arbitrary monoid $S$. We introduce the conception of a semigroup with strongly tight ideal series, and proved that for any infinite cardinal $\lambda$ and any positive integer $n$ the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ has a strongly tight ideal series provides so has $S$. At the finish we show that for every compact Hausdorff semitopological monoid $(S,\tau_{S})$ there exists a unique its compact topological extension $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ in the class of Haudorff semitopological semigroups.

3. Algebraic properties of the semigroup extensions $\mathscr{I}_{\lambda}^{n}(S)$ and $\overline{\mathscr{I}_{\lambda}^{n}}(S)$

The following proposition describes the subset of idempotents of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Proposition 3.1**.**

For every positive integer $i\leqslant n$ a non-zero element \alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is an idempotent if and only if $a_{1}=b_{1},\ldots,a_{i}=b_{i}$ and $s_{1},\ldots,s_{i}\in E(S)$.

Proof.

The implication $(\Leftarrow)$ is trivial.

$(\Rightarrow)$ Suppose that $\alpha_{S}\cdot\alpha_{S}=\alpha_{S}$. Then the definition of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ implies that the symbols $a_{1},\ldots,a_{i}$ are distinct. Similar we get that the symbols $b_{1},\ldots,b_{i}$ are distinct, too. The above arguments and the equality $\alpha_{S}\cdot\alpha_{S}=\alpha_{S}$ imply that $\left\{a_{1},\ldots,a_{i}\right\}=\left\{b_{1},\ldots,b_{i}\right\}$. Assume that $a_{k}\neq b_{k}=a_{l}$ for some integers $k,l\in\{1,\ldots,i\}$, $k\neq l$. Then we have that $a_{l}\neq b_{l}\neq b_{k}$, which contradicts the equality $\alpha_{S}\cdot\alpha_{S}=\alpha_{S}$. The obtained contradiction implies the equalities $a_{1}=b_{1},\ldots,a_{i}=b_{i}$. Now, we get that

[TABLE]

and hence $s_{1}s_{1}=s_{1},\ldots,s_{i}s_{i}=s_{i}$. This completes the proof if the proposition. ∎

Proposition 3.2**.**

For every positive integer $i\leqslant n$ a non-zero element \alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is regular if and only if so are $s_{1},\ldots,s_{i}$ in $S$.

Proof.

The implication $(\Leftarrow)$ is trivial. Indeed, $\alpha_{S}=\alpha_{S}\beta_{S}\alpha_{S}$ for \beta_{S}={\small{\left(\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ t_{1}&\cdots&t_{i}\\ a_{1}&\cdots&a_{i}\end{array}\right)}}, where elements $t_{1},\ldots,t_{i}$ of the semigroup $S$ such that $s_{1}=s_{1}t_{1}s_{1}$, $\ldots,$ $s_{i}=s_{i}t_{i}s_{i}$.

$(\Rightarrow)$ Suppose that $\alpha_{S}$ is a regular element of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$. Then there exists an element \beta_{S}={\small{\left(\begin{array}[]{ccc}c_{1}&\cdots&c_{k}\\ t_{1}&\cdots&t_{k}\\ d_{1}&\cdots&d_{k}\end{array}\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$, $0<k\leqslant n$, such that $\alpha_{S}=\alpha_{S}\cdot\beta_{S}\cdot\alpha_{S}$. Now, this implies that $\{b_{1},\ldots,b_{i}\}\subseteq\{c_{1},\ldots,c_{k}\}$ and hence $k\geqslant i$. Without loss of generality we may assume that $b_{1}=c_{1},\ldots,b_{i}=c_{i}$. Then the equality $\alpha_{S}=\alpha_{S}\cdot\beta_{S}\cdot\alpha_{S}$ and the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ imply that $d_{1}=a_{1},\ldots,d_{i}=a_{i}$ and hence we have that

[TABLE]

This implies that the following equalities $s_{1}=s_{1}t_{1}s_{1}$, $\ldots,$ $s_{i}=s_{i}t_{i}s_{i}$ hold in $S$, which completes the proof of our proposition. ∎

Two elements $a$ and $b$ of a semigroup $S$ are said to be inverses of each other if

[TABLE]

The definition of the semigroup operation in $\mathscr{I}_{\lambda}^{n}(S)$ implies the following proposition.

Proposition 3.3**.**

Let $\lambda$ be a non-zero cardinal, $n$ and $i$ be any positive integers such that $i\leqslant n\leqslant\lambda$. Let $S$ be a semigroup and $a_{1},\ldots,a_{i},b_{1},\ldots,b_{i}\in\lambda$. If the elements $s_{1}$ and $t_{1}$, $\ldots,$ $s_{i}$ and $t_{i}$ are pairwise inverses of each other in $S$ then the elements {\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}} and {\small{\left(\begin{array}[]{ccc}b_{1}&\cdots&b_{i}\\ t_{1}&\cdots&t_{i}\\ a_{1}&\cdots&a_{i}\\ \end{array}\right)}} are pairwise inverses of each other in the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

For arbitrary semigroup $S$, every positive integer $i\leqslant n$, any collection non-empty subsets $A_{1},\ldots,A_{i}$ of $S$, and any two collections of distinct elements $\{a_{1},\ldots,a_{i}\}$ and $\{b_{1},\ldots,b_{i}\}$ of the cardinal $\lambda$ we denote a subset

[TABLE]

of $\mathscr{I}_{\lambda}^{n}(S)$. I the case when $A_{1}=\ldots=A_{i}=A$ in $S$ we put $[A]^{(a_{1},\ldots,a_{i})}_{(b_{1},\ldots,b_{i})}=[A_{1},\ldots,A_{i}]^{(a_{1},\ldots,a_{i})}_{(b_{1},\ldots,b_{i})}$. It is obvious that for every subset $A$ of the semigroup $S$ and any permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ we have that

[TABLE]

Proposition 3.4**.**

Let $\lambda$ be a non-zero cardinal and $n$ be any positive integer $\leqslant\lambda$. Then for arbitrary semigroup $S$, every positive integer $i\leqslant n$ and any collection of distinct elements $\{a_{1},\ldots,a_{i}\}$ of $\lambda$ the direct power $S^{i}$ is isomorphic to a subsemigroup $S^{(a_{1},\ldots,a_{i})}_{(a_{1},\ldots,a_{i})}$ of $\mathscr{I}_{\lambda}^{n}(S)$.

Proof.

The semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that $S^{a_{1},\ldots,a_{i}}_{a_{1},\ldots,a_{i}}$ is a subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$ for any collection of distinct elements $\{a_{1},\ldots,a_{i}\}$ of $\lambda$. An isomorphism $\mathfrak{h}\colon S^{i}\rightarrow S^{(a_{1},\ldots,a_{i})}_{(a_{1},\ldots,a_{i})}$ we define by the formula (s_{1},\ldots,s_{i})\mathfrak{h}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ a_{1}&\cdots&a_{i}\end{array}\right)}}. ∎

Proposition 3.5**.**

For every semigroup $S$, any non-zero cardinal $\lambda$ and any positive integer $n\leqslant\lambda$ the following statements hold:

• $(i)$

$\mathscr{I}_{\lambda}^{n}(S)$* is regular if and only if so is $S$;*

• $(ii)$

$\mathscr{I}_{\lambda}^{n}(S)$* is orthodox if and only if so is $S$;*

• $(iii)$

$\mathscr{I}_{\lambda}^{n}(S)$* is inverse if and only if so is $S$.*

Proof.

Statement $(i)$ follows from Proposition 3.2.

$(ii)$ $(\Leftarrow)$ Suppose that $S$ is an orthodox semigroup. Then statement $(i)$ implies that the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is regular. By Proposition 3.1 every non-zero idempotent of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ has the form {\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ e_{1}&\cdots&e_{i}\\ a_{1}&\cdots&a_{i}\end{array}\right)}}, where $0<i\leqslant n$ and $e_{1},\ldots,e_{i}$ are idempotents of $S$. This implies that the product of two idempotents of $\mathscr{I}_{\lambda}^{n}(S)$ is again an idempotent, and hence the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is orthodox.

$(\Rightarrow)$ Suppose that $\mathscr{I}_{\lambda}^{n}(S)$ is an orthodox semigroup. By Proposition 3.4, $S^{(a)}_{(a)}$ is a subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$ for every $a\in\lambda$ and hence $S^{(a)}_{(a)}$ is orthodox. Then Proposition 3.4 implies the semigroup $S$ is orthodox, too.

$(iii)$ $(\Leftarrow)$ Suppose that $S$ is an inverse semigroup. By statement $(i)$ the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is regular. Then using Proposition 3.1 we get that idempotents commute in $\mathscr{I}_{\lambda}^{n}(S)$ and hence by Theorem 1.17 of [11], $\mathscr{I}_{\lambda}^{n}(S)$ is an inverse semigroup.

$(\Rightarrow)$ Suppose that $\mathscr{I}_{\lambda}^{n}(S)$ is an inverse semigroup. By Proposition 3.4, $S^{(a)}_{(a)}$ is a subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$ for every $a\in\lambda$, and by Proposition 3.3 it is an inverse subsemigroup. Hence by Proposition 3.4, $S$ is an inverse semigroup. ∎

Since any homomorphic image of a regular (resp., orthodox, inverse) semigroup is a regular (resp., orthodox, inverse) semigroup (see [11, Section 7.4] and [29, Lemma 2.2]), Proposition 3.5 implies the following corollary.

Corollary 3.6**.**

For every semigroup $S$, any non-zero cardinal $\lambda$ and any positive integer $n\leqslant\lambda$ the following statements hold:

• $(i)$

$\overline{\mathscr{I}_{\lambda}^{n}}(S)$* is regular if and only if so is $S$;*

• $(ii)$

$\overline{\mathscr{I}_{\lambda}^{n}}(S)$* is orthodox if and only if so is $S$;*

• $(iii)$

$\overline{\mathscr{I}_{\lambda}^{n}}(S)$* is inverse if and only if so is $S$.*

If $S$ is a semigroup, then we shall denote by $\mathscr{R}$, $\mathscr{L}$, $\mathscr{J}$, $\mathscr{D}$ and $\mathscr{H}$ the Green relations on $S$ (see [13] or [11, Section 2.1]):

[TABLE]

Remark 3.7**.**

It is obvious that for non-zero elements \alpha_{S}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}} and \beta_{S}={\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{k}\\ t_{1}&\cdots&t_{k}\\ d_{1}&\cdots&d_{k}\end{array}\!\!\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ any of conditions $\alpha_{S}\mathscr{R}\beta_{S}$, $\alpha_{S}\mathscr{L}\beta_{S}$, $\alpha_{S}\mathscr{D}\beta_{S}$, $\alpha_{S}\mathscr{J}\beta_{S}$, or $\alpha_{S}\mathscr{H}\beta_{S}$ implies the equality $i=k$.

