Sectionally pseudocomplemented posets

TL;DR

This paper extends the concept of sectional pseudocomplementation from lattices to posets, explores their properties, congruences, and completions, and introduces generalized ordinal sums to construct Dedekind-MacNeille completions.

Contribution

It generalizes sectional pseudocomplementation to posets and studies their algebraic properties, congruences, and completion methods, including a new construction technique.

Findings

Sectionally pseudocomplemented lattices form a variety with simple identities.

Every such poset is completely L-semidistributive.

Dedekind-MacNeille completion may not preserve sectional pseudocomplementation, but generalized ordinal sums can be used for construction.

Abstract

The concept of a sectionally pseudocomplemented lattice was introduced by I. Chajda as an extension of relative pseudocomplementation for not necessarily distributive lattices. The typical example of such a lattice is the non-modular lattice N5. The aim of this paper is to extend the concept of sectional pseudocomplementation from lattices to posets. At first we show that the class of sectionally pseudocompelemented lattices forms a variety of lattices which can be described by two simple identities. This variety has nice congruence properties. We summarize properties of sectionally pseudocomplemented posets and show differences to relative pseudocomplementation. We prove that every sectionally pseudocomplemented poset is completely L-semidistributive. We introduce the concept of congruence on these posets and show when the quotient structure becomes a poset again. Finally, we study the Dedekind-MacNeille completion of sectionally pseudocomplemented posets. We show that contrary to the case of relatively pseudocomplemented posets, this completion need not be sectionally pseudocomplemented but we present the construction of a so-called generalized ordinal sum which enables us to construct the Dedekind-MacNeille completion provided the completions of the summands are known.

Equations

$$L(A)$$

$$U(A)$$

$$\begin{array}[]{c|cccccccc}*&a&b&c&d&e&f&g&1\\ \hline\cr a&1&1&c&1&1&1&1&1\\ b&g&1&g&g&1&1&g&1\\ c&f&f&1&f&1&f&1&1\\ d&e&e&e&1&e&1&1&1\\ e&d&f&g&d&1&f&g&1\\ f&a&e&c&g&e&1&g&1\\ g&b&b&e&f&e&f&1&1\\ 1&a&b&c&d&e&f&g&1\end{array}$$

$$\begin{array}[]{c|ccccc}*&0&a&b&c&1\\ \hline\cr 0&1&1&1&1&1\\ a&b&1&b&1&1\\ b&c&a&1&c&1\\ c&b&a&b&1&1\\ 1&0&a&b&c&1\end{array}$$

$$a\vee d$$

$$=(a\vee b)\vee ((a\vee b)\wedge (c\vee b))=a\vee b,$$

$$c\vee d$$

$$=(c\vee b)\vee ((a\vee b)\wedge (c\vee b))=c\vee b$$

$$d\le (a\vee d)\wedge (c\vee d)=(a\vee b)\wedge (c\vee b)=d,$$

$$c\vee b=c\vee d\le a\ast d=a\ast ((a\vee b)\wedge (c\vee b))$$

$$c=c\vee b\le a\ast ((a\vee b)\wedge (c\vee b))=a\ast b$$

$$p(x,y,z):=((x\ast y)\ast z)\wedge ((z\ast y)\ast x).$$

$$p(x,x,y)$$

$$p(y,x,x)$$

$$t_1(x,y):=x\ast y\text{ and }{t}_2(x,y):=y\ast x$$

$$\begin{array}[]{c|ccccccc}*&0&a&b&c&d&e&1\\ \hline\cr 0&1&1&1&1&1&1&1\\ a&b&1&b&1&1&1&1\\ b&c&a&1&c&1&1&1\\ c&b&a&b&1&1&1&1\\ d&0&a&b&c&1&e&1\\ e&0&a&b&c&d&1&1\\ 1&0&a&b&c&d&e&1\end{array}$$

$$a$$

$$U(a,b)$$

$$LU(a,b)$$

$$L(U(a,b),1)$$

$$a\ast b$$

$$L(U(1,a),b)$$

$$L(b)$$

$$b$$

$$\begin{array}[]{c|ccccccc}*&0&a&b&c&d&e&1\\ \hline\cr 0&1&1&1&1&1&1&1\\ a&d&1&1&1&d&1&1\\ b&d&c&1&c&d&1&1\\ c&d&b&b&1&d&1&1\\ d&e&a&b&c&1&e&1\\ e&d&a&b&c&d&1&1\\ 1&0&a&b&c&d&e&1\end{array}$$

$$L(B)$$

$$U(B)$$

$$\text{If }\varnothing \ne M\subseteq L,a,b\in L\text{ and }x\wedge a=b\text{ for all }x\in M\text{ then }(\bigvee M)\wedge a=b.$$

$$\text{If }\varnothing \ne M\subseteq P,a,b\in P\text{ and }L(x,a)=L(b)\text{ for all }x\in M\text{ then }L(U(M),a)=L(b).$$

$$L(b)\subseteq L(U(M),a)\subseteq L(a\ast b,a)=L(U(a,b),a\ast b)=L(b),$$

$$a_4=(a_3\ast a_2)\ast a_2\Theta (a_1\ast a_2)\ast a_2=1\ast a_2=a_2.$$

$$a_5=(a_4\ast a_3)\ast a_3\Theta (a_2\ast a_3)\ast a_3=1\ast a_3=a_3.$$

$$a\Theta 1=a\ast b\Theta 1\ast b=b.$$

$$a=1\ast a=(a\ast a)\ast a\Theta (c\ast a)\ast a=a\ast a=1=a\ast b=(c\ast a)\ast b\Theta (a\ast a)\ast b=1\ast b=b.$$

$$(b\ast c)\ast c\in [(c\ast c)\ast c]\Theta =[1\ast c]\Theta =[c]\Theta =[a]\Theta ,$$

$$Q:=\{x\in P\mid x\text{ is the greatest element of }[x]\Theta \}.$$

Full text

AMS Subject Classification: 06A11, 06D15, 06B23

Keywords: Sectional pseudocomplementation, poset, congruence, Dedekind-MacNeille completion, generalized ordinal sum.

1 Introduction

The concept of relative pseudocomplemented lattices was introduced by R. P. Dilworth in [7]. The usefulness of this concept was shown in numerous papers and books, see e.g. the famous paper [1] by R. Balbes and the monograph [2] by G. Birkhoff. This concept was extended to posets recently by the first and second author in [6]. Relatively pseudocomplemented lattices turn out to be distributive, a property which also holds for relatively pseudocomplemented posets (see [6]). In order to extend relative pseudocomplementation in lattices to the non-distributive case, sectional pseudocomplementation was introduced in [3]. The aim of the present paper is to extend sectional pseudocomplementation to posets which, of course, need not be distributive.

The concept of a sectionally pseudocomplemented lattice was introduced by the first author in [3]. Recall that a lattice $(L,\vee,\wedge)$ is sectionally pseudocomplemented if for all $a,b\in L$ there exists the pseudocomplement of $a\vee b$ with respect to $b$ in $[b,1]$, in other words, there exists a greatest element $c$ of $L$ satisfying $(a\vee b)\wedge c=b$. In this case $c$ is called the sectional pseudocomplement of $a$ with respect to $b$ and it will be denoted by $a*b$.

The aim of this paper is to extend this concept to posets.

2 Properties of sectionally pseudocomplemented posets and lattices

Let $(P,\leq)$ be a poset, $a,b\in P$ and $A,B\subseteq P$. Recall that

[TABLE]

Instead of $L(\{a\})$, $L(\{a,b\})$, $L(A\cup\{a\})$, $L(A\cup B)$, $L(U(A))$ we simply write $L(a)$, $L(a,b)$, $L(A,a)$, $L(A,B)$, $LU(A)$, respectively. Analogously we proceed in similar cases. We also put ${\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(A)=\{x\in P\mid x\leq y\text{ for some }y\in A\}$.

We start with the following definition.

Definition 2.1**.**

A poset $\mathbf{P}=(P,\leq)$ is called sectionally pseudocomplemented if for all $a,b\in P$ there exists a greatest $c\in P$ satisfying $L(U(a,b),c)=L(b)$. This element $c$ is called the sectional pseudocomplement $a*b$ of $a$ with respect to $b$. The poset $\mathbf{P}$ is called strongly sectionally pseudocomplemented if it is sectionally pseudocomplemented, it has a greatest element $1$ and it satisfies the condition $x\leq(x*y)*y$ (which, as we will se later, is equivalent to the identity $x*((x*y)*y)\approx 1$).

