A space with a Lusin $\pi$-base whose square has no Lusin $\pi$-base

TL;DR

This paper constructs a topological space with a Lusin π-base whose Cartesian square lacks such a base, highlighting a unique property of these spaces.

Contribution

It presents the first example of a space with a Lusin π-base whose square does not have a Lusin π-base, revealing new insights into their structural properties.

Findings

Constructed a space with a Lusin π-base but its square lacks one.

Demonstrated the non-preservation of Lusin π-base property under Cartesian square formation.

Provided a counterexample to previous assumptions about Lusin π-bases.

Abstract

We construct a space ${X}$ that has a Lusin $\pi$-base and such that ${X}^{2}$ has no Lusin $\pi$-base.

Equations

$${\mathrm{span}}_L(U,p):-\{\mathrm{length}(a):a\in {}^{<\omega }\omega ,L_a\ni p,\mathsf{and}{\mathrm{shoot}}_L(a)\dot{\in }U\}\subseteq \omega$$

$$r_x^0:-{\mathrm{res}}^0({\Gamma }_x)\in {}^{\omega }\omega ,M_x^0:-M^0({\Gamma }_x)\in [\omega {]}^{\omega },g_x^0:-g^0({\Gamma }_x)\in {}^{M_x^0}\omega ,$$

$$r_x^1:-{\mathrm{res}}^1({\Gamma }_x)\in {}^{\omega }\omega ,M_x^1:-M^0({\Gamma }_x)\in [\omega {]}^{\omega },g_x^1:-g^1({\Gamma }_x)\in {}^{M_x^1}\omega .$$

$$\widehat{\mathbf{S}}_{{r}^{0}_{{x}}}({M}^{0}_{{x}},{g}^{0}_{{x}})\times\widehat{\mathbf{S}}_{{r}^{1}_{{x}}}({M}^{1}_{{x}},{g}^{1}_{{x}})\>\in\>\operatorname{\mathsf{nbhds}}\big{(}\langle{r}^{0}_{{x}},{r}^{1}_{{x}}\rangle,\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\times\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\big{)},$$

$$L_a\ni ⟨r_x^0,r_x^1⟩\text{and}{\mathrm{shoot}}_L(a)\dot{\in }{\hat{\mathbf{S}}}_{r_x^0}(M_x^0,g_x^0)\times {\hat{\mathbf{S}}}_{r_x^1}(M_x^1,g_x^1).$$

$${\mathrm{shoot}}_L(a)/\dot{\in }{\hat{\mathbf{S}}}_{r_x^0}(M_x^0,g_x^0)\times {\hat{\mathbf{S}}}_{r_x^1}(M_x^1,g_x^1),$$

$$\operatorname{\mathsf{shoot}}_{{L}}({b}_{{c}}{\upharpoonright}{k})\>\not\mathop{\dot{\in}}\>{R}_{{c},{n}_{{c}}}\>=\>\big{(}\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}_{{c}}}{\uparrow}\times\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}_{{c}}}\big{)}\>\cup\>\big{(}\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}_{{c}}}\times(\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}_{{c}}}{\uparrow})\big{)}.$$

$${\mathbf{S}}_{c^0}\times {\mathbf{S}}_{c^1}\supseteq L_{b_c}\supseteq {\mathbf{S}}_{a_c^0}\times {\mathbf{S}}_{a_c^1}\supseteq {\mathbf{S}}_{a_c^{0⌢}{n}_c}\times {\mathbf{S}}_{a_c^{1⌢}{n}_c},$$

$$\operatorname{\mathsf{shoot}}_{{L}}({a})\>\mathop{\dot{\in}}\>\big{(}\,\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})\times\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1})\big{)}\,\cap\,(\,\mathbf{S}_{{c}^{0}_{\dot{n}-1}}\!\!\times\mathbf{S}_{{c}^{1}_{\dot{n}-1}})$$

$$\operatorname{\mathsf{shoot}}_{{L}}({a})\>\mathop{\dot{\in}}\>\big{(}\,\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})\cap\mathbf{S}_{{c}^{0}_{\dot{n}-1}}\big{)}\,\times\,\big{(}\,\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1})\cap\mathbf{S}_{{c}^{1}_{\dot{n}-1}}\big{)}.$$

$$q\in {\tilde{\mathbf{S}}}_{r^1\upharpoonright \dot{m}}^{r^1(\dot{m})+1}\subseteq {\tilde{\mathbf{S}}}_{r^1\upharpoonright \dot{m}}={\mathbf{S}}_{r^1\upharpoonright (\mathrm{length}(c_{2\dot{k}+2}^1)-1)}\subseteq {\mathbf{S}}_{r^1\upharpoonright \mathrm{length}(d_{2\dot{k}+1}^1)}={\mathbf{S}}_{d_{2\dot{k}+1}^1}\subseteq {\mathbf{S}}_{d_{\dot{n}}^1}.$$

$${T}^{0}\times{T}^{1}\>\subseteq\>\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}{\uparrow}\times\mathbf{S}_{{d}^{1}_{\dot{n}}}\big{)}\cup\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}\times(\mathbf{S}_{{d}^{1}_{\dot{n}}}{\uparrow})\big{)}.$$

$$\operatorname{\mathsf{shoot}}_{{L}}({a})\>\mathop{\dot{\in}}\>\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}{\uparrow}\times\mathbf{S}_{{d}^{1}_{\dot{n}}}\big{)}\cup\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}\times(\mathbf{S}_{{d}^{1}_{\dot{n}}}{\uparrow})\big{)}.$$

$$\operatorname{\mathsf{shoot}}_{{L}}({b}_{\dot{n}}{\upharpoonright}{k})\>\not\mathop{\dot{\in}}\>\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}{\uparrow}\times\mathbf{S}_{{d}^{1}_{\dot{n}}}\big{)}\cup\big{(}\mathbf{S}_{{d}^{0}_{\dot{n}}}\times(\mathbf{S}_{{d}^{1}_{\dot{n}}}{\uparrow})\big{)}.$$

Full text

1 Introduction

The class of topological spaces with a Lusin $\pi$-base, see Definition 2, was introduced in [1]; this class equals the class of spaces with a $\pi$-tree [2, Remark 11]. In this paper we build a space with a Lusin $\pi$-base whose square has no Lusin $\pi$-base, see Theorem 25.