The following proposition describes the Green relations on the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Proposition 3.8**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal and $n\leqslant\lambda$. Let \alpha_{S}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\!\!\right)}} and \beta_{S}={\small{\left(\!\!\begin{array}[]{ccc}c_{1}&\cdots&c_{i}\\ t_{1}&\cdots&t_{i}\\ d_{1}&\cdots&d_{i}\end{array}\!\!\right)}} be non-zero elements of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$. Then the following conditions hold:

• $(i)$

$\alpha_{S}\mathscr{R}\beta_{S}$* in $\mathscr{I}_{\lambda}^{n}(S)$ if and only if there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $a_{1}=c_{(1)\sigma}$, $\ldots,a_{i}=c_{(i)\sigma}$ and $s_{1}\mathscr{R}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{R}t_{(i)\sigma}$ in $S$;*

• $(ii)$

$\alpha_{S}\mathscr{L}\beta_{S}$* in $\mathscr{I}_{\lambda}^{n}(S)$ if and only if there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $b_{1}=d_{(1)\sigma}$, $\ldots,b_{i}=d_{(i)\sigma}$ and $s_{1}\mathscr{L}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{L}t_{(i)\sigma}$ in $S$;*

• $(iii)$

$\alpha_{S}\mathscr{D}\beta_{S}$* in $\mathscr{I}_{\lambda}^{n}(S)$ if and only if there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{D}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{D}t_{(i)\sigma}$ in $S$;*

• $(iv)$

$\alpha_{S}\mathscr{H}\beta_{S}$* in $\mathscr{I}_{\lambda}^{n}(S)$ if and only if there exists a permutations $\sigma,\rho\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{R}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{R}t_{(i)\sigma}$ and $s_{1}\mathscr{L}t_{(1)\rho}$, $\ldots,s_{i}\mathscr{L}t_{(i)\rho}$ in $S$;*

• $(v)$

$\alpha_{S}\mathscr{J}\beta_{S}$* in $\mathscr{I}_{\lambda}^{n}(S)$ if and only if there exists a permutation $\pi\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{J}t_{(1)\pi}$, $\ldots,s_{i}\mathscr{J}t_{(i)\pi}$ in $S$.*

Proof.

$(i)$ $(\Rightarrow)$ Suppose that $\alpha_{S}\mathscr{R}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. Then there exist non-zero elements \gamma_{S}={\small{\left(\!\!\begin{array}[]{ccc}e_{1}&\cdots&e_{k}\\ u_{1}&\cdots&u_{k}\\ f_{1}&\cdots&f_{k}\end{array}\!\!\right)}} and \delta_{S}={\small{\left(\!\!\begin{array}[]{ccc}g_{1}&\cdots&g_{j}\\ v_{1}&\cdots&v_{j}\\ h_{1}&\cdots&h_{j}\end{array}\!\!\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ such that $\alpha_{S}=\beta_{S}\gamma_{S}$, $\beta_{S}=\alpha_{S}\delta_{S}$, $i\leqslant j\leqslant n$ and $i\leqslant k\leqslant n$. Also, the definition of the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that without loss of generality we may assume that $j=k=i$. Then the equalities $\alpha_{S}=\beta_{S}\gamma_{S}$ and $\beta_{S}=\alpha_{S}\delta_{S}$ imply that $\{a_{1},\ldots,a_{i}\}=\{c_{1},\ldots,c_{i}\}$, $\{b_{1},\ldots,b_{i}\}=\{g_{1},\ldots,g_{i}\}$ and $\{d_{1},\ldots,d_{i}\}=\{e_{1},\ldots,e_{i}\}$. Now, the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that there exist permutations $\sigma,\rho,\zeta\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $a_{1}=c_{(1)\sigma}$, $\ldots,a_{i}=c_{(i)\sigma}$, $d_{1}=e_{(1)\rho}$, $\ldots,d_{i}=e_{(i)\rho}$, and $b_{1}=g_{(1)\zeta},\ldots,b_{i}=g_{(i)\zeta}$, and hence we have that

[TABLE]

and

[TABLE]

Therefore we get that

[TABLE]

Since $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ is a permutation conditions (1) imply that $s_{1}\mathscr{R}t_{(1)\sigma}$, $\ldots,$ $s_{i}\mathscr{R}t_{(i)\sigma}$ in $S$.

$(\Leftarrow)$ Suppose that for $\alpha_{S},\beta_{S}\in\mathscr{I}_{\lambda}^{n}(S)$ there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $a_{1}=c_{(1)\sigma}$, $\ldots,a_{i}=c_{(i)\sigma}$ and $s_{1}\mathscr{R}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{R}t_{(i)\sigma}$ in $S$. Then there exist $u_{1},\ldots,u_{i},v_{1},\ldots,v_{i}\in S^{1}$ such that

[TABLE]

Thus we get that

[TABLE]

and

[TABLE]

and hence $\alpha_{S}\mathscr{R}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$.

The proof of statement $(ii)$ is similar to the proof of $(i)$.

$(iii)$ $(\Rightarrow)$ Suppose that $\alpha_{S}\mathscr{D}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. Then there exist a non-zero element \gamma_{S}={\small{\left(\!\!\begin{array}[]{ccc}e_{1}&\cdots&e_{i}\\ u_{1}&\cdots&u_{i}\\ f_{1}&\cdots&f_{i}\end{array}\!\!\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ such that $\alpha_{S}\mathscr{R}\gamma_{S}$ and $\gamma_{S}\mathscr{L}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. By statement $(i)$ there exists a permutation $\zeta\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $e_{1}=a_{(1)\zeta}$, $\ldots,e_{i}=a_{(i)\zeta}$ and $u_{1}\mathscr{R}s_{(1)\zeta}$, $\ldots,u_{i}\mathscr{R}s_{(i)\zeta}$ in $S$ and by statement $(ii)$ there exists a permutation $\varsigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $f_{1}=d_{(1)\varsigma},\ldots,f_{i}=d_{(i)\varsigma}$ and $u_{1}\mathscr{L}s_{(1)\varsigma}$, $\ldots,u_{i}\mathscr{L}s_{(i)\varsigma}$ in $S$. This implies that $s_{1}\mathscr{D}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{D}t_{(i)\sigma}$ in $S$ for the permutation $\sigma=\zeta\circ\varsigma^{-1}$ of $\{1,\ldots,i\}$.

$(\Leftarrow)$ Suppose that there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{D}t_{(1)\sigma}$, $\ldots,$ $s_{i}\mathscr{D}t_{(i)\sigma}$ in $S$. Then the definition of the relation $\mathscr{D}$ implies that there exist $u_{1},\ldots,u_{i}\in S$ such that $s_{1}\mathscr{R}u_{1}$, $\ldots,$ $s_{i}\mathscr{R}u_{i}$ and $u_{1}\mathscr{L}t_{(1)\sigma}$, $\ldots,$ $u_{i}\mathscr{L}t_{(i)\sigma}$ in $S$. Now, for the element \gamma_{S}={\small{\left(\!\!\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ u_{1}&\cdots&u_{i}\\ d_{(1)\sigma}&\cdots&d_{(i)\sigma}\end{array}\!\!\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ by statements $(i)$ and $(ii)$ we have that $\alpha_{S}\mathscr{R}\gamma_{S}$ and $\gamma_{S}\mathscr{L}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$.

$(iv)$ follows from statements $(i)$ and $(ii)$.

$(v)$ $(\Rightarrow)$ Suppose that $\alpha_{S}\mathscr{J}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. Then there exist non-zero elements \gamma_{S}^{l}={\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{l}&\cdots&e_{k_{l}}^{l}\\ u_{1}^{l}&\cdots&u_{k_{l}}^{l}\\ f_{1}^{l}&\cdots&f_{k_{l}}^{l}\end{array}\!\!\right)}}, \gamma_{S}^{r}={\small{\left(\!\!\begin{array}[]{ccc}e_{1}^{r}&\cdots&e_{k_{r}}^{r}\\ u_{1}^{r}&\cdots&u_{k_{r}}^{r}\\ f_{1}^{r}&\cdots&f_{k_{r}}^{r}\end{array}\!\!\right)}}, \delta_{S}^{l}={\small{\left(\!\!\begin{array}[]{ccc}g_{1}^{l}&\cdots&g_{j_{l}}^{l}\\ v_{1}^{l}&\cdots&v_{j_{l}}^{l}\\ h_{1}^{l}&\cdots&h_{j_{l}}^{l}\end{array}\!\!\right)}} and \delta_{S}^{r}={\small{\left(\!\!\begin{array}[]{ccc}g_{1}^{r}&\cdots&g_{j_{r}}^{r}\\ v_{1}^{r}&\cdots&v_{j_{r}}^{r}\\ h_{1}^{r}&\cdots&h_{j_{r}}^{r}\end{array}\!\!\right)}} of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ such that $\alpha_{S}=\gamma_{S}^{l}\beta_{S}\gamma_{S}^{r}$, $\beta_{S}=\delta_{S}^{l}\alpha_{S}\delta_{S}^{r}$ and $i\leqslant k_{l},k_{r},j_{l},j_{r}\leqslant n$ (see [13] or [14, Section II.1]). Also, the definition of the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that without loss of generality we may assume that $k_{l}=k_{r}=j_{l}=j_{r}=i$. Then the equalities $\alpha_{S}=\gamma_{S}^{l}\beta_{S}\gamma_{S}^{r}$ and $\beta_{S}=\delta_{S}^{l}\alpha_{S}\delta_{S}^{r}$ imply that $\{a_{1},\ldots,a_{i}\}=\{g_{1}^{l},\ldots,g_{i}^{l}\}=\{h_{1}^{l},\ldots,h_{i}^{l}\}$, $\{b_{1},\ldots,b_{i}\}=\{f_{1}^{r},\ldots,f_{i}^{r}\}=\{g_{1}^{r},\ldots,g_{i}^{r}\}$, $\{c_{1},\ldots,c_{i}\}=\{g_{1}^{l},\ldots,g_{i}^{l}\}=\{f_{1}^{l},\ldots,f_{i}^{l}\}$ and $\{d_{1},\ldots,d_{i}\}=\{e_{1}^{r},\ldots,e_{i}^{r}\}=\{h_{1}^{r},\ldots,h_{i}^{r}\}$. Now, the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that there exist permutations $\sigma,\rho,\zeta,\varsigma,\nu,\kappa\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $a_{1}=e^{l}_{(1)\sigma}$, $\ldots,a_{i}=e^{l}_{(i)\sigma}$, $c_{1}=f^{l}_{(1)\rho}$, $\ldots,c_{i}=f^{l}_{(i)\rho}$, $d_{1}=e^{r}_{(1)\zeta}$, $\ldots,d_{i}=e^{r}_{(i)\zeta}$, $c_{1}=g^{l}_{(1)\varsigma}$, $\ldots,c_{i}=g^{l}_{(i)\varsigma}$, $a_{1}=h^{l}_{(1)\nu}$, $\ldots,a_{i}=h^{l}_{(i)\nu}$ and $b_{1}=g^{r}_{(1)\kappa}$, $\ldots,b_{i}=g^{r}_{(i)\kappa}$, and hence we have that

[TABLE]

and

[TABLE]

Then the definition of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ implies the following equalities

[TABLE]

Now, by the equality $\alpha_{S}=\gamma_{S}^{l}\beta_{S}\gamma_{S}^{r}$ we get that

[TABLE]

which implies the following equalities

[TABLE]

Hence for the permutation $\pi=\nu^{-1}\varsigma\rho^{-1}\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ we have that $s_{1}\mathscr{J}t_{(1)\pi}$, $\ldots,s_{i}\mathscr{J}t_{(i)\pi}$ in $S$.