The following example shows that there really exist sectionally pseudocomplemented posets which are not strongly sectionally pseudocomplemented. Hence, we cannot expect that every sectionally pseudocomplemented poset satisfies the condition $x\leq(x*y)*y$.

Example 2.2**.**

The poset visualized in Fig. 1

[TABLE]

is sectionally pseudocomplemented and $*$ has the operation table

[TABLE]

but it is not strongly sectionally pseudocomplemented since $c\not\leq a=f*a=(c*a)*a$.

Recall from [6] or [8] that the relative pseudocomplement of $a$ with respect to $b$ is the greatest $d\in P$ satisfying $L(a,d)\subseteq L(b)$.

We are going to show that every sectionally pseudocomplemented lattice with $1$ is strongly sectionally pseudocomplemented.

The following lemma was proved in [3]. For the convenience of the reader we provide a short proof.

Lemma 2.3**.**

Every sectionally pseudocomplemented lattice $\mathbf{L}=(L,\vee,\wedge,*)$ satisfies $x\vee y\leq(x*y)*y$.

Proof.

Because of $(x\vee y)\wedge y=y$ we have $y\leq x*y$ and hence $((x*y)\vee y)\wedge(x\vee y)\approx(x\vee y)\wedge(x*y)=y$ whence $x\vee y\leq(x*y)*y$. ∎

In a lattice $(P,\vee,\wedge)$ the equation $L(U(a,b),c)=L(b)$ is equivalent to $(a\vee b)\wedge c=b$.

Example 2.4**.**

The poset visualized in Fig. 2

[TABLE]

is a strongly sectionally pseudocomplemented lattice and $*$ has the operation table

[TABLE]

but this poset is not relatively pseudocomplemented since the relative pseudocomplement of $c$ with respect to $a$ does not exist.

It was shown in [3] that the class of sectionally pseudocomplemented lattices forms a variety. However, the defining identities given in [3] are rather complicated. We present some simpler identities as follows.

Theorem 2.5**.**

The class of sectionally pseudocomplemented lattices forms a variety which besides the lattice axioms is determined by the following identities:

1. (i)

$z\vee y\leq x*((x\vee y)\wedge(z\vee y))$, 2. (ii)

$(x\vee y)\wedge(x*y)\approx y$.

Proof.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a lattice and $a,b,c\in L$. First assume $\mathbf{L}$ to be sectionally pseudocomplemented. If $d:=(a\vee b)\wedge(c\vee b)$ then

[TABLE]

and hence

[TABLE]

i.e. $(a\vee d)\wedge(c\vee d)=d$ which shows

[TABLE]

proving (i). Identity (ii) follows from the definition of $*$. Conversely, assume $\mathbf{L}$ to satisfy (i) and (ii). Then $(a\vee b)\wedge(a*b)=b$ according to (ii). If $(a\vee b)\wedge c=b$ then $b\leq c$ and hence $(a\vee b)\wedge(c\vee b)=(a\vee b)\wedge c=b$ whence

[TABLE]

according to (i). This shows that $a*b$ is the sectional pseudocomplement of $a$ with respect to $b$. ∎

We can prove that this variety has very strong congruence properties. Recall that an algebra $\mathbf{A}$ is called arithmetical if $\Theta\circ\Phi=\Phi\circ\Theta$ for all $\Theta,\Phi\in\operatorname{Con}\mathbf{A}$ and if the congruence lattice of $\mathbf{A}$ is distributive. (Here and in the following $\operatorname{Con}\mathbf{A}$ denotes the set of all congruences on $\mathbf{A}$.) Moreover, recall that an algebra $\mathbf{A}$ with $1$ is called weakly regular (see e.g. [4]) if for arbitrary $\Theta,\Phi\in\operatorname{Con}\mathbf{A}$ we have that $[1]\Theta=[1]\Phi$ implies $\Theta=\Phi$. A variety $\mathcal{V}$ (with $1$) is called arithmetical or weakly regular if every of its members has the respective property.

Theorem 2.6**.**

The variety $\mathcal{V}$ of sectionally pseudocomplemented lattices with $1$ is arithmetical and weakly regular.

Proof.

Since every member of $\mathcal{V}$ is a lattice, $\mathcal{V}$ is congruence distributive. Moreover, since for

[TABLE]

we have

[TABLE]

$\mathcal{V}$ is congruence permutable. Finally, since for

[TABLE]

we have that $t_{1}(x,y)=t_{2}(x,y)=1$ is equivalent to $x=y$, $\mathcal{V}$ is weakly regular (cf. [4]). ∎

Evidently, every relatively pseudocomplemented lattice $(L,\vee,\wedge)$ is sectionally pseudocomplemented since for $a,b\in L$ we have $a*b=(a\vee b)\circ b$ where $*$ and $\circ$ denote sectional and relative pseudocomplementation, respectively. (Observe that $(a\vee b)\wedge b=b$ and hence $(a\vee b)\wedge((a\vee b)\circ b)=b$.) The following poset is an example of a sectionally pseudocomplemented poset which is neither a lattice nor relatively pseudocomplemented.

Example 2.7**.**

The poset visualized in Fig. 3

[TABLE]

is strongly sectionally pseudocomplemented and the operation table of $*$ looks as follows:

[TABLE]

Evidently, this poset is not a lattice. However, it is also not relatively pseudocomplemented since the relative pseudocomplement of $c$ with respect to $a$ does not exist.

In the following we list several important properties of sectionally pseudocomplemented posets.

Theorem 2.8**.**

Let $\mathbf{P}=(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$. Then the following hold:

1. (i)

$x\leq y$* if and only if $x*y=1$,* 2. (ii)

$x*x\approx x*1\approx 1$, 3. (iii)

$1*x\approx x$, 4. (iv)

$x*(y*x)\approx 1$, 5. (v)

$x*((y*x)*x)\approx 1$, 6. (vi)

if $x*y=1$ or $y*x=1$ then $x*((x*y)*y)=1$, 7. (vii)

if $x*y=1$ then $(y*z)*(x*z)=1$, 8. (viii)

$L(U(x,y),x*y)\approx L(y)$.

Proof.

Let $a,b,c\in P$.

1. (i)

The following are equivalent:

[TABLE] 2. (ii)

follows from (i). 3. (iii)

The following are equivalent:

[TABLE] 4. (iv)

We have $L(U(a,b),b)=L(b)$ implies $b\leq a*b$. 5. (v)

Because of (iv) we have $L(U(b*a,a),a)=L(b*a,a)=L(a)$ which shows $a\leq(b*a)*a$. 6. (vi)

If $a\leq b$ then $a\leq b=1*b=(a*b)*b$ according to (iii) and (i), and if $b\leq a$ then $L(U(a*b,b),a)=L(a*b,a)=L(U(a,b),a*b)=L(b)$ and hence $a\leq(a*b)*b$. 7. (vii)

If $a\leq b$ then $L(c)\subseteq L(U(a,c),b*c)\subseteq L(U(b,c),b*c)=L(c)$, i.e. $L(U(a,c),b*c)=L(c)$ whence $b*c\leq a*c$. 8. (viii)

follows from the definition of $*$.

∎

Remark 2.9**.**

Assertion (vii) of Theorem 2.8 says that $*$ is antitone in the first variable, i.e. $x\leq y$ implies $y*z\leq x*z$. Contrary to the case of relatively pseudocomplemented posets, sectional pseudocomplementation is not monotone in the second variable. Namely, in Example 2.7 we have $0\leq a$, but $b*0=c\not\leq a=b*a$. However, $*$ need not be monotone in the second variable also in sectionally pseudocomplemented lattices as the following example shows.

Example 2.10**.**

The lattice visualized in Fig. 4

[TABLE]

is sectionally pseudocomplemented and $*$ has the operation table

[TABLE]

Here we have $0<a$ and $b*0=d\parallel c=b*a$.

For every algebra $(A,*,1)$ of type $(2,0)$ and every subset $B$ of $A$ put

[TABLE]

We are going to show that sectionally pseudocomplemented posets can be defined as certain groupoids.