The Baire space $\omega^{\omega},$ the Sorgenfrey line $\mathcal{S},$ and the irrational Sorgenfrey line $\mathcal{I}$ have a Lusin $\pi$-base [1, 3], and all at most countable products of $\omega^{\omega},$ $\hskip 1.0pt\mathcal{S},$ and $\mathcal{I}$ also have a Lusin $\pi$-base [3]. If a space ${X}$ has a Lusin $\pi$-base, then the products ${X}\times\omega^{\omega},$ ${X}\times\mathcal{S},$ and ${X}\times\mathcal{S}^{\omega}$ have a Lusin $\pi$-base [3], and also ${X}\setminus{F}$ has a Lusin $\pi$-base whenever ${F}\subseteq{X}$ is a $\sigma$-compact [2] (but a dense open subset of $X$ can be without a Lusin $\pi$-base).

If a space ${X}$ has a Lusin $\pi$-base, then it can be mapped onto $\omega^{\omega}$ by a continuous one-to-one map and also ${X}$ can be mapped onto an arbitrary nonempty Polish space by a continuous open map [1]. If a space ${X}$ has a Lusin $\pi$-base, then it has a countable $\hskip 1.0pt\pi$-base and a countable pseudo-base (both with clopen members); and also $X$ is a Choquet space (but it can be not strong Choquet even in a separable metrizable case). For each ${M}\subseteq\omega^{\omega}$, there exists a separable metrizable space with a Lusin $\pi$-base that contains a closed subspace homeomorphic to ${M}.$

2 Notation and terminology

We use terminology from [4] and [5]. A space is a topological space. Also we use the following notations:

Notation 1**.**

The symbol $\coloneq$ means ‘‘equals by definition’’; the symbol ${\colon}{\longleftrightarrow}$ is used to show that an expression on the left side is an abbreviation for expression on the right side;

• ✎ 

$\omega$ $\coloneq$ the set of finite ordinals $=$ the set of natural numbers,

so $0=\varnothing\in\omega$ and ${n}=\{0,\ldots,{n}-1\}$ for all ${n}\in\omega;$

• ✎ 

${s}\,$ is a sequence$\quad{\colon}{\longleftrightarrow}\quad{s}$ is a function such that $\mathsf{domain}({s})\in\omega$ or $\mathsf{domain}({s})=\omega;$

• ✎ 

if ${s}$ is sequence, then

$\operatorname{\mathsf{length}}({s})\>\coloneq\>\mathsf{domain}({s});$

• ✎ 

$\langle{s}_{0},\ldots,{s}_{{n}-1}\rangle$ $\coloneq$ the sequence ${s}$ such that $\operatorname{\mathsf{length}}({s})={n}\in\omega$ and ${s}(i)={s}_{i}$ for all ${i}\in{n};$

• ✎ 

$\langle\rangle$ $\coloneq$ the sequence of length 0;

• ✎ 

if ${s}=\langle{s}_{0},\ldots,{s}_{{n}-1}\rangle,$ then

${s}\hskip 0.5pt^{\frown}{x}\>\coloneq\>\langle{s}_{0},\ldots,{s}_{{n}-1},{x}\rangle;$

• ✎ 

${f}{\upharpoonright}{A}$ $\coloneq$ the restriction of function ${f}$ to ${A};$

• ✎ 

${A}\subset{B}\quad{\colon}{\longleftrightarrow}\quad{A}\subseteq{B}\enskip\mathsf{and}\enskip{A}\neq{B};$

• ✎ 

if ${s}$ and ${t}$ are sequences, then

${s}\sqsubseteq{t}\quad{\colon}{\longleftrightarrow}\quad{s}={t}{\upharpoonright}\operatorname{\mathsf{length}}({s})\,,$

${s}\sqsubset{t}\quad{\colon}{\longleftrightarrow}\quad{s}\sqsubseteq{t}\enskip\mathsf{and}\enskip{s}\neq{t}$

(actually, ${s}\sqsubseteq{t}\leftrightarrow{s}\subseteq{t}\;$ and $\;{s}\sqsubset{t}\leftrightarrow{s}\subset{t}$);

• ✎ 

${}^{B}\!{A}$ $\coloneq$ the set of functions from ${B}$ to ${A};$

in particular, {}^{0}{A}=\big{\{}\langle\rangle\big{\}};

• ✎ 

${}^{{<}\omega}{A}\>\coloneq\>\bigcup_{{n}\in\omega}{}^{{n}}{A}\,=\,$ the set of finite sequences in ${A};$

• ✎ 

${A}\,$ is a singleton$\quad{\colon}{\longleftrightarrow}\quad{A}=\{{x}\}$ for some ${x}\,;$

• ✎ 

$\operatorname{\mathsf{nbhds}}({p},{X})$ $\coloneq$ the set of (not necessarily open) neighbourhoods of point ${p}$ in space ${X};$

• ✎ 

$\gamma\,$ is a $\pi$-net for a space ${X}\quad{\colon}{\longleftrightarrow}\quad$for each nonempty open ${U}\subseteq{X},$ there is ${G}\in\gamma$ such that ${G}\subseteq{U},$ and all elements of $\gamma$ are nonempty;

• ✎ 

$\gamma\,$ is a $\pi$-base for a space ${X}\quad{\colon}{\longleftrightarrow}\quad\gamma$ is a $\pi$-net for ${X}$ and all elements of $\gamma$ are open.