$(\Leftarrow)$ Suppose that for elements $\alpha_{S},\beta_{S}\in\mathscr{I}_{\lambda}^{n}(S)$ there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{J}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{J}t_{(i)\sigma}$ in $S$. Then there exist $u_{1},\ldots,u_{i},v_{1},\ldots,v_{i},x_{1},\ldots,x_{i},y_{1},\ldots,y_{i}\in S^{1}$ such that

[TABLE]

Thus, we have that

[TABLE]

and

[TABLE]

and hence we get that $\alpha_{S}\mathscr{J}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. ∎

Remark 3.9**.**

Proposition 3.8$(iv)$ implies that if if there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $s_{1}\mathscr{H}t_{(1)\sigma}$, $\ldots,s_{i}\mathscr{H}t_{(i)\sigma}$ in $S$ then $\alpha_{S}\mathscr{H}\beta_{S}$ in $\mathscr{I}_{\lambda}^{n}(S)$. But Example 3.10 implies that converse statement is not true.

Example 3.10**.**

Let $\lambda$ be any cardinal $\geqslant 2$ and ${\mathscr{C}}(p,q)$ be the bicyclic monoid. The bicyclic monoid ${\mathscr{C}}(p,q)$ is the semigroup with the identity $1$ generated by two elements $p$ and $q$ subjected only to the condition $pq=1$. The distinct elements of ${\mathscr{C}}(p,q)$ are exhibited in the following useful array

[TABLE]

and the semigroup operation on ${\mathscr{C}}(p,q)$ is determined as follows:

[TABLE]

We fix arbitrary distinct elements $a_{1}$ and $a_{1}$ of $\lambda$ and put

[TABLE]

Then we have that

[TABLE]

and hence $\alpha\mathscr{R}\beta$ in $\mathscr{I}_{\lambda}^{n}(S)$, and similar we have that

[TABLE]

and hence $\alpha\mathscr{L}\beta$ in $\mathscr{I}_{\lambda}^{n}(S)$. Thus $\alpha\mathscr{H}\beta$ in $\mathscr{I}_{\lambda}^{n}(S)$, but the elements $qp$ and $q^{2}p^{2}$ are not pairwise $\mathscr{H}$-equivalent to $qp^{2}$ and $q^{2}p$ for any permutation $\sigma\colon\{1,2\}\rightarrow\{1,2\}$.

Recall [28], a semigroup $S$ is said to be:

1. (a)

left stable if for $a,b\in S$, $Sa\subseteq Sab$ implies $Sa=Sab$; 2. (b)

right stable if for $c,d\in S$, $cS\subseteq dcS$ implies $cS=dcS$; 3. (b)

stable if it is both left and right stable.

We observe that in the book [11] there given other definition of a stable semigroup, and these two notion are distinct. A semigroup stable in the sense of Koch and Wallace is always stable in the sense of the book [11], but not conversely (see: [30]). For semigroups with an identity element and for regular semigroups these two definitions of stability coincide.

The following proposition states that the construction of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ preserves left an right stabilities.

Proposition 3.11**.**

For every semigroup $S$, any non-zero cardinal $\lambda$ and any positive integer $n\leqslant\lambda$ the following statements hold:

• $(i)$

$\mathscr{I}_{\lambda}^{n}(S)$* is right stable if and only if so is $S$;*

• $(ii)$

$\mathscr{I}_{\lambda}^{n}(S)$* is left stable if and only if so is $S$;*

• $(iii)$

$\mathscr{I}_{\lambda}^{n}(S)$* is stable if and only if so is $S$.*

Proof.

$(i)$ $(\Leftarrow)$ Suppose the semigroup $S$ is right stable and assume that \alpha_{S}={\small{\left(\begin{array}[]{ccc}a_{1}&\cdots&a_{i}\\ s_{1}&\cdots&s_{i}\\ b_{1}&\cdots&b_{i}\end{array}\right)}} and \beta_{S}={\small{\left(\begin{array}[]{ccc}c_{1}&\cdots&c_{k}\\ t_{1}&\cdots&t_{k}\\ d_{1}&\cdots&d_{k}\end{array}\right)}} are elements of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ such that $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)\subseteq\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$. Then the above inclusion and the definition of the semigroup operation on $\mathscr{I}_{\lambda}^{n}(S)$ imply that $i\leqslant k$ and the following inclusion holds

[TABLE]

Without loss of generality we may assume that $d_{1}=a_{1}$, $\ldots,d_{i}=a_{i}$. Then the inclusion $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)\subseteq\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$ implies that there exists a permutation $\sigma\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ such that $c_{1}=a_{(1)\sigma}$, $\ldots,c_{i}=a_{(i)\sigma}$. Hence by the definition of the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ we get that

[TABLE]

and

[TABLE]

Hence, the inclusion $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)\subseteq\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$ and semigroup operations of $\mathscr{I}_{\lambda}^{n}(S)$ and $S$ imply that $s_{l}S\subseteq t_{(l)\sigma^{-1}}s_{(l)\sigma^{-1}}S$, for every $l\in\{1,\ldots,i\}$. Since the semigroup of all permutations of a finite set is finite, we conclude that there exists a positive integer $j$ such that $\sigma^{j}\colon\{1,\ldots,i\}\rightarrow\{1,\ldots,i\}$ is an identity map and therefore we get that $\sigma^{j-1}=\sigma$. This implies that for every $l\in\{1,\ldots,i\}$ we have that

[TABLE]

Then the right stability of the semigroup $S$ implies the equality $s_{l}S=t_{(l)\sigma^{-1}}t_{(l)\sigma^{-2}}\cdots t_{l}s_{l}S$ and hence we have that $s_{l}S=t_{(l)\sigma^{-1}}s_{(l)\sigma^{-1}}S$, for every $l\in\{1,\ldots,i\}$. Then the inclusion $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)\subseteq\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$ and above formulae imply the following equality $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)=\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$, and hence the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is right stable.

$(\Rightarrow)$ Suppose that the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is right stable and $sS\subseteq tsS$ for $s,t\in S$. We fix an arbitrary $a\in\lambda$ and put \alpha_{S}={\small{\left(\begin{array}[]{c}a\\ s\\ a\end{array}\right)}} and \beta_{S}={\small{\left(\begin{array}[]{c}a\\ t\\ a\end{array}\right)}}. Hence by the definition of the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ we get that

[TABLE]

and

[TABLE]

and hence by the inclusion $sS\subseteq tsS$ we have that $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)\subseteq\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$. Now the right stability of $\mathscr{I}_{\lambda}^{n}(S)$ implies the following equality $\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)=\beta_{S}\alpha_{S}\mathscr{I}_{\lambda}^{n}(S)$. This implies $[sS]^{(a)}_{(p)}=[tsS]^{(a)}_{(p)}$ in $\mathscr{I}_{\lambda}^{n}(S)$ for every $p\in\lambda$, and hence $sS=tsS$.

The proof of statement $(ii)$ is dual to statement $(i)$.

$(iii)$ follows from statements $(i)$ and $(ii)$. ∎

4. On semigroups with a tight ideal series

Fix an arbitrary positive integer $m$ and any $p\in\{0,\ldots,m\}$. Let $A$ be a non-empty set and let $B$ be a non-empty proper subset of $A$. By $\left[B\subset A\right]^{m}_{p}$ we denote all elements $(x_{1},\ldots,x_{m})$ of the power $A^{m}$ which satisfy the following property: at most $p$ coordinates of $(x_{1},\ldots,x_{m})$ belong to $A\setminus B$. It is obvious that $\left[B\subset A\right]^{m}_{m}=A^{m}$ any positive integer $m$, any non-empty set $A$ and any non-empty proper subset $B$ of $A$.

The above definition implies the following two lemmas.

Lemma 4.1**.**

Let $m$ be an arbitrary positive integer and $p\in\{1,\ldots,m\}$. Let $A$ be a non-empty set and let $B$ be a non-empty proper subset of $A$. Then the set $\left[B\subset A\right]^{m}_{p}\setminus\left[B\subset A\right]^{m}_{p-1}$ consists of all elements $(x_{1},\ldots,x_{m})$ of the power $A^{m}$ such that exactly $p$ coordinates of $(x_{1},\ldots,x_{m})$ belong to $A\setminus B$.

Lemma 4.2**.**

Let $m$ be an arbitrary positive integer and $p\in\{0,1,\ldots,m\}$. Let $S$ be a semigroup, $A$ and $B$ be ideals in $S$ such that $B\subset A$. Then $\left[B\subset A\right]^{m}_{p}$ is an ideal of the direct power $S^{m}$.

An subset $D$ of a semigroup $S$ is said to be $\omega$-unstable if $D$ is infinite and $aB\cup Ba\nsubseteq D$ for any $a\in D$ and any infinite subset $B\subseteq D$.

Definition 4.3** ([18]).**

An ideal series (see, for example, [11]) for a semigroup $S$ is a chain of ideals

[TABLE]

We call the ideal series tight if $I_{0}$ is a finite set and $D_{k}=I_{k}\setminus I_{k-1}$ is an $\omega$-unstable subset for each $k=1,\ldots,n$.

It is obvious that for every infinite cardinal $\lambda$ and any positive integer $n$ the semigroup $\mathscr{I}_{\lambda}^{n}$ has a tight ideal series. A finite direct product of semigroups with tight ideal series is a semigroup with a tight ideal series and a homomorphic image of a semigroup with a tight ideal series with finite preimages is a semigroup with a tight ideal series too [18].

A subset $D$ of a semigroup $S$ is said to be strongly $\omega$-unstable if $D$ is infinite and $aB\cup Bb\nsubseteq D$ for any $a,b\in D$ and any infinite subset $B\subseteq D$. It is obvious that a subset $D$ of a semigroup $S$ is strongly $\omega$-unstable then $D$ is $\omega$-unstable.