Theorem 2.11**.**

An algebra $(A,*,1)$ of type $(2,0)$ can be organized into a sectionally pseudocomplemented poset with $1$ if and only if the following hold:

1. (i)

$x*x\approx 1$** 2. (ii)

$x*y=y*x=1\Rightarrow x=y$, 3. (iii)

$x*y=y*z=1\Rightarrow x*z=1$, 4. (iv)

$L(U(x,y),x*y)=L(y)$, 5. (v)

$L(U(x,y),z)=L(y)\Rightarrow z*(x*y)=1$.

Proof.

The necessity of the conditions is clear. Conversely, assume (i) – (v) to hold. Define a binary relation $\leq$ on $A$ by $x\leq y$ if $x*y=1$ ($x,y\in A$). Now

(i) implies reflexivity of $\leq$,

(ii) implies antisymmetry of $\leq$,

(iii) implies transitivity of $\leq$,

(iv) and (v) imply that $x*y$ is the sectional pseudocomplement of $x$ with respect to $y$.

Hence $(A,\leq)$ is a sectionally pseudocomplemented poset with sectional pseudocomplementation $*$. ∎

Recall that a lattice $(L,\vee,\wedge)$ is called completely meet-semidistributive if the following holds:

[TABLE]

For posets, we modify this concept as follows.

Definition 2.12**.**

A poset $(P,\leq)$ is called completely $L$-semidistributive if the following holds:

[TABLE]

Theorem 2.13**.**

Let $\mathbf{P}=(P,\leq,*)$ be a sectionally pseudocomplemented poset. Then $\mathbf{P}$ is completely $L$-semidistributive.

Proof.

If $\emptyset\neq M\subseteq P,a,b\in P$ and $L(x,a)=L(b)$ for all $x\in M$ then $b\leq a$ and therefore $L(U(a,b),x)=L(a,x)=L(b)$ for all $x\in M$ and hence $x\leq a*b$ for all $x\in M$ whence $a*b\in U(M)$ which finally implies

[TABLE]

i.e. $L(U(M),a)=L(b)$. ∎

3 Congruences in sectionally pseudocomplemented posets

Theorem 2.8 (i) shows that in a sectionally pseudocomplemented poset $(P,\leq,*,1)$ with $1$, $\leq$ is uniquely determined by $*$. Let $(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$ and $\Theta\in\operatorname{Con}(P,*)$. We are interested in the question when $(P/\Theta,\leq^{\prime})$ is a poset where $\leq^{\prime}$ is defined by $[x]\Theta\leq^{\prime}[y]\Theta$ if $[x]\Theta*[y]\Theta=[1]\Theta$ ($x,y\in P$). We will see that this is the case if $\Theta$ is convex, i.e. every class of $\Theta$ is a convex subset of $(P,\leq)$.

First we show that if $(P,\leq,*,1)$ is a finite sectionally pseudocomplemented poset with $1$ then $(P,*)$ has convex congruences.

In the following lemma and theorem we frequently use Theorem 2.8 (vi).

Lemma 3.1**.**

Let $(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$ satisfying the Ascending Chain Condition, let $a,b\in P$ and $\Theta\in\operatorname{Con}(P,*)$ and assume $a<b<(b*a)*a$ and $a\mathrel{\Theta}(b*a)*a$. Then $a\mathrel{\Theta}b$.

Proof.

Assume $(a,b)\notin\Theta$. Put $a_{1}:=a$, $a_{2}:=b$ and $a_{n}:=(a_{n-1}*a_{n-2})*a_{n-2}$ for $n\geq 3$. Then $a_{1}<a_{2}<a_{3}$ and $a_{3}\mathrel{\Theta}a_{1}$. Now

[TABLE]

Moreover, $a_{3}\leq a_{4}$. Now $a_{3}=a_{4}$ would imply $a_{1}\mathrel{\Theta}a_{3}=a_{4}\mathrel{\Theta}a_{2}$, a contradiction. This shows $a_{3}<a_{4}$. Now

[TABLE]

Moreover, $a_{4}\leq a_{5}$. Now $a_{4}=a_{5}$ would imply $a_{1}\mathrel{\Theta}a_{3}\mathrel{\Theta}a_{5}=a_{4}\mathrel{\Theta}a_{2}$, a contradiction. This shows $a_{4}<a_{5}$. Going on in this way we would obtain an infinite strictly ascending chain $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<\cdots$ contradicting the Ascending Chain Condition. This shows $a\mathrel{\Theta}b$. ∎

Hence, we conclude

Theorem 3.2**.**

Let $(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$ satisfying the Ascending Chain Condition and let $\Theta\in\operatorname{Con}(P,*)$. Then $\Theta$ is convex.

Proof.

Assume $a,b,c\in P$, $a<b<c$ and $(a,c)\in\Theta$. Then $b\leq(b*a)*a$. If $b=(b*a)*a$ then $b=(b*a)*a\mathrel{\Theta}(b*c)*a=1*a=a$. If $b<(b*a)*a$ then $a\mathrel{\Theta}b$ according to Lemma 3.1. ∎

Corollary 3.3**.**

If $(P,\leq,*,1)$ is a finite sectionally pseudocomplemented poset with $1$ then $(P,*)$ has convex congruences.

For the infinite case we have the following result.

Theorem 3.4**.**

Let $(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$ such that $x,y\in P$, $x<y<1$, $x\not\prec y$ and $x<y*x$ together imply $\Theta(x,y)=P^{2}$. Then $(P,*)$ has convex congruences.

Proof.

Let $\Theta\in\operatorname{Con}(P,*)$ and $a,b,c\in P$ and assume $a<b<c$ and $(a,c)\in\Theta$. If $c=1$ then

[TABLE]

If $c<1$ and $a<c*a$ then $\Theta(a,c)=P^{2}$ and hence $\Theta=P^{2}$ which implies $a\mathrel{\Theta}b$. If $c<1$ and $a=c*a$ then

[TABLE]

This shows that $\Theta$ is convex. ∎

Let $(P,\leq,*,1)$ be a sectionally pseudocomplemented poset with $1$ and $\Theta\in\operatorname{Con}(P,*)$. We define a binary relation $\leq^{\prime}$ on $P/\Theta$ by $[x]\Theta\leq^{\prime}[y]\Theta$ if $[x]\Theta*[y]\Theta=[1]\Theta$ ($x,y\in P$). Now we can prove

Theorem 3.5**.**

Let $(P,\leq,*,1)$ be a strongly sectionally pseudocomplemented poset and let $a,b\in P$ and $\Theta$ a convex congruence on $(A,*)$. Then the following hold:

1. (i)

If $[a]\Theta\leq^{\prime}[b]\Theta$ then there exists some $d\in[b]\Theta$ with $a\leq d$, 2. (ii)

if $a\leq b$ then $[a]\Theta\leq^{\prime}[b]\Theta$, 3. (iii)

$(P/\Theta,\leq^{\prime})$* is a poset.*

Proof.

1. (i)

Put $d:=(a*b)*b$. Then $d=(a*b)*b\mathrel{\Theta}1*b=b$ and $a\leq(a*b)*b=d$. 2. (ii)

If $a\leq b$ then $a*b=1$ according to Theorem 2.8 and hence $a*b\mathrel{\Theta}1$, i.e. $[a]\Theta\leq^{\prime}[b]\Theta$. 3. (iii)

Obviously, $\leq^{\prime}$ is reflexive. Now assume $[a]\Theta\leq^{\prime}[b]\Theta\leq^{\prime}[a]\Theta$. Then, by (i), there exists some $d\in[b]\Theta$ with $a\leq d$. Because of $[d]\Theta=[b]\Theta\leq^{\prime}[a]\Theta$ there exists some $e\in[a]\Theta$ with $d\leq e$. Since $a\leq d\leq e$, $(a,e)\in\Theta$ and $\Theta$ is convex we conclude $(a,d)\in\Theta$ showing $[a]\Theta=[d]\Theta=[b]\Theta$ proving antisymmetry of $\leq^{\prime}$. Finally, let $c\in P$ and assume $[a]\Theta\leq^{\prime}[b]\Theta\leq^{\prime}[c]\Theta$. Then, by (i) there exists some $f\in[b]\Theta$ with $a\leq f$ and because of $[f]\Theta=[b]\Theta\leq^{\prime}[c]\Theta$ some $g\in[c]\Theta$ with $f\leq g$. Because of $a\leq f\leq g$ we have $a\leq g$ which implies $[a]\Theta\leq^{\prime}[g]\Theta=[c]\Theta$ by (ii), proving transitivity of $\leq^{\prime}$.