Recall that [6] a Lusin scheme on a set $X$ is a family ${L}=\langle{L}_{a}\rangle_{{a}\in{}^{{<}\omega}\omega}$ of subsets of ${X}$ such that

• (L0)

${L}_{a}\supseteq{L}_{{a}\hskip 0.5pt^{\frown}{n}}\quad$ for all ${a}\in{}^{<\hskip 0.2pt\omega}\omega,$ ${n}\in\omega\enskip$ and

• (L1)

${L}_{{a}\hskip 0.5pt^{\frown}{n}}\cap{L}_{{a}\hskip 0.5pt^{\frown}{m}}=\varnothing\quad$ for all ${a}\in{}^{<\hskip 0.2pt\omega}\omega$ and ${n}\neq{m}\in\omega.$

Definition 2**.**

• ✎ 

A Lusin scheme ${L}$ on a set $X$ is strict  iff

• (L2)

${L}_{\scriptscriptstyle\langle\rangle}={X},$

• (L3)

${L}_{a}=\bigcup_{n\in\omega}{L}_{{a}\hskip 0.5pt^{\frown}{n}}\quad$ for all ${a}\in{}^{{<}\omega}\omega,\enskip$ and

• (L4)

$\bigcap_{n\in\omega}{L}_{{p}{\upharpoonright}{n}}$ is a singleton for all ${p}\in{}^{\omega}\omega.$

• ✎ 

A Lusin scheme ${L}$ on a space ${X}$ is open  iff

• (L5)

each ${L}_{a}$ is an open subset of ${X}.$

• ✎ 

A Lusin $\hskip 1.0pt\pi$-base for a space $X$ is an open strict Lusin scheme ${L}$ on $X$ such that

• (L6)

$\forall{p}\in{X}\enskip\forall{U}\in\operatorname{\mathsf{nbhds}}({p},{X})$

\exists{a}\in{}^{<\hskip 0.2pt\omega}\omega\enskip\exists{k}\in\omega\

• ➢ 

${L}_{a}\ni{p}\enskip$ and

• ➢ 

$\bigcup_{{j}\geqslant{k}}{L}_{{a}\hskip 0.5pt^{\frown}{j}}\subseteq{U}.$

Remark 3**.**

Suppose that ${L}$ is a strict Lusin scheme. Then, for all ${a},{b}\in{}^{{<}\omega}\omega,$

• (1)

${L}_{{a}}\neq\varnothing;$

• (2)

${L}_{{a}}\supset{L}_{{b}}\ \longleftrightarrow\ {a}\sqsubset{b};$

• (3)

${L}_{{a}}\cap{L}_{{b}}\neq\varnothing\ \longleftrightarrow\ ({a}\sqsubseteq{b}\textsf{ or }{a}\sqsupseteq{b}).$ ∎

Remark 4**.**

Suppose that ${L}$ is a Lusin $\pi$-base for a space ${X}.$ Then the family $\{{L}_{{a}}:{a}\in{}^{{<}\omega}\omega\}$ is a $\pi$-base for ${X}.$ ∎

Notation 5**.**

• ✎ 

$\mathbf{S}$ $\coloneq$ the standard Lusin scheme $\coloneq$ the Lusin scheme $\langle\mathbf{S}_{a}\rangle_{{a}\in{}^{{<}\omega}\omega}$ such that

$\mathbf{S}_{{a}}=\{{p}\in{}^{\omega}\omega:{a}\sqsubseteq{p}\}\quad$ for all ${a}\in{}^{{<}\omega}\omega.$

• ✎ 

${\omega^{\omega}}$ $\coloneq$ the Baire space $\coloneq$ the space $\langle{}^{\omega}\omega,\tau_{\mathsf{prod}}\rangle,$ where $\tau_{\mathsf{prod}}$ is the Tychonoff product topology with $\omega$ carrying the discrete topology.

Remark 6**.**

• (1)

The family $\{\mathbf{S}_{{a}}:{a}\in{}^{{<}\omega}\omega\}$ is a base for the Baire space ${\omega^{\omega}}.$

• (2)

The standard Lusin scheme $\mathbf{S}$ is a Lusin $\pi$-base for ${\omega^{\omega}}.$ ∎

Notation 7**.**

We denote by $\vartriangleleft$ the strict lexicographic order on ${}^{\omega}\omega;$ that is, for all ${p},{q}\in{}^{\omega}\omega,$

• ✎ 

${p}\vartriangleleft{q}\quad{\colon}{\longleftrightarrow}\quad\exists{n},{i},{j}\in\omega$ such that

• ➢ 

${p}{\upharpoonright}{n}={q}{\upharpoonright}{n},$

• ➢ 

$({p}{\upharpoonright}{n})\hskip 0.5pt^{\frown}{i}\sqsubseteq{p},$

• ➢ 

$({q}{\upharpoonright}{n})\hskip 0.5pt^{\frown}{j}\sqsubseteq{q},\quad$ and

• ➢ 

${i}<{j}.$

• ✎ 

${p}{\downfilledspoon}\>\coloneq\>\{{p}\}\cup\{{q}\in{}^{\omega}\omega:{p}\vartriangleleft{q}\}.$

• ✎ 

${A}{\uparrow}\>\coloneq\>\bigcup\{{p}{\downfilledspoon}:{p}\in{A}\}.$

Remark 8**.**

Suppose that ${a}\in{}^{{<}\omega}\omega$ and ${n},{m}\in\omega.$ Then

• ➢ 

$\mathbf{S}_{{a}\hskip 0.5pt^{\frown}{n}}\cap(\mathbf{S}_{{a}\hskip 0.5pt^{\frown}{m}}{\uparrow})=\varnothing\quad$ whenever ${n}<{m}.$ ∎

3 Construction of a space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$

In this section, for each family $\mathcal{F}=\langle\mathcal{F}_{{p}}\rangle_{{p}\in{}^{\omega}\omega}$ of free filters on a $\omega,$ we construct a topology $\tau_{\scriptscriptstyle\mathcal{F}}$ on ${}^{\omega}\omega$ such that the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ has a Lusin $\pi$-base, see Example 14. In the next section we build a family $\mathcal{F}$ in such a way that the square of $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ has no Lusin $\pi$-base.