Definition 4.4**.**

We call the ideal series $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{n}=S$ strongly tight if $I_{0}$ is a finite set and $D_{k}=I_{k}\setminus I_{k-1}$ is a strongly $\omega$-unstable subset for each $k=1,\ldots,n$.

An example of a semigroup with a strongly tight ideal series gives the following proposition.

Proposition 4.5**.**

Let $\lambda$ be any infinite cardinal and $n$ be any positive integer. Then

[TABLE]

is the strongly tight ideal series in the semigroup $\mathscr{I}_{\lambda}^{n}$.

Proof.

The definition of the semigroup $\mathscr{I}_{\lambda}^{n}$ implies that $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{n}$ is an ideal series in $\mathscr{I}_{\lambda}^{n}$.

Fix an arbitrary integer $i=1,\ldots,n$. For any infinite subset $B$ of $\mathscr{I}_{\lambda}^{i}\setminus\mathscr{I}_{\lambda}^{i-1}$ at least one of the following families of sets

[TABLE]

is infinite. Then the definition of the semigroup operation in $\mathscr{I}_{\lambda}^{n}$ implies that $\alpha B\nsubseteq\mathscr{I}_{\lambda}^{i}\setminus\mathscr{I}_{\lambda}^{i-1}$ in the case when the set $\mathfrak{d}(B)$ is infinite, and $B\beta\nsubseteq\mathscr{I}_{\lambda}^{i}\setminus\mathscr{I}_{\lambda}^{i-1}$ in the case when the set $\mathfrak{r}(B)$ is infinite, for any $\alpha,\beta\in\mathscr{I}_{\lambda}^{i}\setminus\mathscr{I}_{\lambda}^{i-1}$. ∎

Later for an arbitrary non-empty set $A$, any positive integer $n$ and any $i\in\{1,\ldots,n\}$ by $\pi_{i}\colon A^{n}\to A$, $(a_{1},\ldots,a_{n})\mapsto a_{i}$ we shall denote the projection on $i$-th factor of the power $A^{n}$.

Proposition 4.6**.**

Let $n$ be a positive integer $\geqslant 2$ and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the strongly tight ideal series for a semigroup $S$. Then the following series

[TABLE]

is a strongly tight ideal series for the direct power $S^{n}$.

Proof.

It is obvious that $I_{0}^{n}$ is a finite ideal of $S^{n}$. Also by Lemma 4.2 all subsets in series (2) are ideals in $S^{n}$.

Fix any $k\in\{1,\ldots,m\}$ and any $p\in\{1,\ldots,n\}$. We claim that the sets

[TABLE]

are strongly $\omega$-unstable in $S^{n}$. Indeed, fix an arbitrary infinite subset $B\subseteq[I_{k-1}\subset I_{k}]^{n}_{p}\setminus[I_{k-1}\subset I_{k}]^{n}_{p-1}$ and any points $a=(a_{1},\ldots,a_{n}),b=(b_{1},\ldots,b_{n})\in[I_{k-1}\subset I_{k}]^{n}_{p}\setminus[I_{k-1}\subset I_{k}]^{n}_{p-1}$. Then there exists a coordinate $i\in\{1,\ldots,n\}$ such that the set $\pi_{i}(B)\subseteq I_{k}\setminus I_{k-1}$ is infinite. If $a_{i}\notin I_{k}\setminus I_{k-1}$ or $b_{i}\notin I_{k}\setminus I_{k-1}$ then $(a_{i}\cdot\pi_{i}(B)\cup\pi_{i}(B)\cdot b_{i})\cap I_{k}\setminus I_{k-1}=\varnothing$, and hence $aB\cup Bb\nsubseteq[I_{k-1}\subset I_{k}]^{n}_{p}\setminus[I_{k-1}\subset I_{k}]^{n}_{p-1}$. If $a_{i},b_{i}\in I_{k}\setminus I_{k-1}$ then $(a_{i}\cdot\pi_{i}(B)\cup\pi_{i}(B)\cdot b_{i})\nsubseteq I_{k}\setminus I_{k-1}$, because the set $I_{k}\setminus I_{k-1}$ is strongly $\omega$-unstable in $S$, and hence $aB\cup Bb\nsubseteq[I_{k-1}\subset I_{k}]^{n}_{p}\setminus[I_{k-1}\subset I_{k}]^{n}_{p-1}$. The proof of the statement that the set $[I_{k-1}\subset I_{k}]^{n}_{1}\setminus I_{k-1}^{n}$ is strongly $\omega$-unstable in $S^{n}$ is similar. ∎

Later we fix an arbitrary positive integer $n$. Then for any semigroup $S$ and any positive integer $k\leqslant n$ since $\mathscr{I}_{\lambda}^{k}(S)$ is a subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$ by $\iota\colon\mathscr{I}_{\lambda}^{k}(S)\to\mathscr{I}_{\lambda}^{n}(S)$ we denote this natural embedding. Similar arguments imply that, without loss of generality for any subsemigroup $T$ of $S$ and any positive integer $k\leqslant n$ since $\mathscr{I}_{\lambda}^{k}(T)$ is a subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$ by $\iota\colon\mathscr{I}_{\lambda}^{k}(T)\to\mathscr{I}_{\lambda}^{n}(S)$ we denote this natural embedding.

Let $A\neq\varnothing$ and $k$ be any positive integer. A subset $B\subseteq A^{k}$ is said to be $k$-symmetric if the following condition holds: $(b_{1},\ldots,b_{k})\in B$ implies $\left(b_{(1)\sigma},\ldots,b_{(k)\sigma}\right)\in B$ for every permutation $\sigma\colon\{1,\ldots,k\}\to\{1,\ldots,k\}$.

Remark 4.7**.**

We observe that every element of the tight ideal series (2) is $m$-symmetric in $S^{n}$, and moreover the sets $[I_{k-1}\subset I_{k}]^{n}_{p}\setminus[I_{k-1}\subset I_{k}]^{n}_{p-1}$ and $[I_{k-1}\subset I_{k}]^{n}_{1}\setminus I_{k-1}^{n}$ is $m$-symmetric in $S^{n}$, too, for all $k\in\{1,\ldots,m\}$ and $p\in\{1,\ldots,n\}$.

We need the following construction.

Construction 4.8**.**

Let $\lambda$ be cardinal $\geqslant 1$, $n$ be any positive integer, $k$ be any positive integer $\leqslant\min\{n,\lambda\}$ and $S$ be a semigroup. For any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$ we define a map $\mathfrak{f}^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}\colon S^{k}\rightarrow S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ by the formula

[TABLE]

For any non-empty subset $A\subseteq S^{k}$ and any positive integer $k\leqslant n$ we determine the following subsets

[TABLE]

and

[TABLE]

of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

The following lemma follows from the definition of $k$-symmetric sets.

Lemma 4.9**.**

Let $\lambda$ be cardinal $\geqslant 1$, $k$ be any positive integer $\leqslant\lambda$ and $S$ be a semigroup. Let $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ be arbitrary ordered collections of $k$ distinct elements of $\lambda^{k}$. If $A\neq\varnothing$ is a $k$-symmetric subset of $S^{k}$ then $(A)\mathfrak{f}^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}=(A)\mathfrak{f}^{(a_{(1)\sigma},\ldots,a_{(k)\sigma})}_{(b_{(1)\sigma},\ldots,b_{(k)\sigma})}$ for every permutation $\sigma\colon\{1,\ldots,k\}\to\{1,\ldots,k\}$.

Theorem 4.10**.**

Let $\lambda$ be an infinite cardinal and $n$ be a positive integer. If $S$ is a finite semigroup then

[TABLE]

is the strongly tight ideal series for the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Proof.

It is obvious that for every $i=0,1,\ldots,n$ the set $I_{i}$ is an ideal in $\mathscr{I}_{\lambda}^{n}(S)$ and moreover the set $I_{0}$ is finite.

Fix an arbitrary $i=1,\ldots,n$ and any infinite subset $B\subseteq I_{i}\setminus I_{i-1}$. Since the semigroup $S$ is finite, every infinite subset $X$ of $I_{i}\setminus I_{i-1}$ intersects infinitely many sets of the form $S^{(a_{1},\ldots,a_{i})}_{(b_{1},\ldots,b_{i})}$. Then the definition of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ implies that at least one of the following families of sets

[TABLE]

is infinite. Then the definition of the semigroup operation in $\mathscr{I}_{\lambda}^{n}(S)$ implies that $\alpha B\nsubseteq I_{i}\setminus I_{i-1}$ in the case when the set $\mathfrak{d}(B)$ is infinite, and $B\beta\nsubseteq I_{i}\setminus I_{i-1}$ in the case when the set $\mathfrak{r}(B)$ is infinite, for any $\alpha,\beta\in I_{i}\setminus I_{i-1}$. ∎

Theorem 4.11**.**

Let $\lambda$ be an infinite cardinal, $n$ be a positive integer and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the strongly tight ideal series for a semigroup $S$. Then the following series

[TABLE]

is a strongly tight ideal series for the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Proof.

The definition of the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ and Lemma 4.2 imply that all subsets in series (3) are ideals in $\mathscr{I}_{\lambda}^{n}(S)$.

Since $I_{0}$ is a finite ideal in $S$, the following equalities

[TABLE]

and the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ imply that

[TABLE]

are strongly $\omega$-unstable subsets in $\mathscr{I}_{\lambda}^{n}(S)$.

Next we shall show that the set $J_{k,p}\setminus J_{k,p-1}$ is strongly $\omega$-unstable in $\mathscr{I}_{\lambda}^{n}(S)$ for all $k=1,\ldots,n$ and $p=1,\ldots,km$.

Fix any infinite subset $B$ of $J_{k,p}\setminus J_{k,p-1}$ and any $\alpha,\beta\in J_{k,p}\setminus J_{k,p-1}$. If $\mathop{\textsf{{d}}}(B)\neq\mathop{\textsf{{r}}}(\alpha)$ then the semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that $\alpha B\nsubseteq J_{k,p}\setminus J_{k,p-1}$. Similar, if $\mathop{\textsf{{d}}}(\beta)\neq\mathop{\textsf{{r}}}(B)$ then $B\beta\nsubseteq J_{k,p}\setminus J_{k,p-1}$.

Next we suppose that $\mathop{\textsf{{d}}}(B)=\mathop{\textsf{{r}}}(\alpha)$, $\mathop{\textsf{{d}}}(\beta)=\mathop{\textsf{{r}}}(B)$,

[TABLE]

for some $s_{1},\ldots,s_{k},t_{1},\ldots,t_{k}\in S$ and ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$, $(b_{1},\ldots,b_{k})$, $(c_{1},\ldots,c_{k})$, $(d_{1},\ldots,d_{k})$ of $\lambda^{k}$. Then the set $B$ consists of elements of the form

[TABLE]

where $x_{1},\ldots,x_{k}\in S$ and $\sigma\colon\{1,\ldots,k\}\to\{1,\ldots,k\}$ is a permutation.