∎

Lemma 3.6**.**

Let $\mathbf{P}=(P,\leq,*,1)$ be a strongly sectionally pseudocomplemented poset, $a\in P$ and $\Theta\in\operatorname{Con}(P,*)$. Then $[a]\Theta$ is up-directed. Hence, if $\mathbf{P}$ satisfies the Ascending Chain Condition, then $[a]\Theta$ has a greatest element.

Proof.

If $b,c\in[a]\Theta$ then

[TABLE]

$b\leq(b*c)*c$ since $\mathbf{P}$ is strongly sectionally pseudocomplemented, and $c\leq(b*c)*c$ according to Theorem 2.8 (v). ∎

From Theorem 3.5 we have: If $(P,\leq,*,1)$ is a strongly sectionally pseudocomplemented poset, $\Theta$ a convex congruence on $(P,*)$, $a$ the greatest element of $[a]\Theta$ and $b$ the greatest element of $[b]\Theta$ then $a\leq b$ if and only if $[a]\Theta\leq^{\prime}[b]\Theta$.

Now we solve the problem for which $\Theta\in\operatorname{Con}(P,*)$ the quotient $P/\Theta$ is again sectionally pseudocomplemented. We can state a sufficient condition.

Definition 3.7**.**

Let $\mathbf{P}=(P,\leq,*)$ be a sectionally pseudocomplemented poset and $\Theta\in\operatorname{Con}(P,*)$. We say that $\Theta$ is strong if the following holds: If $a,b\in P$, $a$ is the greatest element of $[a]\Theta$ and $b$ the greatest element of $[b]\Theta$ then $a*b$ is the greatest element of $[a*b]\Theta$. In this case we define $[a]\Theta*^{\prime}[b]\Theta:=[a*b]\Theta$.

It is easy to see that if $\Theta$ is strong, $a,b,c,d\in P$, $a\leq b$ and $c$ is the greatest element of $[a]\Theta$ and $d$ the greatest element of $[b]\Theta$ then $c\leq d$.

Theorem 3.8**.**

Let $\mathbf{P}=(P,\leq,*,1)$ be a strongly sectionally pseudocomplemented poset and $\Theta\in\operatorname{Con}(P,*)$ a strong congruence. If $\mathbf{P}$ satisfies the Ascending Chain Condition then $(P/\Theta,\leq^{\prime},*^{\prime},[1]\Theta)$ is a strongly sectionally pseudocomplemented poset.

Proof.

Since $\mathbf{P}$ satisfies the Ascending Chain Condition we know from Theorems 3.2 and 3.5 that $\Theta$ is convex and $(P/\Theta,\leq^{\prime})$ a poset. Moreover, from Lemma 3.6 we have that any congruence class of $\Theta$ has a greatest element. Put

[TABLE]

Then $(Q,*,1)$ is a subalgebra of $(P,*,1)$. Assume $a,b\in Q$. Since $\mathbf{P}$ is a strongly sectionally pseudocomplemented poset and $\Theta$ is strong we have $a,b\leq(a*b)*b$ and $a*b,(a*b)*b\in Q$. This implies

[TABLE]

Note that the first inclusion follows from the fact that $b\leq a*b$ and $b\in L_{Q}U_{Q}(a,b)$. The second inclusion follows from the fact that $L_{Q}U_{Q}(a,b)\subseteq L_{Q}U_{Q}((a*b)*b)$. The first equality follows from $L_{Q}U_{Q}((a*b)*b)=L_{Q}((a*b)*b)$. The second equality follows from the definition of $L_{Q}$. Since $\mathbf{P}$ is sectionally pseudocomplemented we have the next equality.

Now, let $c\in Q$ be such that $L_{Q}(U_{Q}(a,b),c)=L_{Q}(b)$. We have $U_{Q}(a,b)=U(a,b)\cap Q\subseteq U(a,b)$. Hence $LU(a,b)\subseteq LU_{Q}(a,b)={\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(L_{Q}U_{Q}(a,b))$. The last equality follows from the fact that $L_{Q}U_{Q}(a,b)\subseteq LU_{Q}(a,b)$ which yields ${\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(L_{Q}U_{Q}(a,b))\subseteq{\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(LU_{Q}(a,b))=LU_{Q}(a,b)$ and from the fact that $y\in LU_{Q}(a,b)$ implies $y\leq x\in L_{Q}U_{Q}(a,b)$ where $x$ is the greatest element of $[y]\Theta$.

We obtain $L(b)\subseteq L(U(a,b),c)\subseteq{\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(L_{Q}U_{Q}(a,b)\cap L_{Q}(c))={\left\downarrow\vbox to7.58606pt{}\right.\kern-1.2pt}(L_{Q}(b))=L(b)$ since $b,c\in Q$. Since $\mathbf{P}$ is a sectionally pseudocomplemented poset we have $c\leq a*b$, i.e., $(Q,*,1)$ is sectionally pseudocomplemented. Since $(Q,*,1)$ is a subalgebra of $(P,*,1)$ we have that $(Q,*,1)$ is strongly sectionally pseudocomplemented. Moreover, $x\mapsto[x]\Theta$ an isomorphism from $(Q,\leq,*,1)$ to $(P/\Theta,\leq^{\prime},*^{\prime},[1]\Theta)$ and hence $(P/\Theta,\leq^{\prime},*^{\prime},[1]\Theta)$ is also a strongly sectionally pseudocomplemented poset. ∎

The following lemma shows that in a strongly sectionally pseudocomplemented poset all principal congruences are given by the congruences of the form $\Theta(c,1)$.

Lemma 3.9**.**

Let $(P,\leq,*,1)$ be a strongly sectionally pseudocomplemented poset, assume that every congruence on $(P,*)$ is convex and let $a,b\in P$ with $a\leq b$ . Then $\Theta(b*a,1)=\Theta(a,b)$.

Proof.

Since $(a,b)\in\Theta(a,b)$ yields $(b*a,1)=(b*a,b*b)\in\Theta(a,b)$, we have $\Theta(b*a,1)\subseteq\Theta(a,b)$. Conversely, $(b*a,1)\in\Theta(b*a,1)$ yields $((b*a)*a,a)=((b*a)*a,1*a)\in\Theta(b*a,1)$ which because of $a\leq b\leq(b*a)*a$ and the convexity of $\Theta(b*a,1)$ implies $(a,b)\in\Theta(b*a,1)$, i.e. $\Theta(a,b)\subseteq\Theta(b*a,1)$. ∎

4 Completion of sectionally pseudocomplemented

posets

Now we consider the Dedekind-MacNeille completion of sectionally pseudocomplemented posets.

It was shown by Y. S. Pawar ([8]) that the Dedekind-MacNeille completion $\operatorname{\mathbf{DM}}(\mathbf{Q})$ of a relatively pseudocomplemented poset $\mathbf{Q}$ is relatively pseudocomplemented and that the relative pseudocomplementation in $\operatorname{DM}(\mathbf{Q})$ extends the relative pseudocomplementation in $\mathbf{Q}$ if $\mathbf{Q}$ is canonically embedded into $\operatorname{\mathbf{DM}}(\mathbf{Q})$. In contrast to this the Dedekind-MacNeille completion of a strongly sectionally pseudocomplemented poset $\mathbf{P}$ need not be sectionally pseudocomplemented, even if $\mathbf{P}$ is finite and has a greatest element.

Example 4.1**.**

Though the poset visualized in Fig. 5

[TABLE]

is strongly sectionally pseudocomplemented with

[TABLE]

as the operation table for $*$, its Dedekind-MacNeille completion visualized in Fig. 6

[TABLE]

is not sectionally pseudocomplemented since $a*0$ does not exist.

For our next investigations, we introduce the following useful concepts.