First we reformulate condition (L6) of the definition of a Lusin $\pi$-base, see Definition 2 and Remark 10, by using the notion of a shoot:

Notation 9**.**

Suppose that ${L}$ is a Lusin scheme, ${a}\in{}^{{<}\omega}\omega,$ and ${k}\in\omega.$ Then

• ✎ 

$\widetilde{{L}}^{{k}}_{{a}}\>\coloneq\>\bigcup_{{j}\geqslant{k}}{L}_{{a}\hskip 0.5pt^{\frown}{j}};$

• ✎ 

\operatorname{\mathsf{shoot}}_{{L}}({a})\>\coloneq\>\big{\{}\widetilde{{L}}^{{k}}_{{a}}:{k}\in\omega\big{\}};

• ✎ 

$\gamma\mathop{\dot{\in}}{U}\quad{\colon}{\longleftrightarrow}\quad\exists{G}\in\gamma\>[{G}\subseteq{U}].$

Remark 10**.**

The clause (L6) of the definition of a Lusin $\pi$-base is equivalent to the following:

• (L6’)

$\forall{p}\in{X}\enskip\forall{U}\in\operatorname{\mathsf{nbhds}}({p},{X})$

\exists{a}\in{}^{<\hskip 0.2pt\omega}\omega\ such that

• ➢ 

${L}_{a}\ni{p}\enskip$ and

• ➢ 

$\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{U}.$ ∎

It is convenient to call $\widetilde{{L}}^{{k}}_{{a}}$ a ${k}$-tail of ${L}_{{a}},$ so that the $\operatorname{\mathsf{shoot}}_{{L}}({a})$ is the set of tails of ${L}_{{a}}$ and $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{U}$ iff some tail of ${L}_{{a}}$ is contained in ${U}.$ Note also that the family $\operatorname{\mathsf{shoot}}_{{L}}({a})$ is closed under finite intersections. We will use the following simple properties of shoots:

Remark 11**.**

Suppose that ${L}$ is a strict Lusin scheme and ${a}\in{}^{{<}\omega}\omega.$ Then

• (1)

$\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{L}_{{a}}.$

• (2)

If $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{A}$ and ${A}\subseteq{B}$, then $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{B}.$

• (3)

If $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{A}$ and $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{B}$, then $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{A}\cap{B}.$

• (4)

If ${A}\cap{B}=\varnothing,$ then $\operatorname{\mathsf{shoot}}_{{L}}({a})\not\mathop{\dot{\in}}{A}$ or $\operatorname{\mathsf{shoot}}_{{L}}({a})\not\mathop{\dot{\in}}{B}.$ ∎

Next we define sets $\widehat{\mathbf{S}}_{{p}}({M},{g}),$ which will form basic neighbourhoods at points ${p}\in{}^{\omega}\omega$ for topology $\tau_{\scriptscriptstyle\mathcal{F}}$ in Example 14. Recall that $\mathbf{S}_{a}=\{{p}\in{}^{\omega}\omega:{a}\sqsubseteq{p}\},$ $\widetilde{\mathbf{S}}^{{k}}_{{a}}=\bigcup_{{j}\geqslant{k}}\mathbf{S}_{{a}\hskip 0.5pt^{\frown}{j}},$ and $[\omega]^{\omega}$ is the set of infinite subsets of $\omega.$

Notation 12**.**

Suppose that ${p}\in{}^{\omega}\omega,$ ${M}\in[\omega]^{\omega},$ and ${g}\colon{M}\to\omega.$ Then

• ✎ 

$\displaystyle\widehat{\mathbf{S}}_{{p}}({M},{g})\>\coloneq\>\{{p}\}\,\cup\bigcup_{{m}\in{M}}\widetilde{\mathbf{S}}^{{g}({m})}_{{p}{\upharpoonright}{m}}.$

Remark 13**.**

Suppose that ${p}\in{}^{\omega}\omega,$ ${M}\in[\omega]^{\omega},$ ${g}\colon{M}\to\omega,$ and ${g}({m})>{p}({m})$ for all ${m}\in{M}.$ Then:

• (1)

$\widetilde{\mathbf{S}}^{{g}({m})}_{{p}{\upharpoonright}{m}}\,\subseteq\;{\mathbf{S}}_{{p}{\upharpoonright}{m}}\!\setminus{\mathbf{S}}_{{p}{\upharpoonright}({m}+1)}\quad$ for all ${m}\in{M}.$

• (2)

$\widehat{\mathbf{S}}_{{p}}({M},{g})\,\subseteq\,{p}{\downfilledspoon}\,.$ ∎

Recall that $\gamma$ is a filter on $\omega$ iff $\gamma$ is a family of nonempty subsets of $\omega$ such that (1) ${A}\cap{B}\in\gamma$ for all ${A},{B}\in\gamma$ and (2) ${C}\in\gamma$ whenever $\omega\supseteq{C}\supseteq{A}\in\gamma.$ A filter $\gamma$ is free iff $\bigcap\{{A}:{A}\in\gamma\}=\varnothing.$ Now we construct a space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle,$ which has $\mathbf{S}$ as a Lusin $\pi$-base:

Example 14**.**

Suppose that $\mathcal{F}=\langle\mathcal{F}_{{p}}\rangle_{{p}\in{}^{\omega}\omega}$ is an indexed family of free filters on $\omega.$ Then

• ✎ 

$\tau_{\scriptscriptstyle\mathcal{F}}\>\coloneq\>$ the topology on ${}^{\omega}\omega$ generated by the subbase \big{\{}\,\widehat{\mathbf{S}}_{{p}}({M},{g})\>:\>{p}\in{}^{\omega}\omega,\>{M}\in\mathcal{F}_{{p}},\ \mathsf{and}\ {g}\colon{M}\to\omega\big{\}}.

The space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ has the following properties:

• (F1)

For each point ${p}\in{}^{\omega}\omega,$ the family \big{\{}\,\widehat{\mathbf{S}}_{{p}}({M},{g}):{M}\in\mathcal{F}_{{p}}\enskip\mathsf{and}\enskip{g}\colon{M}\to\omega\big{\}} is a neighbourhood base for $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ at ${p}.$

• (F2)

The standard Lusin scheme $\mathbf{S}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle.$ ∎

It is interesting to note one additional property of the above space, although we do not use it in this paper. In [3] we found sufficient conditions under which the product of spaces that have a Lusin $\pi$-base also has a Lusin $\pi$-base; these conditions were formulated by using the following terminology: For a Lusin scheme ${L}$ on a space ${X},$ a point ${p}\in{X}$ and its neighbourhood ${U}\subseteq{X},$ the span of ${U}$ at ${p}$ is defined as follows:

[TABLE]

(so that clause (L6’) of Remark 10 is equivalent to the assertion that every $\operatorname{\mathsf{span}}_{{L}}({U},{p})$ is nonempty). If ${L}$ is a Lusin $\pi$-base for ${X}$ and ${p}\in{X},$ then the family $\{\operatorname{\mathsf{span}}_{{L}}({U},{p}):{U}\in\operatorname{\mathsf{nbhds}}({p},{X})\}$ has the finite intersection property [3, Lemma 3.6], so we can define the $\operatorname{\mathsf{root}}_{{L}}({p},{X})$ to be the filter on $\omega$ generated by this family. A theorem in [3] says that if some spaces have Lusin $\pi$-bases with ‘‘good’’ roots, then the product of these spaces has a Lusin $\pi$-base.