First we consider the case when $J_{k,p}=J_{k,jk}=\overline{\left[I_{j}^{k}\right]}^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$ for some $j=1,\ldots,m$. Then

[TABLE]

and $B\subseteq\left[I_{j}^{k}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$. Since the set $B$ is infinite, there exists $b_{i_{0}}\in\{b_{1},\ldots,b_{k}\}$ such that there exist infinitely many $\gamma\in B$ such that $\mathbf{d}(\gamma)\ni b_{i_{0}}$. Without loss of generality we may assume that $b_{i_{0}}=b_{1}$. We put $B_{0}=\left\{\gamma\in B\colon b_{1}\in\mathbf{d}(\gamma)\right\}$. Then the set $B_{0}$ is infinite and hence the set

[TABLE]

is infinite, too. The above implies that there exists a permutation $\sigma_{0}$ of $\{1,\ldots,k\}$ such that infinitely many elements of the form \left(\begin{array}[]{ccc}b_{1}&\cdots&b_{k}\\ x_{1}&\cdots&x_{k}\\ c_{(1)\sigma_{0}}&\cdots&c_{(k)\sigma_{0}}\\ \end{array}\right) belong to $B_{0}$. Since $s_{1},t_{(1)\sigma_{0}}\in I_{j}\setminus I_{j-1}$ and the set $I_{j}\setminus I_{j-1}$ is strongly $\omega$-unstable we obtain that $a_{1}\cdot B_{0}^{S}\cup B_{0}^{S}\cdot t_{(1)\sigma_{0}}\nsubseteq I_{j}\setminus I_{j-1}$, and hence the set $\left[I_{j}^{k}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$ is strongly $\omega$-unstable, as well.

Next we consider the case $J_{k,p}=J_{n,(j-1)k+q}=\overline{\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]}^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$ for some $j=1,\ldots,m$. Then

[TABLE]

and $B\subseteq\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$. Since the set $B$ is infinite, without loss of generality we may assume that $B$ contains an infinite subset $B_{0}$ which consists of elements of the form

[TABLE]

where $x_{1},\ldots,x_{q}\in I_{j}\setminus I_{j-1}$ and $x_{q+1},\ldots,x_{k}\in I_{j-1}\setminus I_{j-2}$ for some ordered collections of $k$ distinct elements $(b_{1},\ldots,b_{k})$ and $(c_{1},\ldots,c_{k})$ of $\lambda^{k}$. Fix arbitrary elements

[TABLE]

of the set $B$. If $s_{u}\notin I_{j}\setminus I_{j-1}$ for some $u\in\{1,\ldots,q\}$ or $s_{v}\notin I_{j-1}\setminus I_{j-2}$ for some $v\in\{q+1,\ldots,k\}$ then $\alpha B_{0}\nsubseteq\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$. Similarly, $t_{u}\notin I_{j}\setminus I_{j-1}$ for some $u\in\{1,\ldots,q\}$ or $t_{v}\notin I_{j-1}\setminus I_{j-2}$ for some $v\in\{q+1,\ldots,k\}$ then $B_{0}\beta\nsubseteq\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$. Hence later we shall assume that $s_{u}\in I_{j}\setminus I_{j-1}$ for all $u\in\{1,\ldots,q\}$, $s_{v}\in I_{j-1}\setminus I_{j-2}$ for all $v\in\{q+1,\ldots,k\}$, $t_{u}\in I_{j}\setminus I_{j-1}$ for all $u\in\{1,\ldots,q\}$ and $t_{v}\in I_{j-1}\setminus I_{j-2}$ for all $v\in\{q+1,\ldots,k\}$. Since the set $B_{0}$ is infinite there exists ${i_{0}}\in\{1,\ldots,k\}$ such that there exist infinitely many $\gamma\in B_{0}$ such that $\mathbf{d}(\gamma)\ni b_{i_{0}}$. We put $B_{1}=\left\{\gamma\in B_{0}\colon b_{i_{0}}\in\mathbf{d}(\gamma)\right\}$. Since the set $B_{1}$ is inifinite the following statements hold:

• $(1)$

if $i_{0}\in\{1,\ldots,q\}$ then $s_{i_{0}}A\cup At_{i_{0}}\nsubseteq I_{j}\setminus I_{j-1}$, where

[TABLE]

because the set $I_{j}\setminus I_{j-1}$ is strongly $\omega$-unstable in $S$;

• $(2)$

if $i_{0}\in\{q+1,\ldots,k\}$ then $s_{i_{0}}A\cup At_{i_{0}}\nsubseteq I_{j-1}\setminus I_{j-2}$, where

[TABLE]

because the set $I_{j-1}\setminus I_{j-2}$ is strongly $\omega$-unstable in $S$.

Both above statements imply that $\alpha B_{1}\cup B_{1}\gamma\nsubseteq\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$ and hence $\alpha B\cup B_{1}\gamma\nsubseteq\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$, i.e., the set $\left[[I_{j-1}\subset I_{j}]^{k}_{q}\right]^{(\ast)_{k}}_{\mathscr{I}_{\lambda}^{n}(S)}$ is strongly $\omega$-unstable in $\mathscr{I}_{\lambda}^{n}(S)$. This completed the proof of the theorem. ∎

Theorem 4.11 implies the following

Corollary 4.12**.**

Let $\lambda$ be an infinite cardinal, $n$ be a positive integer and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the strongly tight ideal series for a semigroup $S$. Then the ideal series (3) is tight for the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

The proof of the following theorem is similar to Theorem 4.11.

Theorem 4.13**.**

Let $\lambda$ be a finite cardinal, $n$ be a positive integer $\leqslant\lambda$ and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the strongly tight ideal series for a semigroup $S$. Then the following series

[TABLE]

is a strongly tight ideal series for the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Theorem 4.13 implies the following

Corollary 4.14**.**

Let $\lambda$ be a finite cardinal, $n$ be a positive integer $\leqslant\lambda$ and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the strongly tight ideal series for a semigroup $S$. Then the ideal series (3) is tight for the semigroup $\mathscr{I}_{\lambda}^{n}(S)$.

Let $\mathfrak{S}$ be a class of semitopological semigroups. A semigroup $S\in\mathfrak{S}$ is called $H$-closed in $\mathfrak{S}$, if $S$ is a closed subsemigroup of any semitopological semigroup $T\in\mathfrak{S}$ which contains $S$ both as a subsemigroup and as a topological space. The $H$-closed topological semigroups were introduced by Stepp in [32], and there they were called maximal semigroups. An algebraic semigroup $S$ is called:algebraically complete in $\mathfrak{S}$, if $S$ with any Hausdorff topology $\tau$ such that $(S,\tau)\in\mathfrak{S}$ is $H$-closed in $\mathfrak{S}$. We observe that many distinct types of $H$-closedness of topological and semitopological semigroups studied in [1]–[10], [16]–[21], [24], [26].

By Proposition 10 from [18] every inverse semigroup $S$ with a tight ideal series is algebraically complete in the class of Hausdorff semitopological inverse semigroups with continuous inversion. Hence Proposition 3.5 and Theorems 4.11, 4.13 imply the following

Theorem 4.15**.**

Let $S$ be an inverse semigroup which admits a strongly tight ideal series. Then for every non-zero cardinal $\lambda$ and any positive integer $n\leqslant\lambda$ the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ is algebraically complete in the class of Hausdorff semitopological inverse semigroups with continuous inversion.

We remark that in the case when $n=1$ the construction of $\mathscr{I}_{\lambda}^{1}(S)$ preserves the property to be a semigroup with a tight ideal series, and this follows from the following theorem.

Theorem 4.16**.**

Let $\lambda$ be any non-zero cardinal, $n$ be a positive integer $n\leqslant\lambda$ and let $I_{0}\subseteq I_{1}\subseteq I_{2}\subseteq\cdots\subseteq I_{m}=S$ be the tight ideal series for a semigroup $S$. Then the following series

[TABLE]

is a tight ideal series for the semigroup $\mathscr{I}_{\lambda}^{1}(S)$ in the case when $\lambda$ is infinite, and

[TABLE]

is a tight ideal series for the semigroup $\mathscr{I}_{\lambda}^{1}(S)$ in the case when $\lambda$ is finite.

Proof.

We consider the case when cardinal $\lambda$ is infinite. In the other case the proof is similar.

The semigroup operation of $\mathscr{I}_{\lambda}^{1}(S)$ implies that the the set $J_{k}$ is ideal in $\mathscr{I}_{\lambda}^{1}(S)$ for an arbitrary integer $k\in\{0,1,\ldots,m+1\}$.

Fix an arbitrary $k\in\{1,\ldots,m+1\}$. Then for any infinite subset $B$ of $J_{k}\setminus J_{k-1}$ and any \alpha=\left(\begin{array}[]{c}a\\ s\\ b\\ \end{array}\right)\in J_{k}\setminus J_{k-1} the following statements hold.

• $(1)$

If $B\cap S^{(i)}_{(i)}$ is infinite for some $i\in\lambda$ then $B\cap S^{(i)}_{(i)}\subseteq[I_{k-1}\setminus I_{k_{2}}]^{(i)}_{(i)}$. Hence, the semigroup operation of $\mathscr{I}_{\lambda}^{1}(S)$ implies that $\alpha B\cup B\alpha\nsubseteq J_{k}\setminus J_{k-1}$ in the case when $a=b=i$, because the set $I_{k-1}\setminus I_{k_{2}}$ is $\omega$-unstable in $S$. Otherwise $0\in\alpha B\cup B\alpha\nsubseteq J_{k}\setminus J_{k-1}$.

• $(2)$

In the other case the semigroup operation of $\mathscr{I}_{\lambda}^{1}(S)$ implies that $0\in\alpha B\cup B\alpha\nsubseteq J_{k}\setminus J_{k-1}$.

Both above statements imply that the set $J_{k}\setminus J_{k-1}$ is $\omega$-unstable in $\mathscr{I}_{\lambda}^{1}(S)$, which completes the proof of the theorem. ∎

5. On a semitopological semigroup $\mathscr{I}_{\lambda}^{n}(S)$

For any element \alpha=\left(\begin{array}[]{ccc}i_{1}&\ldots&i_{k}\\ j_{1}&\ldots&j_{k}\\ \end{array}\right) of the semigroup $\mathscr{I}_{\lambda}^{n}$ and any $s\in S$ we denote \alpha[s]=\left(\begin{array}[]{ccc}i_{1}&\ldots&i_{k}\\ s&\ldots&s\\ j_{1}&\ldots&j_{k}\\ \end{array}\right) which is the element of $\mathscr{I}_{\lambda}^{n}(S)$. Later in this case we shall say that $\alpha[s]$ is the $s$-extension of $\alpha$ or $\alpha$ is the $\mathscr{I}_{\lambda}^{n}$-restriction of $\alpha[s]$.