Definition 4.2**.**

Let $(I,\leq)$ be a chain with greatest element $\top$ and smallest element $\bot$. Let $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ be a family of posets such that $\mathbf{P}_{\top}$ has a greatest element $1$ and such that the following hold:

1. (i)

If $i\in I$ then there are $a,b\in P_{i}$ with $a<b$, 2. (ii)

if $i,j,k\in I$ and $i<k<j$ then $P_{i}\cap P_{j}=\emptyset$, 3. (iii)

if $i,j\in I$ and $i<j$ then $|P_{i}\cap P_{j}|\leq 1$, 4. (iv)

*if $i,j\in I$, $i<j$ and $P_{i}\cap P_{j}=\{a\}$ then $P_{i}=L_{P_{i}}(a)$ and $P_{j}=U_{P_{j}}(a)$. *

Put $P=\bigcup\limits_{i\in I}P_{i}$. For $a,b\in P$, say $a\in P_{i}$ and $b\in P_{j}$ with $i,j\in I$, define

[TABLE]

We call $\mathbf{P}=(P,\leq)$ the generalized ordinal sum of $\mathbf{P}_{i},i\in I$.

It is elementary that $\mathbf{P}$ is a poset with a greatest element $1$.

Now, we can state some sufficient conditions under which the Dedekind-MacNeille completion of a sectionally pseudocomplemented poset is sectionally pseudocomplemented. By [9] the Dedekind-MacNeille completion of a poset $\mathbf{P}$ is (up to isomorphism) any complete lattice $\mathbf{Q}$ into which $\mathbf{P}$ can be supremum-densely and infimum-densely embedded (i.e., for every element $x\in Q$ there exist $M,N\subseteq P$ such that $x=\bigvee\varphi(M)=\bigwedge\varphi(N)$, where $\varphi\colon P\to Q$ is the embedding). We usually identify $P$ with $\varphi(P)$. In this sense $\mathbf{Q}$ preserves all infima and suprema existing in $\mathbf{P}$.

Now we turn our attention to a notion of a DM-yoked family of a generalized ordinal sum. The importance of this concept is based on the fact that under natural assumptions (which are e.g. satisfied for a finite index set $I$) the Dedekind-MacNeille completion of a generalized ordinal sum will be isomorphic to a generalized ordinal sum of the respective DM-yoked family.

Definition 4.3**.**

Let $\mathbf{P}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I$. We say that a family $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ of posets is a DM-yoked family of $\mathbf{P}$ if the following conditions are satisfied:

1. (y1)

$\mathbf{P}_{i}$* is a sub-poset of $\mathbf{Q}_{i}$ such that $\mathbf{Q}_{i}$ is the Dedekind-MacNeille completion of $\mathbf{P}_{i}$ for every $i\in I$,* 2. (y2)

if $i,j,k\in I$ and $i<k<j$ then $Q_{i}\cap Q_{j}=\emptyset$, 3. (y3)

if $i,j\in I$ and $i<j$ then $|Q_{i}\cap Q_{j}|\leq 1$, 4. (y4)

if $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{i}$ has a greatest element and $\mathbf{P}_{j}$ has a smallest element then $Q_{i}\cap Q_{j}=\emptyset$, 5. (y5)

if $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{i}$ does not have a greatest element and $\mathbf{P}_{j}$ has a smallest element $0_{P_{j}}$ then $0_{P_{j}}$ is the greatest element of $\mathbf{Q}_{i}$, 6. (y6)

if $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{j}$ does not have a smallest element and $\mathbf{P}_{i}$ has a greatest element $1_{P_{i}}$ then $1_{P_{i}}$ is the smallest element of $\mathbf{Q}_{j}$, 7. (y7)

if $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{j}$ does not have a smallest element and $\mathbf{P}_{i}$ does not have a greatest element then the greatest element $1_{Q_{i}}$ of $\mathbf{Q}_{i}$ is the smallest element $0_{Q_{j}}$ of $\mathbf{Q}_{j}$, 8. (y8)

if $i,j\in I$, $i<j$ and $Q_{i}\cap Q_{j}=\{a\}$ then $a$ is the greatest element of $\mathbf{Q}_{i}$ and the smallest element of $\mathbf{Q}_{j}$.

The question when there exists a DM-yoked family for a given poset $\mathbf{P}$ which is a generalized ordinal sum of posets $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ is positively answered in the following series of lemmas under the natural assumption that $P_{j}\cap(\operatorname{\mathbf{DM}}(P_{i})\times\{i\})=\emptyset$ for all $i,j\in I$.

We will first need the following definition.

Definition 4.4**.**

Let $\mathbf{P}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I$. We say that a family $\mathbf{R}_{i}=(R_{i},\leq_{i}),i\in I,$ of posets is a DM-related family of $\mathbf{P}$ if the following conditions are satisfied:

1. (r1)

$\mathbf{P}_{i}$* is a sub-poset of $\mathbf{R}_{i}$ such that $\mathbf{R}_{i}$ is the Dedekind-MacNeille completion of $\mathbf{P}_{i}$ for every $i\in I$,* 2. (r2)

if $i\in I$ and $x\in R_{i}$ then $x\in R_{i}\setminus P_{i}$ if and only if $x=(A,i)$ and $A$ is a non-principal cut of $\operatorname{\mathbf{DM}}(\mathbf{P}_{i})$.

Lemma 4.5**.**

Let $\mathbf{P}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ such that $P_{i}\cap(\operatorname{\mathbf{DM}}(P_{i})\times\{i\})=\emptyset$. Then a DM-related family $\mathbf{R}_{i}=(R_{i},\leq_{i}),i\in I,$ of $\mathbf{P}$ exists.

Proof.

Let $i\in I$. We put

[TABLE]

We have $P_{i}\cap(\{A\in\operatorname{\mathbf{DM}}(P_{i})\mid A\ \text{is not a principal cut in}\ \operatorname{\mathbf{DM}}(P_{i})\}\times\{i\})=\emptyset$. Define a mapping $\kappa_{i}\colon\operatorname{\mathbf{DM}}(P_{i})\to R_{i}$ as follows:

[TABLE]

for all $A\in\operatorname{\mathbf{DM}}(P_{i})$. Clearly, $\kappa_{i}$ is a bijection. Let $x,y\in R_{i}$. We define $x\leq_{i}y$ if and only if $\kappa_{i}^{-1}(x)\subseteq\kappa_{i}^{-1}(x)$. Then $(R_{i},\leq_{i}),i\in I,$ is a poset containing $P_{i}$ isomorphic with the poset $\operatorname{\mathbf{DM}}(\mathbf{P}_{i})$. Hence the family $\mathbf{R}_{i}=(R_{i},\leq_{i}),i\in I,$ is DM-related. ∎

Lemma 4.6**.**

Let $\mathbf{P}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ such that $P_{j}\cap(\operatorname{\mathbf{DM}}(P_{i})\times\{i\})=\emptyset$ for all $i,j\in I$. Then a DM-yoked family $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ of $\mathbf{P}$ exists.

Proof.

Let $\mathbf{R}_{i}=(R_{i},\leq_{i}),i\in I,$ be the DM-related family of $\mathbf{P}$ which exists by

Lemma 4.5. We will proceed in two steps.

Step 1: Let $i\in I$. Assume that there exists some $j\in I$ with $i\prec j$ and $P_{i}\cap P_{j}=\emptyset$. If $\mathbf{P}_{i}$ does not have a greatest element we put $S_{i}:=(R_{i}\setminus\{1_{R_{i}}\})\cup\{0_{R_{j}}\}$ such that $0_{R_{j}}$ will be the greatest element of $S_{i}$ and the order $\leq_{i}$ on $S_{i}$ restricted to $R_{i}\setminus\{1_{R_{i}}\}$ will be the restriction of the order on $\mathbf{R}_{i}$. Clearly $P_{i}\subseteq S_{i}$ and $\mathbf{P}_{i}$ is a sub-poset of $\mathbf{S}_{i}$. If $\mathbf{P}_{i}$ does have a greatest element, we put $\mathbf{S}_{i}:=\mathbf{R}_{i}$. If $P_{i}\cap P_{j}=\{a\}$ then $a=1_{P_{i}}=0_{P_{j}}$. Hence also $a=1_{R_{i}}=0_{R_{j}}$ and we put $\mathbf{S}_{i}:=\mathbf{R}_{i}$. If there is no $j\in I$ such that $i\prec j$, we put again $\mathbf{S}_{i}:=\mathbf{R}_{i}$.