Now, the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ from Example 14 has the following property:

• (F3)

\operatorname{\mathsf{root}}_{\hskip 1.0pt\mathbf{S}}\big{(}{p},\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\big{)}=\mathcal{F}_{{p}}  for all ${p}\in{}^{\omega}\omega.$

Therefore we can build a space with a Lusin $\pi$-base that has any roots we wish.

4 Construction of a space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ whose square

has no Lusin $\pi$-base

In this section we prove Theorem 25, the main result of the article, which says that there exists a space with a Lusin $\pi$-base whose square has no Lusin $\pi$-base. To build such a space we construct a family $\mathcal{F}$ of free filters on $\omega$ in such a way that the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle^{2}$ has no Lusin $\pi$-base. We construct this family in the proof of Lemma 24, so Theorem 25 is a corollary to this lemma. To make the logic of construction more transparent, we state Lemmas 17, 19, 20, 22 without proofs and prove them in Section 5.

Notation 15**.**

• ✎ 

A game is an ordered pair of sequences \Gamma=\big{\langle}\langle{c}_{{n}}\rangle_{{n}\in\omega},\langle{d}_{{n}}\rangle_{{n}\in\omega}\big{\rangle} such that, for all ${n}\in\omega,$

• ➢ 

${c}_{{n}}=\langle{c}^{0}_{{n}},{c}^{1}_{{n}}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$

• ➢ 

${d}_{{n}}=\langle{d}^{0}_{{n}},{d}^{1}_{{n}}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$

• ➢ 

${c}^{0}_{{n}}\sqsubset{d}^{0}_{{n}}\sqsubset{c}^{0}_{{n}+1},\,$ and $\,{c}^{1}_{{n}}\sqsubset{d}^{1}_{{n}}\sqsubset{c}^{1}_{{n}+1}.$

• ✎ 

The result of a game $\Gamma$ is the pair \big{\langle}\operatorname{\mathsf{res}}^{0}(\Gamma),\operatorname{\mathsf{res}}^{1}(\Gamma)\big{\rangle}\in{}^{\omega}\omega\times{}^{\omega}\omega such that, for all ${n}\in\omega,$

• ➢ 

${c}^{0}_{{n}}\sqsubseteq\operatorname{\mathsf{res}}^{0}(\Gamma)\,$ and ${c}^{1}_{{n}}\sqsubseteq\operatorname{\mathsf{res}}^{1}(\Gamma).$

• ✎ 

A strategy is an ordered pair $\sigma=\langle\sigma^{0},\sigma^{1}\rangle$ of functions $\sigma^{0},\sigma^{1}\colon{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega\to{}^{{<}\omega}\omega$ such that for each ${c}=\langle{c}^{0},{c}^{1}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$

• ➢ 

${c}^{0}\sqsubset\sigma^{0}({c})$ and ${c}^{1}\sqsubset\sigma^{1}({c}).$

• ✎ 

A game $\Gamma$ follows a strategy $\sigma$ iff  for all ${n}\in\omega,$

• ➢ 

${d}^{0}_{{n}}=\sigma^{0}({c}_{{n}})$ and ${d}^{1}_{{n}}=\sigma^{1}({c}_{{n}}).$

Remark 16**.**

Suppose that \Gamma=\big{\langle}\langle{c}_{{n}}\rangle_{{n},{m}\in\omega},\langle{d}_{{n}}\rangle_{{n}\in\omega}\big{\rangle} is a game. Then, for all ${i}\in{2}=\{0,1\}$ and ${n},{m}\in\omega,$

• (1)

$\mathbf{S}_{{c}^{{i}}_{{n}}}\,\supset\>\mathbf{S}_{{d}^{{i}}_{{n}}}\,\supset\>\mathbf{S}_{{c}^{{i}}_{{n}+1}}\ni\>\operatorname{\mathsf{res}}^{{i}}(\Gamma);$

• (2)

${c}^{i}_{{n}}=\operatorname{\mathsf{res}}^{{i}}(\Gamma){\upharpoonright}\operatorname{\mathsf{length}}({c}^{i}_{{n}});$

• (3)

${d}^{i}_{{n}}=\operatorname{\mathsf{res}}^{{i}}(\Gamma){\upharpoonright}\operatorname{\mathsf{length}}({d}^{i}_{{n}});$

• (4)

$\operatorname{\mathsf{length}}({c}^{i}_{{n}+1})-1\geqslant\operatorname{\mathsf{length}}({d}^{i}_{{n}});$

• (5)

${c}^{{i}}_{{n}}\sqsubseteq{c}^{{i}}_{{m}}\ \longleftrightarrow\ {n}\leqslant{m};$

• (6)

${d}^{{i}}_{{n}}\sqsubseteq{d}^{{i}}_{{m}}\ \longleftrightarrow\ {n}\leqslant{m}.$ ∎

Lemma 17**.**

For each strategy $\sigma,$ there exists a family $\langle\Gamma_{{z}}\rangle_{{z}\in{}^{\omega}{2}}$ of games that follow $\sigma$ such that $\operatorname{\mathsf{res}}^{i}(\Gamma_{{z}})\neq\operatorname{\mathsf{res}}^{{j}}(\Gamma_{{y}})$ for all $\langle{z},{i}\rangle\neq\langle{y},{j}\rangle\in{}^{\omega}{2}\times{2}.$