Proposition 5.1**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\mathscr{I}_{\lambda}^{n}(S)$ be a Hausdorff semitopological semigroup. Then for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$ and any element $\alpha_{S}\in S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ there exists an open neighbourhood $U(\alpha_{S})$ of $\alpha_{S}$ such that

• •

$U(\alpha_{S})\cap\mathscr{I}_{\lambda}^{k-1}(S)=\varnothing$* and $U(\alpha_{S})\cap\mathscr{I}_{\lambda}^{k}(S)\subseteq S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ in the case when $k\geqslant 2$,*

• •

$0\notin U(\alpha_{S})$* and $U(\alpha_{S})\cap\mathscr{I}_{\lambda}^{1}(S)\subseteq S^{(a_{1})}_{(b_{1})}$ in the case when $k=1$.*

Thus $\mathscr{I}_{\lambda}^{k}(S)$ is a closed subsemigroup of $\mathscr{I}_{\lambda}^{n}(S)$.

Proof.

Fix an arbitrary $k\leqslant n$ and an arbitrary \alpha_{S}=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}&\ldots&s_{k}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right)\in S^{a_{1},\ldots,a_{k}}_{b_{1},\ldots,b_{k}}. It is obvious that $\varepsilon_{1}[1_{S}]\cdot\alpha_{S}\cdot\varepsilon_{2}[1_{S}]=\alpha_{S}$, where

[TABLE]

and $1_{S}$ is the unit element of $S$.

Simple calculation implies that

[TABLE]

We observe that $eT$ and $Te$ are closed subset in an arbitrary Hausdorff semitopological semigroup $T$ for any its idempotent $e$. Since for any idempotent $\varepsilon\in\mathscr{I}_{\lambda}^{n}$ the set ${\downarrow}\varepsilon=\{\iota\in E(\mathscr{I}_{\lambda}^{n})\colon\iota\leqslant\varepsilon\}$ is finite, the set

[TABLE]

is closed in $\mathscr{I}_{\lambda}^{n}(S)$. Fix an arbitrary open neighbourhood $W(\alpha_{S})$ of $\alpha_{S}$ such that $W(\alpha_{S})\cap A_{\alpha_{S}}=\varnothing$. The separate continuity of the semigroup operation on $\mathscr{I}_{\lambda}^{n}(S)$ implies that there exist an open neighbourhood $U(\alpha_{S})$ of $\alpha_{S}$ such that $\varepsilon_{1}[1_{S}]\cdot U(\alpha_{S})\cdot\varepsilon_{2}[1_{S}]\subseteq W(\alpha_{S})$. The neighbourhood $U(\alpha_{S})$ is requested. Indeed, if there exists $\beta_{S}\in\mathscr{I}_{\lambda}^{k}(S)\setminus S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ then $\varepsilon_{1}[1_{S}]\cdot\beta_{S}\cdot\varepsilon_{2}[1_{S}]\in A_{\alpha_{S}}$. ∎

Remark 5.2**.**

We observe that in Proposition 5.1 we may assume that the neighbourhood $U(\alpha_{S})$ may be chosen with the following property $\varepsilon_{1}[1_{S}]\cdot U(\alpha_{S})\cdot\varepsilon_{2}[1_{S}]\subseteq S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$.

Proposition 5.3**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\mathscr{I}_{\lambda}^{n}(S)$ be a Hausdorff semitopological semigroup. Then for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$, $(b_{1},\ldots,b_{k})$, $(c_{1},\ldots,c_{k})$, and $(d_{1},\ldots,d_{k})$ of $\lambda^{k}$ the subspaces $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ and $S^{(c_{1},\ldots,c_{k})}_{(d_{1},\ldots,d_{k})}$ are homeomorphic, and moreover $S^{(a_{1},\ldots,a_{k})}_{(a_{1},\ldots,a_{k})}$ and $S^{(c_{1},\ldots,c_{k})}_{(c_{1},\ldots,c_{k})}$ are topologically isomorphic subsemigroups of $\mathscr{I}_{\lambda}^{n}(S)$.

Proof.

Since $\mathscr{I}_{\lambda}^{n}(S)$ is a semitopological semigroup, the restrictions of the following maps

[TABLE]

and

[TABLE]

on the subspaces $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ and $S^{(c_{1},\ldots,c_{k})}_{(d_{1},\ldots,d_{k})}$, respectively, are mutually inverse, and hence $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ and $S^{(c_{1},\ldots,c_{k})}_{(d_{1},\ldots,d_{k})}$ are homeomorphic subspaces in $\mathscr{I}_{\lambda}^{n}(S)$. Also, it is obvious that in the case of subsemigroups $S^{(a_{1},\ldots,a_{k})}_{(a_{1},\ldots,a_{k})}$ and $S^{(c_{1},\ldots,c_{k})}_{(c_{1},\ldots,c_{k})}$ so defined restrictions of maps are topological isomorphisms. ∎

For any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$ we define a map

[TABLE]

Proposition 5.1 implies the following corollary.

Corollary 5.4**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\mathscr{I}_{\lambda}^{n}(S)$ be a Hausdorff semitopological semigroup. Then the set

[TABLE]

is open in $\mathscr{I}_{\lambda}^{n}(S)$ for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$.

We recall that a topological space $X$ is said to be

• •

compact if each open cover of $X$ has a finite subcover;

• •

H-closed if $X$ is a closed subspace of every Hausdorff topological space in which it contained.

It is well known that every Hausdorff compact space is H-closed, and every regular H-closed topological space is compact (see [12, 3.12.5]).

Lemma 5.5**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\mathscr{I}_{\lambda}^{n}(S)$ be a Hausdorff semitopological semigroup. If $S^{(a)}_{(b)}$ is a closed subset of $\mathscr{I}_{\lambda}^{n}(S)$ for any $a,b\in\lambda$ then $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ is a closed subspace of $\mathscr{I}_{\lambda}^{n}(S)$ for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$.

Proof.

For any $a,b\in\lambda$ the map

[TABLE]

is continuous, because $\mathscr{I}_{\lambda}^{n}(S)$ is a semitopological semigroup. This and Proposition 5.1 imply that

[TABLE]

a closed subspace of $\mathscr{I}_{\lambda}^{n}(S)$. ∎

Since a continuous image of a compact (an H-closed) space is compact (H-closed) (see [12, Chapter 3]), Proposition 5.3 and Lemma 5.5 imply the following corollary.

Corollary 5.6**.**

Let $S$ be a monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\mathscr{I}_{\lambda}^{n}(S)$ be a Hausdorff semitopological semigroup. If the set $S^{(a)}_{(b)}$ is H-closed (compact) in $\mathscr{I}_{\lambda}^{n}(S)$ for some $a,b\in\lambda$ then $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ is a closed subspace of $\mathscr{I}_{\lambda}^{n}(S)$ for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$.

Definition 5.7**.**

Let $\mathfrak{S}$ be a class of semitopological semigroups. Let $\lambda\geqslant 1$ be a cardinal, $n$ be a positive integer $\leqslant\lambda$, and $(S,\tau)\in\mathfrak{S}$. Let $\tau_{\mathscr{I}}$ be a topology on $\mathscr{I}_{\lambda}^{n}(S)$ such that

• a)

$\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)\in\mathfrak{S}$;

• b)

the topological subspace $\left(S_{(a)}^{(a)},\tau_{B}|_{S_{\alpha,\alpha}}\right)$ is naturally homeomorphic to $(S,\tau)$ for some $a\in{\lambda}$, i.e., the map $\mathfrak{H}\colon S\to\mathscr{I}_{\lambda}^{n}(S)$, s\mapsto\left(\begin{array}[]{c}a\\ s\\ a\\ \end{array}\right) is a topological embedding.

Then $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ is called a topological $\mathscr{I}_{\lambda}^{n}$-extension of $(S,\tau)$ in $\mathfrak{S}$.

Lemma 5.8**.**

Let $(S,\tau)$ be a semitopological monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ be a topological $\mathscr{I}_{\lambda}^{n}$-extension of $(S,\tau)$ in the class of semitopological semigroups. Let $U_{1}(s_{1}),\ldots,U_{k}(s_{k})$ be open neighbourhoods of the points $s_{1},\ldots,s_{k}$ in $(S,\tau)$, respectively. Then the following sets

[TABLE]

and

[TABLE]

are open neighbourhoods of the points

[TABLE]

in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$, respectively, for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$.

Proof.

Since $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ be a topological $\mathscr{I}_{\lambda}^{n}$-extension of $(S,\tau)$ in the class of Hausdorff semitopological semigroups, there exist open neighbourhoods $W_{1},\ldots,W_{k}$ of of the points \left(\begin{array}[]{c}a_{1}\\ s_{1}\\ b_{1}\\ \end{array}\right),\cdots,\left(\begin{array}[]{c}a_{k}\\ s_{k}\\ b_{k}\\ \end{array}\right) in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$, respectively, such that

[TABLE]

Then the requested statement of the lemma follows from the separate continuity of the semigroup operation in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$. ∎

Theorem 5.9**.**

Let $(S,\tau)$ be a Hausdorff compact semitopological monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ be a compact topological $\mathscr{I}_{\lambda}^{n}$-extension of $(S,\tau)$ in the class of Hausdorff semitopological semigroups. Then the subspace $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ is compact and moreover it is homeomorphic to the power $S^{k}$ with the product topology by the mapping

[TABLE]

for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$.

Proof.

Since the monoid $(S,\tau)$ is compact Corollary 5.6 implies that $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ a closed subset of of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$. Then compactness of of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ implies that $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ is compact, as well.

It is obvious that the above defined map $\mathfrak{H}\colon S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}\to S^{k}$ is a bijection. Also, Lemma 5.8 implies that the map $\mathfrak{H}$ is continuous, and it is a homeomorphism, because $S^{k}$ and $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ are compacta. ∎

Proposition 5.1 and Theorem 5.9 imply the following corollary.

Corollary 5.10**.**

Let $(S,\tau)$ be a Hausdorff compact semitopological monoid, $\lambda$ be any non-zero cardinal, $n$ be an arbitrary positive integer $\leqslant\lambda$, $0<k\leqslant n$ and $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ be a compact topological $\mathscr{I}_{\lambda}^{n}$-extension of $(S,\tau)$ in the class of Hausdorff semitopological semigroups. Then $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ is an open-and-closed subset of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ for any ordered collections of $k$ distinct elements $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $\lambda^{k}$, and the space $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}\right)$ is the topological sum of such sets with isolated zero.

Remark 5.11**.**

Since by Theorem  of [21] an infinite semigroup of matrix units and hence an infinite semigroup $\mathscr{I}_{\lambda}^{n}$ do not embed into compact Hausdorff topological semigroups, Corollary 5.10 describes compact topological $\mathscr{I}_{\lambda}^{n}$-extensions of compact semigroups $(S,\tau)$ in the class of Hausdorff topological semigroups.