Step 2: Let $j\in I$. Assume there exists some $i\in I$ with $i\prec j$ and $P_{i}\cap P_{j}=\emptyset$. If $\mathbf{P}_{j}$ does not have a smallest element, we put $Q_{j}:=(S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\}$ such that $1_{S_{i}}$ will be the smallest element of $Q_{j}$ and the order $\leq_{j}$ on $Q_{j}$ restricted to $S_{j}\setminus\{0_{S_{j}}\}$ will be the restriction of the order on $\mathbf{S}_{j}$. Clearly $P_{j}\subseteq Q_{j}$ and $\mathbf{P}_{j}$ is a sub-poset of $\mathbf{Q}_{j}$. Otherwise we always put $\mathbf{Q}_{j}:=\mathbf{S}_{j}$.

Let us now check that $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ is a DM-yoked family of $\mathbf{P}$.

1. (y1):

This follows immediately from the definition of $\mathbf{Q}_{i}$. 2. (y2):

Let $i,j,k\in I$ and assume $i<k<j$. We always have $P_{i}\cap P_{j}=\emptyset$ and hence also $R_{i}\cap R_{j}=\emptyset$.

Step 1: Assume that $a\in S_{i}\cap S_{j}$ for some $a$. Then $a\not\in R_{i}$ or $a\not\in R_{j}$. Suppose first that $a\not\in R_{i}$. Then there exists some $l\in I$ with $i\prec l\leq k<j$ and $P_{i}\cap P_{l}=\emptyset$ and hence $a=0_{R_{l}}<1_{R_{l}}$. Now either $a=0_{P_{l}}$ or $a=(b,l)=0_{R_{l}}\not\in P_{l}$ for some $b$. Since $a\in S_{j}$, it can be only of a form $0_{P_{j}}$ or $0_{P_{m}}$ for $j\prec m$. But both cases are not possible (in the first case we would obtain $0_{P_{j}}=1_{P_{l}}=0_{P_{l}}$, in the second $P_{l}\cap P_{m}\not=\emptyset$). Assume now that $a\not\in R_{j}$. Then there exists some $m\in I$ such that $j\prec m$ and $a=0_{R_{m}}$. Since $a\in S_{i}$, it can be only of a form $1_{P_{i}}$ or $0_{R_{n}}$ for $i\prec n$. Since $i\prec n\leq k<j\prec m$, this is again not possible. Hence $S_{i}\cap S_{j}=\emptyset$.

Step 2: Suppose $a\in Q_{i}\cap Q_{j}$ for some $a$. Then $a\not\in S_{i}$ or $a\not\in S_{j}$. Assume now $a\not\in S_{i}$. Then there exists some $l\in I$ with $l\prec i<k<j$ and $P_{l}\cap P_{i}=\emptyset$ and hence $a=1_{S_{l}}\not\in P_{l}$ (otherwise we would have $a\in P_{j}$ which is not possible or $a=0_{P_{m}}$ for some $m\in I$ with $j\prec m$ which is also not possible). We conclude that either $a=(b,l)$ for some element $b$ or $a=0_{R_{i}}\in S_{i}$, a contradiction in the last case. Hence $a=(b,l)=1_{S_{l}}$. Since $a\in Q_{j}$, we have $a\not\in P_{j}$, i.e., $a=(c,j)$ for some element $c$ (which is not possible) or $a=1_{S_{n}}$ for some $n\prec j$ with $l\prec i<k\leq n<j$ (which is not possible) or $a=0_{P_{m}}$ for some $m\in I$ with $j\prec m$ (which is also not possible). Suppose $a\not\in S_{j}$. Then there exists some $n\in I$ with $i<k\leq n\prec j$ and $a=1_{S_{n}}\not\in P_{n}$ (otherwise we would have $a\in P_{i}$ which is not possible since then $a=0_{P_{n}}$ or $a=1_{P_{r}}$ for some $r\in I$ with $r\prec i$ which is not possible since $P_{r}\cap P_{n}=\emptyset$ or $a=0_{P_{q}}$ for some $q\in I$ with $i\prec q\leq k\leq n$ in which case $0_{P_{q}}=0_{1_{q}}$, a contradiction). Hence $a=1_{S_{n}}=(c,n)$ for some element $c$. Since $a\in Q_{i}$, we have that either there exists some $r\in I$ with $r\prec i$ such that $a=1_{S_{r}}$, i.e., either $a=(d,r)$ or $a=(e,i)$ for some elements $d,r$, a contradiction to $r<i<n$, or there exists some $q\in I$ with $i\prec q\leq k\leq n$ such that $a=0_{S_{q}}$ in which case $q=k=n$ and $a=1_{S_{n}}=0_{S_{n}}$, a contradiction. Hence $Q_{i}\cap Q_{j}=\emptyset$. 3. (y3):

Let $i,j\in I$ with $i<j$. If $i\not\prec j$ we know from (y2) that $|Q_{i}\cap Q_{j}|=0$. Hence we may assume $i\prec j$. Let $a\in Q_{i}\cap Q_{j}$. It is enough to show that $a=1_{Q_{i}}=0_{Q_{j}}$ (which will give us also (y8)). Assume first $P_{i}\cap P_{j}\not=\emptyset$. Then $1_{P_{i}}=0_{P_{j}}$, $S_{i}=R_{i}$ and $Q_{j}=S_{j}$. Hence either $Q_{j}=R_{j}$ or there exists some $k\in I$ such that $j\prec k$, $P_{j}\cap P_{k}=\emptyset$ and ${\mathbf{P}}_{j}$ does not have a greatest element. Similarly, either $Q_{i}=R_{i}$ or there exists some $h\in I$ such that $h\prec i$, $P_{h}\cap P_{i}=\emptyset$ and ${\mathbf{P}}_{i}$ does not have a smallest element. We distinguish the following four cases:

$Q_{j}=R_{j}$ and $Q_{i}=R_{i}$: Since $a\in Q_{i}\cap Q_{j}$ we have $a\in P_{i}\cap P_{j}$, i.e., $a=1_{P_{i}}=0_{P_{j}}$. We have $a=1_{Q_{i}}=0_{Q_{j}}$.

$Q_{j}=R_{j}$ and there exists some $h\in I$ such that $h\prec i\prec j$, $P_{h}\cap P_{i}=\emptyset$ and ${\mathbf{P}}_{i}$ does not have a smallest element: Then either $a\in S_{i}=R_{i}$ (which yields that $a=1_{P_{i}}=0_{P_{j}}$) or $a=1_{S_{h}}$. If $a=1_{S_{h}}$ we have either $a=1_{P_{h}}\in P_{h}\cap P_{j}=\emptyset$ or $a=0_{S_{i}}=0_{R_{i}}\in R_{j}$, i.e., $a\in P_{i}\cap P_{j}$, a contradiction.

$Q_{i}=R_{i}$ and there exists some $k\in I$ such that $j\prec k$, $P_{j}\cap P_{k}=\emptyset$ and ${\mathbf{P}}_{j}$ does not have a greatest element: Since $a\in Q_{j}=S_{j}$ we have either $a\in P_{j}\cap P_{i}$, i.e., $a=1_{P_{i}}=0_{P_{j}}$ or $a=0_{R_{k}}$. If $a=0_{R_{k}}$ then either $a=0_{P_{k}}\in P_{k}\cap P_{i}=\emptyset$ or $a=0_{R_{k}}\in R_{k}\setminus P_{k}$, $a=(b,k)\not\in R_{i}=Q_{i}$, a contradiction.

There exist $h,k\in I$ such that $h\prec i\prec j\prec k$, $P_{h}\cap P_{i}=\emptyset=P_{j}\cap P_{k}$, ${\mathbf{P}}_{j}$ does not have a greatest element and ${\mathbf{P}}_{i}$ does not have a smallest element: We have $a\in Q_{i}\cap S_{j}$. Hence ($a\in R_{j}$ or $a=0_{R_{k}}$) and ($a\in R_{i}$ or $a=1_{R_{h}}$). We have four cases. Three of them can be settled as above. So assume that ($a=0_{R_{k}}$) and ($a=1_{R_{h}}$). If $a=0_{P_{k}}$ or $a=1_{P_{h}}$, we have $a\in P_{k}\cap P_{h}=\emptyset$, a contradiction. Hence $a=0_{R_{k}}\in R_{k}\setminus P_{k}$ and $a=1_{R_{h}}\in R_{h}\setminus P_{h}$, a contradiction to $h<k$.