Recall that a family $\gamma$ is a $\pi$-net for a space ${X}$ iff $\varnothing\nin\gamma$ and each nonempty open ${U}\subseteq{X}$ contains some ${G}\in\gamma$ as a subset. Recall also that ${\omega^{\omega}}$ is the Baire space $\langle{}^{\omega}\omega,\tau_{\mathsf{prod}}\rangle.$

Notation 18**.**

A candidate is a strict Lusin scheme ${L}$ on the set ${}^{\omega}\omega\times{}^{\omega}\omega$ such that $\{{L}_{{b}}:{b}\in{}^{{<}\omega}\omega\}$ is a $\pi$-net for the space ${\omega^{\omega}}\times{\omega^{\omega}}$ and each ${L}_{{b}}$ has nonempty interior in ${\omega^{\omega}}\times{\omega^{\omega}}.$

Lemma 19**.**

Suppose that the standard Lusin scheme $\mathbf{S}$ is a Lusin $\pi$-base for a space $\langle{}^{\omega}\omega,\tau\rangle$ and $\,{L}\!$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle\times\langle{}^{\omega}\omega,\tau\rangle.$ Then ${L}$ is a candidate.

Lemma 20**.**

For each candidate ${L},$ there exists a strategy $\sigma$ that possesses the following property:

• (★)

For each ${c}=\langle{c}^{0},{c}^{1}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$

there is ${b}\in{}^{{<}\omega}\omega$ such that

• ➢ 

$\mathbf{S}_{{c}^{0}}\times\mathbf{S}_{{c}^{1}}\>\supseteq\>{L}_{{b}}\>\supseteq\>\mathbf{S}_{\sigma^{0}({c})}\times\mathbf{S}_{\sigma^{1}({c})}\quad$* and*

• ➢ 

\operatorname{\mathsf{shoot}}_{{L}}({b}{\upharpoonright}{k})\>\not\mathop{\dot{\in}}\>\big{(}\mathbf{S}_{\sigma^{0}({c})}{\uparrow}\times\mathbf{S}_{\sigma^{1}({c})}\big{)}\cup\big{(}\mathbf{S}_{\sigma^{0}({c})}\times(\mathbf{S}_{\sigma^{1}({c})}{\uparrow})\big{)}\quad* for all ${k}<\operatorname{\mathsf{length}}({b}).$*

Notation 21**.**

Suppose that ${L}$ is a candidate. Then we denote by $\sigma_{\scriptscriptstyle{L}}$ some (fixed) strategy $\sigma$ such that property (★) holds.

Lemma 22**.**

For each game $\Gamma,$ there are sets ${M}^{0},{M}^{1}\in[\omega]^{\omega}$ and functions ${g}^{0}\colon{M}^{0}\to\omega,$ ${g}^{1}\colon{M}^{1}\to\omega$ that possess the following property:

• (◆)

For each candidate ${L}$ and each ${a}\in{}^{{<}\omega}\omega,$

• ➢ 

if game $\Gamma$ follows strategy $\sigma_{\scriptscriptstyle{L}}$ and $\,{L}_{{a}}\ni\langle\operatorname{\mathsf{res}}^{0}(\Gamma),\operatorname{\mathsf{res}}^{1}(\Gamma)\rangle,$

• ➢ 

then $\ \operatorname{\mathsf{shoot}}_{{L}}({a})\>\not\mathop{\dot{\in}}\>\widehat{\mathbf{S}}_{\operatorname{\mathsf{res}}^{0}(\Gamma)}({M}^{0},{g}^{0})\times\widehat{\mathbf{S}}_{\operatorname{\mathsf{res}}^{1}(\Gamma)}({M}^{1},{g}^{1}).$

Notation 23**.**

Suppose that $\Gamma$ is a game. Then we denote by ${M}^{0}(\Gamma),$ ${M}^{1}(\Gamma),$ ${g}^{0}(\Gamma),$ ${g}^{1}(\Gamma)$ some (fixed) sets ${M}^{0},{M}^{1}$ and functions ${g}^{0},{g}^{1}$ such that property (◆) holds.

Lemma 24**.**

There exists an indexed family $\mathcal{F}=\langle\mathcal{F}_{{p}}\rangle_{{p}\in{}^{\omega}\omega}$ of free filters on $\omega$ such that the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\times\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ (see Example 14) has no Lusin $\pi$-base.

Proof*.*

We may assume that $\{\sigma({x}):{x}\in{}^{\omega}{2}\}$ is the set of all strategies, because the cardinality of this set equals the cardinality of ${}^{\omega}{2}.$ Using Lemma 17, we can build (by transfinite recursion on ${}^{\omega}{2},$ well-ordered in the type of its cardinality) an indexed family $\langle\Gamma_{{x}}\rangle_{{x}\in{}^{\omega}{2}}$ of games such that

• (A1)

game $\Gamma_{{x}}$ follows strategy $\sigma({x})\quad$ for all ${{x}\in{}^{\omega}{2}}\quad$ and

• (A2)

$\operatorname{\mathsf{res}}^{{i}}(\Gamma_{{x}})\neq\operatorname{\mathsf{res}}^{{j}}(\Gamma_{{y}})\quad$ for all $\langle{x},{i}\rangle\neq\langle{y},{j}\rangle\in{}^{\omega}{2}\times{2}.$

Next, using Notation 23, for every ${{x}\in{}^{\omega}{2}},$ we set

[TABLE]

[TABLE]

Now we define an indexed family $\mathcal{F}=\langle\mathcal{F}_{{p}}\rangle_{{p}\in{}^{\omega}\omega}$ of free filters on $\omega.$ Since all ${r}^{{i}}_{{x}}$ are different by (A2), we may, for each pair $\langle{x},{i}\rangle\in{}^{\omega}{2}\times{2},$ take $\mathcal{F}_{{r}^{i}_{{x}}}$ to be an arbitrary free filter on $\omega$ such that $\mathcal{F}_{{r}^{i}_{{x}}}\ni{M}^{{i}}_{{x}}.$ Otherwise, if ${p}$ differs from all ${r}^{i}_{{x}},$ set $\mathcal{F}_{{p}}$ to be an arbitrary free filter on $\omega.$