Example 5.12**.**

Let $(S,\tau_{S})$ be a compact Hausdorff semitopological monoid. On the semigroup $\mathscr{I}_{\lambda}^{n}(S)$ we define a topology $\tau_{\mathscr{I}}^{\mathbf{c}}$ in the following way. Put

[TABLE]

for any $k=1,\ldots,n$, and

[TABLE]

The topology $\tau_{\mathscr{I}}^{\mathbf{c}}$ on $\mathscr{I}_{\lambda}^{n}(S)$ is generated by the family

[TABLE]

as a subbase.

Remark 5.13**.**

Lemma 5.8 and the definition of the topology $\tau_{\mathscr{I}}^{\mathbf{c}}$ on $\mathscr{I}_{\lambda}^{n}(S)$ implies that the following statements hold.

• (1)

For any $k=1,\ldots,n$ and every ordered collection $(a_{1},\ldots,a_{k})$ and $(b_{1},\ldots,b_{k})$ of $k$ distinct elements of $\lambda^{k}$ the set ${\Uparrow}S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ is closed in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$.

• (2)

For any element \alpha_{S}=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}&\ldots&s_{k}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right) of $\mathscr{I}_{\lambda}^{n}(S)$ and any open neighbourhoods $U_{1}(s_{1}),\ldots,U_{k}(s_{k})$ of the points $s_{1},\ldots,s_{k}$ in $(S,\tau)$ the set ${\Uparrow}\left[U_{1}(s_{1}),\ldots,U_{k}(s_{k})\right]^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{l_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{l_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{p}_{1},\ldots,a^{p}_{l_{p}})}_{(b^{p}_{1},\ldots,b^{p}_{l_{p}})}\right)$ such that $\alpha_{S}\notin{\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{l_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{l_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{p}_{1},\ldots,a^{p}_{l_{p}})}_{(b^{p}_{1},\ldots,b^{p}_{l_{p}})}$, is an open neighbourhood of the point $\alpha_{S}$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. Here we have that $\{a_{1},\ldots,a_{k}\}\subsetneqq\left\{a^{j}_{1},\ldots,a^{j}_{l_{j}}\right\}$ and $\{b_{1},\ldots,b_{k}\}\subsetneqq\left\{b^{j}_{1},\ldots,b^{j}_{l_{j}}\right\}$ for all $j=1,\ldots,p$.

Theorem 5.14**.**

If $(S,\tau_{S})$ is a compact Hausdorff semitopological monoid then $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is a compact Hausdorff semitopological semigroup.

Proof.

It is obvious that the topology $\tau_{\mathscr{I}}^{\mathbf{c}}$ is Hausdorff.

By the Alexander Subbase Theorem (see [12, 3.12.2]) it is sufficient to show that every open cover of $\mathscr{I}_{\lambda}^{n}(S)$ which consists of elements of the subbase $\mathscr{P}^{\mathbf{c}}$ has a finite subcover.

We shall show that the space $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is compact by induction. In the case when $n=1$, Corollary 13 from [23] implies that the space $\left(\mathscr{I}_{\lambda}^{1}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is compact. Next we shall show the step of induction: $\left(\mathscr{I}_{\lambda}^{k-1}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is compact implies $\left(\mathscr{I}_{\lambda}^{k}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is compact, too, for $k=2,\ldots,n$. Without loss of generality we my assume that $k=n$.

Let $\mathscr{U}$ be an arbitrary open cover of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ which consists of elements of the subbase $\mathscr{P}^{\mathbf{c}}$. The assumption of induction implies that there exists a finite subfamily $\mathscr{U}_{n-1}$ of $\mathscr{U}$ which is a subcover of $\mathscr{I}_{\lambda}^{n-1}(S)$. Fix an arbitrary element $V_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus{\Uparrow}S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}\in\mathscr{U}_{n-1}$ which contains the zero [math] of $\mathscr{I}_{\lambda}^{n}(S)$. Then $p\in\{1,\ldots,n\}$.

We observe that an arbitrary element $U_{0}$ of the family $\left\{\mathscr{P}_{k}^{\mathbf{c}}(0)\colon k=1,\ldots,n\right\}$ contains the set $S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}$ if and only if $U_{0}\cap S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}\neq\varnothing$. This implies that only one of the following conditions holds:

• (1)

there does not exist an element of $\mathscr{U}_{n-1}$ from the family $\left\{\mathscr{P}_{k}^{\mathbf{c}}(0)\colon k=1,\ldots,n\right\}$ which contains the set $S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}$;

• (2)

there exists $W_{0}\in\mathscr{U}_{n-1}\cap\left\{\mathscr{P}_{k}^{\mathbf{c}}(0)\colon k=1,\ldots,n\right\}$ such that $S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}\subseteq W_{0}$.

Suppose that condition $(1)$ holds. First we consider the case when $p<n$. By Theorem 5.9, $S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}$ is compact, and hence there exists finitely many elements ${\Uparrow}\left[U(s_{1})\right]^{(c_{1})}_{(d_{1})},\ldots,{\Uparrow}\left[U(s_{m})\right]^{(c_{m})}_{(d_{m})}$ in $\mathscr{U}_{n-1}\cap\mathscr{P}^{\mathbf{c}}\setminus\left\{\mathscr{P}_{k}^{\mathbf{c}}(0)\colon k=1,\ldots,n\right\}$ such that $S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}\subseteq{\Uparrow}\left[U(s_{1})\right]^{(c_{1})}_{(d_{1})}\cup\cdots\cup{\Uparrow}\left[U(s_{m})\right]^{(c_{m})}_{(d_{m})}$. It is obvious that $\left\{U_{0},{\Uparrow}\left[U(s_{1})\right]^{(c_{1})}_{(d_{1})},\ldots,{\Uparrow}\left[U(s_{m})\right]^{(c_{m})}_{(d_{m})}\right\}$ is a finite cover of $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$.

Next, we consider case $p=n$. We identify the set $S^{(a_{1},\ldots,a_{n})}_{(b_{1},\ldots,b_{n})}$ and the power $S^{n}$ by the mapping

[TABLE]

The semigroup operation of $\mathscr{I}_{\lambda}^{n}(S)$ implies that ${\Uparrow}\left[U(s)\right]^{(c)}_{(d)}\cap S^{(a_{1},\ldots,a_{n})}_{(b_{1},\ldots,b_{n})}\neq\varnothing$ if and only if $c=a_{i}$ and $d=b_{i}$ for some $i=1,\ldots,n$. Then by (8) for every $i=1,\ldots,n$ we have that

[TABLE]

Then the subbase $\mathscr{P}^{\mathbf{c}}$ on $\mathscr{I}_{\lambda}^{n}(S)$ and the map (8) determine the product topology on $S^{n}$ from the space $S$, and hence the space $S^{n}$ is compact.

Suppose that $S^{(a_{1},\ldots,a_{n})}_{(b_{1},\ldots,b_{n})}$ is not compact. Then $S^{(a_{1},\ldots,a_{n})}_{(b_{1},\ldots,b_{n})}$ has a cover $\mathscr{W}$ which consists of the open sets of the form ${\Uparrow}\left[U(s)\right]^{(c)}_{(d)}$ and $\mathscr{W}$ does not have a finite subcover. Then the cover $\mathscr{W}_{S^{n}}$ of $S^{n}$ which determines by formula (9) from the family $\mathscr{W}$ has no finite subcover, too. This contradicts the compactness of $S^{n}$.

Hence in case $(1)$ the cover $\mathscr{U}$ of $\mathscr{I}_{\lambda}^{n}(S)$ has a finite subcover.

Suppose that condition $(2)$ holds. Then $W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus{\Uparrow}S^{(c_{1},\ldots,c_{q})}_{(d_{1},\ldots,d_{q})}$ with $q\leqslant n$. If ${\Uparrow}S^{(c_{1},\ldots,c_{q})}_{(d_{1},\ldots,d_{q})}\cap{\Uparrow}S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}=\varnothing$ then $\{V_{0},W_{0}\}$ is a cover of $\mathscr{I}_{\lambda}^{n}(S)$. In the other case there exists a smallest positive integer $p_{1}$ such that $\max\{p+1,q\}\leqslant p_{1}\leqslant n$ and two ordered $p_{1}$-collections of distinct elements $(e_{1},\ldots,e_{p_{1}})$ and $(f_{1},\ldots,f_{p_{1}})$ of the power $\lambda^{p_{1}}$ such that ${\Uparrow}S^{(c_{1},\ldots,c_{q})}_{(d_{1},\ldots,d_{q})}\cap{\Uparrow}S^{(a_{1},\ldots,a_{p})}_{(b_{1},\ldots,b_{p})}={\Uparrow}S^{(e_{1},\ldots,e_{p_{1}})}_{(f_{1},\ldots,f_{p_{1}})}$. Then for the open set $U_{1}=U_{0}\cup W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus{\Uparrow}S^{(e_{1},\ldots,e_{p_{1}})}_{(f_{1},\ldots,f_{p_{1}})}$ either condition $(1)$ or condition $(2)$ holds.

Since $p+1\leqslant p_{1}\leqslant n$, we repeating finitely many items the above procedure we get that the space $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is compact.

Next we shall show that the topology $\tau_{\mathscr{I}}^{\mathbf{c}}$ is shift-continuous on $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. We consider all possible cases.

$(i)$ $0\cdot 0=0$. Then for any open neighbourhood $U_{0}$ of zero in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ we have that

[TABLE]

$(ii)$ $\alpha\cdot 0=0$. Then for any open neighbourhoods $U_{0}$ and $U_{\alpha}$ of zero and $\alpha$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$, respectively, we have that

[TABLE]

Let $W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{p_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{p_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{k}_{1},\ldots,a^{k}_{p_{k}})}_{(b^{k}_{1},\ldots,b^{k}_{p_{k}})}\right)$ be an arbitrary basic neighbourhood of [math] in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. Without loss of generality we may assume that $p_{1},\ldots,p_{k}\leqslant|\mathbf{d}(\alpha)|$. Put

[TABLE]

Then the family $\mathbf{B}$ is finite and $\alpha\cdot U_{0}\subseteq W_{0}$ for $U_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\bigcup_{S^{(a)}_{(b)}\in\mathbf{B}}{\Uparrow}S^{(a)}_{(b)}$.

$(iii)$ $0\cdot\alpha=0$. Then for any open neighbourhoods $U_{0}$ and $U_{\alpha}$ of zero and $\alpha$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$, respectively, we have that

[TABLE]

Let $W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{p_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{p_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{k}_{1},\ldots,a^{k}_{p_{k}})}_{(b^{k}_{1},\ldots,b^{k}_{p_{k}})}\right)$ be an arbitrary basic neighbourhood of [math] in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. Without loss of generality we may assume that $p_{1},\ldots,p_{k}\leqslant|\mathbf{d}(\alpha)|$. Put

[TABLE]

Then the family $\mathbf{B}$ is finite and $U_{0}\cdot\alpha\subseteq W_{0}$ for $U_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\bigcup_{S^{(a)}_{(b)}\in\mathbf{B}}{\Uparrow}S^{(a)}_{(b)}$.