Assume now $P_{i}\cap P_{j}=\emptyset$. Then $R_{i}\cap R_{j}=\emptyset$. Clearly $S_{i}\cap S_{j}\subseteq\{0_{R_{j}}\}$. Assume first $S_{i}\cap S_{j}=\emptyset$. Then $Q_{i}\cap Q_{j}\subseteq\{1_{S_{i}}\}=\{1_{R_{i}}\}$. If $Q_{i}\cap Q_{j}=\{1_{S_{i}}\}=\{1_{R_{i}}\}$ then $1_{Q_{i}}=1_{S_{i}}=0_{Q_{j}}$. Suppose now $S_{i}\cap S_{j}=\{0_{R_{j}}\}$. Then $0_{R_{j}}=0_{S_{j}}=1_{S_{i}}$. If ${\mathbf{P}}_{j}$ does have a smallest element then $S_{j}=Q_{j}$ and $Q_{i}=S_{i}$ or $Q_{i}=(S_{i}\setminus\{0_{S_{i}}\})\cup\{1_{S_{h}}\}$ for some $h\in I$ with $h\prec i\prec j$. In the first case $Q_{i}\cap Q_{j}=S_{i}\cap S_{j}=\{0_{R_{j}}\}=\{0_{P_{j}}\}=\{0_{Q_{j}}\}$. But $1_{S_{i}}=1_{Q_{i}}$. Hence $Q_{i}\cap Q_{j}=\{1_{Q_{i}}\}$ as well. In the second case $Q_{i}\cap Q_{j}=((S_{i}\setminus\{0_{S_{i}})\cup\{1_{S_{h}}\})\cap S_{j}\subseteq(S_{i}\cup S_{h})\cap S_{j}\subseteq(S_{i}\cup R_{h})\cap S_{j}\subseteq\{0_{R_{j}}\}$. Moreover, $1_{S_{i}}=1_{Q_{i}}\in Q_{i}$ and $1_{S_{i}}=0_{Q_{j}}\in Q_{j}$. Hence $Q_{i}\cap Q_{j}=\{0_{Q_{j}}\}=\{1_{Q_{i}}\}$.

Assume ${\mathbf{P}}_{j}$ does not have a smallest element. Then $Q_{j}=(S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\}$ and $Q_{i}=S_{i}$ or $Q_{i}=(S_{i}\setminus\{0_{S_{i}})\cup\{1_{S_{h}}\}$ for some $h\in I$ with $h\prec i\prec j$. In the first case $Q_{i}\cap Q_{j}=S_{i}\cap((S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\})=\{0_{R_{j}}\}=\{1_{S_{i}}\}$. Moreover, $0_{Q_{j}}=1_{S_{i}}=1_{Q_{i}}$. In the second case $Q_{i}\cap Q_{j}=((S_{i}\setminus\{0_{S_{i}})\cup\{1_{S_{h}}\})\cap((S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\})=\{1_{S_{i}}\}$. As above, $1_{S_{i}}=1_{Q_{i}}\in Q_{i}$ and $1_{S_{i}}=0_{Q_{j}}\in Q_{j}$.

Summarizing, we obtain that always $Q_{i}\cap Q_{j}=\emptyset$ or $Q_{i}\cap Q_{j}=\{a\}$ in which case $a=1_{Q_{i}}=0_{Q_{j}}$. 4. (y4):

Assume $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{i}$ has a greatest element $1_{P_{i}}$ and $\mathbf{P}_{j}$ has a smallest element $0_{P_{j}}$. Then $R_{i}\cap R_{j}=\emptyset$, $S_{i}=R_{i}$ and $Q_{j}=S_{j}$. Assume $a\in Q_{i}\cap Q_{j}$. Then $a\in Q_{i}\cap S_{j}$. We have either $a=1_{S_{h}}$ for some $h\in I$ with $h\prec i$ or $a\in S_{i}=R_{i}$, and either $a\in R_{j}$ or $a=0_{R_{k}}$ for some $k\in I$ with $j\prec k$. We can assume $a\in S_{i}=R_{i}$ or $a\in R_{h}\cup R_{i}$ for some $h\prec i<j$ and $a\in S_{j}\subseteq R_{j}\cup R_{k}$. Moreover, $R_{h}\cap(R_{j}\cup R_{k})=\emptyset$ and $R_{i}\cap(R_{j}\cup R_{k})=\emptyset$, a contradiction. We conclude $Q_{i}\cap Q_{j}=\emptyset$. 5. (y5):

Suppose $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{i}$ does not have a greatest element and $\mathbf{P}_{j}$ has a smallest element $0_{P_{j}}$. We have $R_{i}\cap R_{j}=\emptyset$, $Q_{j}=S_{j}$ and $S_{i}=\{0_{R_{j}}\}\cup(R_{i}\setminus\{1_{R_{i}}\})$. Clearly, $0_{R_{j}}\in Q_{i}$ and $0_{R_{j}}\in S_{j}=Q_{j}$. Hence $0_{P_{j}}=0_{R_{j}}\in Q_{i}\cap Q_{j}$ is the greatest element of $\mathbf{Q}_{i}$. 6. (y6):

Assume $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{j}$ does not have a smallest element and $\mathbf{P}_{i}$ has a greatest element $1_{P_{i}}$. We obtain $S_{i}=R_{i}$, $Q_{j}=(S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\}$ and $R_{i}\cap R_{j}=\emptyset$. We have $1_{P_{i}}=1_{R_{i}}=1_{S_{i}}\in Q_{j}$ and $1_{S_{i}}\in Q_{i}$. Hence $1_{P_{i}}$ is the smallest element of $\mathbf{Q}_{j}$. 7. (y7):

Assume $i,j\in I$, $i\prec j$, $P_{i}\cap P_{j}=\emptyset$, $\mathbf{P}_{j}$ does not have a smallest element and $\mathbf{P}_{i}$ does not have a greatest element. Then $S_{i}=\{0_{R_{j}}\}\cup(R_{i}\setminus\{1_{R_{i}}\})$ and $Q_{j}=(S_{j}\setminus\{0_{S_{j}}\})\cup\{1_{S_{i}}\}$. Since $1_{S_{i}}=0_{R_{j}}=0_{Q_{j}}$ and $1_{S_{i}}=1_{Q_{i}}$, we obtain that the greatest element $1_{Q_{i}}$ of $\mathbf{Q}_{i}$ is the smallest element $0_{Q_{j}}$ of $\mathbf{Q}_{j}$.

∎

Finally, we can state our results on Dedekind-MacNeille completion of posets which are the generalized ordered sum of their parts $(P_{i},\leq_{i}),i\in I$.

Theorem 4.7**.**

Let $\mathbf{P}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ and $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ be a DM-yoked family of $\mathbf{P}$. Then the generalized ordinal sum $\mathbf{Q}=(Q,\leq)$ of $\mathbf{Q}_{i}=(Q_{i},\subseteq),i\in I,$ exists. If any non-empty subset of $I$ has a maximal element then $\operatorname{\mathbf{DM}}(\mathbf{P})\cong\mathbf{Q}$.

Proof.

First let us check that the assumptions of Definition 4.2 are satisfied. Clearly, $\mathbf{Q}_{i},i\in I,$ is a family of posets such that $\mathbf{Q}_{\top}$ has a greatest element $1$ (we may identify $1$ with $1_{P_{\top}}$) and condition (i) is satisfied by (y1). Condition (ii) follows from (y2) and condition (iii) from (y3). From (y8) we obtain condition (iv). Hence the generalized ordinal sum $\mathbf{Q}=(Q,\leq)$ of $\mathbf{Q}_{i}=(Q_{i},\subseteq),i\in I,$ exists.

Assume now that any non-empty subset of $I$ has a maximal element and $I$ has a smallest element $\bot$. Let us show $\mathbf{Q}\cong\operatorname{\mathbf{DM}}(\mathbf{P})$. Note that we have, for every $i\in I$, order isomorphisms $\varphi_{i}\colon Q_{i}\to\operatorname{DM}(P_{i})$ and $\psi_{i}\colon\operatorname{DM}(P_{i})\to Q_{i}$ defined by $\varphi_{i}(a):=L_{P_{i}}U_{P_{i}}(\{x\in P_{i}\mid x\leq a\})$ and $\psi_{i}(B):=\bigvee_{Q_{i}}(B\cap P_{i})$ for all $a\in Q_{i}$ and $B\in\operatorname{DM}(P_{i})$, and $\varphi_{i}\circ\psi_{i}=\mbox{id}_{\operatorname{DM}(P_{i})}$, $\psi_{i}\circ\varphi_{i}=\mbox{id}_{Q_{i}}$.