We must prove that the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\times\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ has no Lusin $\pi$-base. Suppose, on the contrary, that ${L}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle\times\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle.$ Since the standard Lusin scheme $\mathbf{S}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle$ (see (F2) of Example 14), it follows by Lemma 19 that ${L}$ is candidate, so there is ${x}\in{}^{\omega}{2}$ such that $\sigma({x})=\sigma_{\scriptscriptstyle{L}}$ (see Notation 21). Since ${M}^{0}_{{x}}\in\mathcal{F}_{{r}^{0}_{{x}}}$ and ${g}^{0}_{{x}}\colon{M}^{0}_{{x}}\to\omega,$ it follows by (F1) of Example 14 that $\widehat{\mathbf{S}}_{{r}^{0}_{{x}}}({M}^{0}_{{x}},{g}^{0}_{{x}})$ is a neighbourhood of point ${r}^{0}_{{x}}$ in the space $\langle{}^{\omega}\omega,\tau_{\scriptscriptstyle\mathcal{F}}\rangle;$ similarly, $\widehat{\mathbf{S}}_{{r}^{1}_{{x}}}({M}^{1}_{{x}},{g}^{1}_{{x}})$ is a neighbourhood of ${r}^{1}_{{x}}$ in the same space. Then

[TABLE]

so, by Remark 10, there is ${a}\in{}^{{<}\omega}\omega$ such that

[TABLE]

On the other hand, game $\Gamma_{{x}}$ follows strategy $\sigma({x})=\sigma_{\scriptscriptstyle{L}}$ by (A1) and ${L}_{{a}}\ni\langle{r}^{0}_{{x}},{r}^{1}_{{x}}\rangle=\langle\operatorname{\mathsf{res}}^{0}(\Gamma_{{x}}),\operatorname{\mathsf{res}}^{1}(\Gamma_{{x}})\rangle,$ so we get a contradiction with

[TABLE]

which follows from the choice of ${M}^{0}_{{x}},$ ${M}^{1}_{{x}},$ ${g}^{0}_{{x}},$ and ${g}^{1}_{{x}}.$ ∎

It looks like we did not use Lemmas 20, 22 in the above proof; actually, these lemmas were needed to make Notations 21, 23 correct. As a corollary to Lemma 24 and (F2) of Example 14, we get the main result of the article:

Theorem 25**.**

There exists a space with a Lusin $\pi$-base whose square has no Lusin $\pi$-base. ∎

5 Proofs of Lemmas

Proof of Lemma 17.

This lemma says that for each strategy $\sigma,$ there exists a family $\langle\Gamma_{{z}}\rangle_{{z}\in{}^{\omega}{2}}$ of games that follow $\sigma$ such that $\operatorname{\mathsf{res}}^{i}(\Gamma_{{z}})\neq\operatorname{\mathsf{res}}^{{j}}(\Gamma_{{y}})$ for all $\langle{z},{i}\rangle\neq\langle{y},{j}\rangle\in{}^{\omega}{2}\times{2}.$

In this proof we use the following notation: for ${a},{b}\in{}^{{<}\omega}\omega,$ ${a}\parallel{b}$ means ${a}\nsqsubseteq{b}$ and ${a}\nsqsupseteq{b}.$ It is not hard to build $\check{c}^{i}_{{u}},\check{d}^{i}_{{u}}\in{}^{{<}\omega}\omega$ for all ${u}\in{}^{{<}\omega}{2}$ and ${i}\in{2}$ such that the following holds:

• (B1)

$\check{d}^{i}_{{u}}=\sigma^{i}(\langle\check{c}^{0}_{{u}},\check{c}^{1}_{{u}}\rangle)\quad$ for all ${u}\in{}^{{<}\omega}{2}$ and ${i}\in{2};$

• (B2)

$\check{d}^{i}_{{u}}\sqsubset\check{c}^{i}_{{u}\hskip 0.5pt^{\frown}{0}}\,$ and $\check{d}^{i}_{{u}}\sqsubset\check{c}^{i}_{{u}\hskip 0.5pt^{\frown}{1}}\quad$ for all ${u}\in{}^{{<}\omega}{2}$ and ${i}\in{2};$

• (B3)

if ${n}\in\omega,$ then

$\check{c}^{i}_{v}\parallel\check{c}^{j}_{w}\quad$ for all $\langle{v},{i}\rangle\neq\langle{w},{j}\rangle\in{}^{{n}}{2}\times{2}.$

Since $\sigma$ is a strategy, it follows from (B1)–(B2) that for each ${z}\in{}^{\omega}{2},$ there is a unique game \Gamma=\big{\langle}\langle{c}_{{n}}\rangle_{{n}\in\omega},\langle{d}_{{n}}\rangle_{{n}\in\omega}\big{\rangle} such that ${c}_{{n}}=\langle\check{c}^{0}_{{z}{\upharpoonright}{n}},\check{c}^{1}_{{z}{\upharpoonright}{n}}\rangle$ and ${d}_{{n}}=\langle\check{d}^{0}_{{z}{\upharpoonright}{n}},\check{d}^{1}_{{z}{\upharpoonright}{n}}\rangle$ for all ${n}\in\omega.$ We denote this game by $\Gamma_{{z}}.$ Each game $\Gamma_{{z}}$ follows strategy $\sigma$ by (B1). Now, suppose that $\langle{z},{i}\rangle\neq\langle{y},{j}\rangle\in{}^{\omega}{2}\times{2}.$ There is ${n}\in\omega$ such that $\langle{z}{\upharpoonright}{n},{i}\rangle\neq\langle{y}{\upharpoonright}{n},{j}\rangle,$ so $\check{c}^{{i}}_{{z}{\upharpoonright}{n}}\parallel\check{c}^{{j}}_{{y}{\upharpoonright}{n}}$ by (B3). Then $\operatorname{\mathsf{res}}^{i}(\Gamma_{{z}})\neq\operatorname{\mathsf{res}}^{{j}}(\Gamma_{{y}})$ because $\operatorname{\mathsf{res}}^{i}(\Gamma_{{z}})\sqsupseteq\check{c}^{{i}}_{{z}{\upharpoonright}{n}}$ and $\operatorname{\mathsf{res}}^{{j}}(\Gamma_{{y}})\sqsupseteq\check{c}^{{j}}_{{y}{\upharpoonright}{n}}.$ ∎

Proof of Lemma 19.