$(iv)$ $\alpha\cdot\beta=0$. Fix an arbitrary open neighbourhood $W_{0}$ of [math] in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. Without loss of generality we may assume that $W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\left({\Uparrow}S^{(a_{1})}_{(b_{1})}\cup\cdots\cup{\Uparrow}S^{(a_{k})}_{(b_{k})}\right)$. Since $\alpha\cdot\beta=0$ we have that $\mathbf{r}(\alpha)\cap\mathbf{d}(\beta)=\varnothing$. We put

[TABLE]

and

[TABLE]

Let $S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ and $S^{(c_{1},\ldots,c_{p})}_{(d_{1},\ldots,d_{p})}$, $1\leqslant k,p\leqslant n$, such that $\alpha\in S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}$ and $\beta\in S^{(c_{1},\ldots,c_{p})}_{(d_{1},\ldots,d_{p})}$. Then the families $\mathbf{B}_{\alpha}$ and $\mathbf{B}_{\beta}$ are finite, and hence by Remark 5.13(2) the sets $V_{\alpha}=S^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}\setminus\bigcup_{S^{(a)}_{(b)}\in\mathbf{B}_{\alpha}}{\Uparrow}S^{(a)}_{(b)}$ and $V_{\beta}=S^{(c_{1},\ldots,c_{p})}_{(d_{1},\ldots,d_{p})}\setminus\bigcup_{S^{(a)}_{(b)}\in\mathbf{B}_{\beta}}{\Uparrow}S^{(a)}_{(b)}$ are open neighbourhoods of the points $\alpha$ and $\beta$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$, respectively, such that

[TABLE]

$(v)$ $\alpha\cdot\beta=\gamma\neq 0$ and $\mathbf{r}(\alpha)=\mathbf{d}(\beta)$. Without loss of generality we may assume that \alpha=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}&\ldots&s_{k}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right) and \beta=\left(\begin{array}[]{ccc}b_{1}&\ldots&b_{k}\\ t_{1}&\ldots&t_{k}\\ c_{1}&\ldots&c_{k}\\ \end{array}\right), and hence we have that \gamma=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}t_{1}&\ldots&s_{k}t_{k}\\ c_{1}&\ldots&c_{k}\\ \end{array}\right). Then for any open neighbourhood $U_{\gamma}={\Uparrow}\left[U_{1}(s_{1}t_{1}),\ldots,U_{k}(s_{k}t_{k})\right]^{(a_{1},\ldots,a_{k})}_{(c_{1},\ldots,c_{k})}\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{l_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{l_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{p}_{1},\ldots,a^{p}_{l_{p}})}_{(b^{p}_{1},\ldots,b^{p}_{l_{p}})}\right)$ of $\gamma$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ we have that

[TABLE]

and

[TABLE]

where $V_{1}(s_{1}),\ldots,V_{k}(s_{k}),V_{1}(t_{1}),\ldots,V_{k}(t_{k})$ are open neighbourhoods of the points $s_{1},\ldots,s_{k},t_{1},\ldots,t_{k}$ in $(S,\tau_{S})$, respectively, such that

[TABLE]

$(vi)$ $\alpha\cdot\beta=\gamma\neq 0$ and $\mathbf{r}(\alpha)\subsetneqq\mathbf{d}(\beta)$. Without loss of generality we may assume that \alpha=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}&\ldots&s_{k}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right) and \beta=\left(\begin{array}[]{cccccc}b_{1}&\ldots&b_{k}&b_{k+1}&\ldots&b_{k+j}\\ t_{1}&\ldots&t_{k}&t_{k+1}&\ldots&t_{k+j}\\ c_{1}&\ldots&c_{k}&c_{k+1}&\ldots&c_{k+j}\\ \end{array}\right), where $1\leqslant j\leqslant n-k,$ and hence we have that \gamma=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}t_{1}&\ldots&s_{k}t_{k}\\ c_{1}&\ldots&c_{k}\\ \end{array}\right). Then for any open neighbourhood $U_{\gamma}={\Uparrow}\left[U_{1}(s_{1}t_{1}),\ldots,U_{k}(s_{k}t_{k})\right]^{(a_{1},\ldots,a_{k})}_{(c_{1},\ldots,c_{k})}\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{l_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{l_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{p}_{1},\ldots,a^{p}_{l_{p}})}_{(b^{p}_{1},\ldots,b^{p}_{l_{p}})}\right)$ of the point $\gamma$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ we have that

[TABLE]

where $V_{1}(t_{1}),\ldots,V_{k}(t_{k})$ are open neighbourhoods of the points $t_{1},\ldots,t_{k}$ in $(S,\tau_{S})$, respectively, such that

[TABLE]

Fix an arbitrary open neighbourhood $U_{\gamma}$ of the point $\gamma$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$. Then Lemma 5.8 implies that without loss of generality we may assume that

[TABLE]

for some $x_{1},\ldots,x_{p}\in\lambda\setminus\{a_{1},\ldots,a_{k}\}$ and $y_{1},\ldots,y_{p}\in\lambda\setminus\{c_{1},\ldots,c_{k}\}$. We put

[TABLE]

It is obvious that the family $\mathbf{B}_{\alpha}$ is finite. Then $V_{\alpha}\cdot\beta\subseteq U_{\gamma}$ for

[TABLE]

where $V_{1}(s_{1}),\ldots,V_{k}(s_{k})$ are open neighbourhoods of the points $s_{1},\ldots,s_{k}$ in $(S,\tau_{S})$, respectively, such that

[TABLE]

$(vii)$ $\alpha\cdot\beta=\gamma\neq 0$ and $\mathbf{d}(\beta)\subsetneqq\mathbf{r}(\alpha)$. In this case the proof of separate continuity of the semigroup operation is similar to case $(vi)$.

$(viii)$ $\alpha\cdot\beta=\gamma\neq 0$, $\mathbf{d}(\gamma)\subsetneqq\mathbf{d}(\alpha)$ and $\mathbf{r}(\gamma)\subsetneqq\mathbf{r}(\beta)$. Without loss of generality we may assume that \alpha=\left(\begin{array}[]{cccccc}a_{1}&\ldots&a_{k}&a_{k+11}&\ldots&a_{k+m}\\ s_{1}&\ldots&s_{k}&s_{k+11}&\ldots&s_{k+m}\\ b_{1}&\ldots&b_{k}&b_{k+11}&\ldots&b_{k+m}\\ \end{array}\right), \beta=\left(\begin{array}[]{cccccc}b_{1}&\ldots&b_{k}&b_{k+1}&\ldots&b_{k+j}\\ t_{1}&\ldots&t_{k}&t_{k+1}&\ldots&t_{k+j}\\ c_{1}&\ldots&c_{k}&c_{k+1}&\ldots&c_{k+j}\\ \end{array}\right) and \gamma=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}t_{1}&\ldots&s_{k}t_{k}\\ c_{1}&\ldots&c_{k}\\ \end{array}\right), where $1\leqslant j,m\leqslant n-k$. We put \varepsilon=\left(\begin{array}[]{ccc}b_{1}&\ldots&b_{k}\\ 1_{S}&\ldots&1_{S}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right), where $1_{S}$ is the unit element of $S$. It is obvious that $\gamma=\alpha\cdot\varepsilon\cdot\beta$. Hence, in this case the separate continuity of the semigroup operation at the point $\alpha\cdot\beta$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ follows from cases $(vi)$ and $(vii)$.

The previous statements of this section imply that $\tau_{\mathscr{I}}^{\mathbf{c}}\subseteq\tau_{\mathscr{I}}$ for any compact shift-continuous Hausdorff topology $\tau_{\mathscr{I}}$ on $\mathscr{I}_{\lambda}^{n}(S)$, and hence $\tau_{\mathscr{I}}^{\mathbf{c}}$ is the unique requested compact shift-continuous Hausdorff topology on $\mathscr{I}_{\lambda}^{n}(S)$. ∎

Corollary 5.15**.**

If $(S,\tau_{S})$ is a compact Hausdorff semitopological inverse monoid with continuous inversion then $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ is a compact Hausdorff semitopological inverse semigroup with continuous inversion.

Proof.

Since $W_{0}^{-1}=\mathscr{I}_{\lambda}^{n}(S)\setminus\left({\Uparrow}S_{(a^{1}_{1},\ldots,a^{1}_{p_{1}})}^{(b^{1}_{1},\ldots,b^{1}_{p_{1}})}\cup\cdots\cup{\Uparrow}S_{(a^{k}_{1},\ldots,a^{k}_{p_{k}})}^{(b^{k}_{1},\ldots,b^{k}_{p_{k}})}\right)$ for an arbitrary basic neighbourhood $W_{0}=\mathscr{I}_{\lambda}^{n}(S)\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{p_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{p_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{k}_{1},\ldots,a^{k}_{p_{k}})}_{(b^{k}_{1},\ldots,b^{k}_{p_{k}})}\right)$ of zero, inversion is continuous at zero in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$.

Also, for an arbitrary element \alpha=\left(\begin{array}[]{ccc}a_{1}&\ldots&a_{k}\\ s_{1}&\ldots&s_{k}\\ b_{1}&\ldots&b_{k}\\ \end{array}\right) of $\mathscr{I}_{\lambda}^{n}(S)$ and any its open neighbourhood $V_{\alpha}={\Uparrow}\left[V_{1}(s_{1}),\ldots,V_{k}(s_{k})\right]^{(a_{1},\ldots,a_{k})}_{(b_{1},\ldots,b_{k})}\setminus\left({\Uparrow}S^{(a^{1}_{1},\ldots,a^{1}_{l_{1}})}_{(b^{1}_{1},\ldots,b^{1}_{l_{1}})}\cup\cdots\cup{\Uparrow}S^{(a^{p}_{1},\ldots,a^{p}_{l_{p}})}_{(b^{p}_{1},\ldots,b^{p}_{l_{p}})}\right)$ we have that $\left(V_{\alpha}\right)^{-1}\subseteq U_{\alpha^{-1}}$ for the neighbourhood $U_{\alpha^{-1}}={\Uparrow}\left[U_{1}(s^{-1}_{1}),\ldots,V_{k}(s^{-1}_{k})\right]_{(a_{1},\ldots,a_{k})}^{(b_{1},\ldots,b_{k})}\setminus$ of $\alpha^{-1}$ in $\left(\mathscr{I}_{\lambda}^{n}(S),\tau_{\mathscr{I}}^{\mathbf{c}}\right)$ with

[TABLE]

This completes the proof of the corollary. ∎