We define mappings $\varphi\colon Q\to\operatorname{DM}(P)$ and $\psi\colon\operatorname{DM}(P)\to Q$ as follows:

[TABLE]

where $j:=\max\limits_{B\cap P_{m}\neq\emptyset}m$ ($a\in Q$, $B\in\operatorname{DM}(P)$). Clearly, $\varphi$ and $\psi$ are well-defined and order-preserving. Recall also that $\varphi(a)=\varphi_{k}(a)\cup\bigcup\limits_{m<k}P_{m}$ where $k:=\max\limits_{a\in Q_{m}}m$. We have

[TABLE]

for all $a\in Q$, here $k:=\max\limits_{a\in Q_{m}}m$. Let $B\in\operatorname{DM}(P)$. If $B=\emptyset$ and $B=LU(B)$ then $\varphi(\psi(\emptyset))=\varphi(\psi_{\bot}(\emptyset))=\varphi_{\bot}(0_{\operatorname{DM}(P_{\bot})})$. Assume $x\in\varphi_{\bot}(0_{\operatorname{DM}(P_{\bot})})$. Then $x$ is the smallest element of $\mathbf{P}_{\bot}$, i.e., $x\leq p$ for all $p\in P$, i.e., $x\in B$, a contradiction. Hence $\varphi(\psi(\emptyset))=\emptyset$. Suppose now that $B\neq\emptyset$ and put $j:=\max\limits_{B\cap P_{m}\neq\emptyset}m$. Then

[TABLE]

∎

Now we show that the construction of a generalized ordinal sum preserves the property of sectional pseudocomplementation.

Theorem 4.8**.**

Let $\mathbf{P}=(P,\leq)$ be a generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ assume that $(P_{i},\leq,*_{i})$ are sectionally pseudocomplemented for all $i\in I$, that $j\in I$, $L_{j}(P_{j})=\emptyset$ implies $U_{s}(P_{s})=\emptyset$ where $s:=\max\limits_{m<j}m$ and that any non-empty subset of $I$ has a maximal element. Then ${\mathbf{P}}$ is sectionally pseudocomplemented.

Proof.

Let $i,j\in I$, $a\in P_{i}$ and $b\in P_{j}$ such that $i$ and $j$ are maximal with this property. We put

[TABLE]

We prove that $a*b$ is the sectional pseudocomplement of $a$ and $b$ in $\mathbf{P}$.

Case 1. $a\leq b$.

We have $L(U(a,b),1)=L(b,1)=L(b)$.

Case 2. $a\not\leq b$ and $i=j$.

We have

[TABLE]

Now assume $L(U(a,b),c)=L(b)$, $c\in P_{k}$, $k\in I$. Then $b\leq c$ which implies $k\geq i$. Now $k>i$ would imply

[TABLE]

a contradiction. Hence $k=i$ and

[TABLE]

which implies $c\leq a*_{i}b$.

Case 3. $a\not\leq b$ and $i>j$.

We have $L(U(a,b),b)=L(a,b)=L(b)$. Now assume $L(U(a,b),c)=L(b)$, $c\in P_{k}$, $k\in I$. Then $b\leq c$ which implies $k\geq j$ and $L(b)=L(a,c)$.

Case 3a. $i>k$.

We have $c\leq b$ which yields $b=c$.

Case 3b. $i<k$.

We have $L(b)=L(a,c)=L(a)$, i.e., $a\leq b$, a contradiction.

Case 3c. $i=k$.

We obtain $k>j$. Since $L(b)=L(a,c)$, $b=1_{j}$ is the greatest element of $\mathbf{P}_{j}$. If $\mathbf{P}_{i}$ has a smallest element $0_{i}$ then necessarily $b=0_{i}\in P_{i}$, a contradiction to the assumption that $j$ is the maximal index from $I$ with $b\in P_{j}$. Hence $\mathbf{P}_{i}$ does not have a smallest element. Put $r:=\max\limits_{m<k}m$. Assume first that $j<r<k=i$. Then there exist by Definition 4.2 (i) elements $x,y\in P_{r}$ with $b\leq x<y\leq a,c$. We conclude $y\in L(a,c)$, $y\not\leq b$, a contradiction. Suppose now that $j=r<k=i$. Since $L_{i}(P_{i})=\emptyset$ and $b$ is the greatest element of $\mathbf{P}_{j}$ we obtain $\{b\}=U_{r}(P_{r})=\emptyset$, a contradiction. Hence the only possible case is 3a which yields $b=c$ and finally that ${\mathbf{P}}$ is sectionally pseudocomplemented. ∎

Altogether, we can summarize our results as follows.

Corollary 4.9**.**

Let ${\mathbf{P}}=(P,\leq)$ be the generalized ordinal sum of $\mathbf{P}_{i}=(P_{i},\leq_{i}),i\in I,$ such that $P_{j}\cap(\operatorname{\mathbf{DM}}(P_{i})\times\{i\})=\emptyset$ for all $i,j\in I$. Let $(\operatorname{DM}(\mathbf{P}_{i}),\leq_{i},*_{i})$ be sectionally pseudocomplemented for all $i\in I$ and assume that any non-empty subset of $I$ has a maximal element and that $j\in I$ and $L_{j}(P_{j})=\emptyset$ imply $U_{s}(P_{s})=\emptyset$ where $s:=\max\limits_{m<j}m$. Then ${\operatorname{\mathbf{DM}}(\mathbf{P})}$ is sectionally pseudocomplemented.

Proof.

From Lemma 4.6 we obtain that there exists a DM-yoked family $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ of $\mathbf{P}$ such that $\mathbf{Q}_{i}$ is isomorphic to a sectionally pseudocomplemented poset $\operatorname{\mathbf{DM}}(\mathbf{P}_{i})$ for every $i\in I$. From Theorem 4.7 we know that $\operatorname{\mathbf{DM}}(\mathbf{P})$ is isomorphic to the generalized ordinal sum $\mathbf{Q}$ of the DM-yoked family $\mathbf{Q}_{i}=(Q_{i},\leq_{i}),i\in I,$ of ${\mathbf{P}}$. Since every $\mathbf{Q}_{i}$ is sectionally pseudocomplemented we have from Theorem 4.8 that $\mathbf{Q}$ and hence also ${\operatorname{\mathbf{DM}}(\mathbf{P})}$ are sectionally pseudocomplemented. ∎

The situation described in Theorem 4.8 and Corollary 4.9 can be illustrated by the following example.

Example 4.10**.**

*Consider the sectionally pseudocomplemented poset $\mathbf{P}$ from

Example 2.7. It is evident that $\mathbf{P}$ is the generalized ordinal sum of the sectionally pseudocomplemented posets $\mathbf{P}_{1}=(P_{1},\leq)=(\{0,a,b,c\},\leq)$ and $\mathbf{P}_{2}=(P_{2},\leq)=(\{d,e,1\},\leq)$. Of course, $P_{1}\cap P_{2}=\emptyset$. Hence the conditions of Definition 4.2 are satisfied. Then $\operatorname{\mathbf{DM}}(\mathbf{P}_{1})$ is the lattice N5 and $\operatorname{\mathbf{DM}}(\mathbf{P}_{2})$ the four-element Boolean algebra, i.e. both are sectionally pseudocomplemented lattices. The generalized ordinal sum of $\operatorname{\mathbf{DM}}(\mathbf{P}_{1})$ and $\operatorname{\mathbf{DM}}(\mathbf{P}_{2})$ is visualized in Fig. 7. According to Theorem 4.8 it is again sectionally pseudocomplemented.*

The Dedekind-MacNeille completion of the poset visualized in Fig. 3 is visualized in Fig. 7

[TABLE]

Here $L(x)$ is abbreviated by $x$ for $x\in\{0,a,b,c,d,e,1\}$ and $f$ is an abbreviation of $L(d,e)$. The operation table of $*$ in $\operatorname{\mathbf{DM}}(\mathbf{P})$ looks as follows:

[TABLE]

According to Corollary 4.9, $\operatorname{\mathbf{DM}}(\mathbf{P})$ is just the generalized ordinal sum of $\operatorname{\mathbf{DM}}(\mathbf{P}_{1})$ and $\operatorname{\mathbf{DM}}(\mathbf{P}_{2})$.