This lemma says that if $\mathbf{S}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle$ and $\,{L}\!$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle\times\langle{}^{\omega}\omega,\tau\rangle,$ then ${L}$ is a strict Lusin scheme on the set ${}^{\omega}\omega\times{}^{\omega}\omega,$ $\{{L}_{{b}}:{b}\in{}^{{<}\omega}\omega\}$ is a $\pi$-net for the space ${\omega^{\omega}}\times{\omega^{\omega}},$ and each ${L}_{{b}}$ has nonempty interior in ${\omega^{\omega}}\times{\omega^{\omega}}.$

Recall that ${\omega^{\omega}}=\langle{}^{\omega}\omega,\tau_{\mathsf{prod}}\rangle.$ Since $\mathbf{S}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle,$ it follows that each $\mathbf{S}_{{a}}$ is open in $\langle{}^{\omega}\omega,\tau\rangle,$ so

• (C)

$\tau\supseteq\tau_{\mathsf{prod}}$

because $\{\mathbf{S}_{{a}}:{a}\in{}^{{<}\omega}\omega\}$ is a base for ${\omega^{\omega}}$ by (1) Remark 6. Since ${L}$ is a Lusin $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle\times\langle{}^{\omega}\omega,\tau\rangle,$ ${L}$ is a strict Lusin scheme on ${}^{\omega}\omega\times{}^{\omega}\omega.$ The family $\{{L}_{{b}}:{b}\in{}^{{<}\omega}\omega\}$ is a $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle\times\langle{}^{\omega}\omega,\tau\rangle$ by Remark 4, so $\{{L}_{{b}}:{b}\in{}^{{<}\omega}\omega\}$ is a $\pi$-net for ${\omega^{\omega}}\times{\omega^{\omega}}$ by (C). Now, for each ${L}_{{b}}$ there are nonempty ${U}^{{0}},{U}^{{1}}\in\tau$ such that ${L}_{{b}}\supseteq{U}^{{0}}\times{U}^{{1}}$ because ${L}_{{b}}$ is nonempty and open in $\langle{}^{\omega}\omega,\tau\rangle\times\langle{}^{\omega}\omega,\tau\rangle.$ Then there are ${a}^{{0}},{a}^{{1}}\in{}^{{<}\omega}\omega$ such that ${U}^{{0}}\supseteq\mathbf{S}_{{a}^{{0}}}$ and ${U}^{{1}}\supseteq\mathbf{S}_{{a}^{{1}}}$ because $\{\mathbf{S}_{{a}}:{a}\in{}^{{<}\omega}\omega\}$ is a $\pi$-base for $\langle{}^{\omega}\omega,\tau\rangle$ by Remark 4. Therefore ${L}_{{b}}\supseteq\mathbf{S}_{{a}^{{0}}}\times\mathbf{S}_{{a}^{{1}}},$ so the interior of ${L}_{{b}}$ in ${\omega^{\omega}}\times{\omega^{\omega}}$ contains $\mathbf{S}_{{a}^{{0}}}\times\mathbf{S}_{{a}^{{1}}},$ which in not empty. ∎

Proof of Lemma 20.

This lemma says that for each candidate ${L},$ there exists a strategy $\sigma$ that possesses the following property:

• (★)

For each ${c}=\langle{c}^{0},{c}^{1}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$

there is ${b}\in{}^{{<}\omega}\omega$ such that

• ➢ 

$\mathbf{S}_{{c}^{0}}\times\mathbf{S}_{{c}^{1}}\>\supseteq\>{L}_{{b}}\>\supseteq\>\mathbf{S}_{\sigma^{0}({c})}\times\mathbf{S}_{\sigma^{1}({c})}\quad$ and

• ➢ 

\operatorname{\mathsf{shoot}}_{{L}}({b}{\upharpoonright}{k})\>\not\mathop{\dot{\in}}\>\big{(}\mathbf{S}_{\sigma^{0}({c})}{\uparrow}\times\mathbf{S}_{\sigma^{1}({c})}\big{)}\cup\big{(}\mathbf{S}_{\sigma^{0}({c})}\times(\mathbf{S}_{\sigma^{1}({c})}{\uparrow})\big{)}\quad for all ${k}<\operatorname{\mathsf{length}}({b}).$

Suppose that ${L}$ is a candidate; we must find a pair $\sigma=\langle\sigma^{0},\sigma^{1}\rangle$ of functions $\sigma^{0},\sigma^{1}\colon{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega\to{}^{{<}\omega}\omega$ such that (★) holds. Let ${c}=\langle{c}^{0},{c}^{1}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega.$ The set $\mathbf{S}_{{c}^{0}}\times\mathbf{S}_{{c}^{1}}$ is nonempty and open in the space ${\omega^{\omega}}\times{\omega^{\omega}},$ so there is ${b}_{{c}}\in{}^{{<}\omega}\omega$ such that ${L}_{{b}_{{c}}}\subseteq\mathbf{S}_{{c}^{0}}\times\mathbf{S}_{{c}^{1}}$ because $\{{L}_{{b}}:{b}\in{}^{{<}\omega}\omega\}$ is a $\pi$-net for ${\omega^{\omega}}\times{\omega^{\omega}}.$ Then there are ${a}_{{c}}^{0},{a}_{{c}}^{1}\in{}^{{<}\omega}\omega$ such that $\mathbf{S}_{{a}_{{c}}^{0}}\times\mathbf{S}_{{a}_{{c}}^{1}}\subseteq{L}_{{b}_{{c}}}$ because ${L}_{{b}_{{c}}}$ has nonempty interior in ${\omega^{\omega}}\times{\omega^{\omega}}$ and $\{\mathbf{S}_{{a}}:{a}\in{}^{{<}\omega}\omega\}$ is a base for ${\omega^{\omega}}.$ For each ${n}\in\omega,$ put

• ✎ 

{R}_{{c},{n}}\>\coloneq\>\big{(}\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}}{\uparrow}\times\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}}\big{)}\>\cup\>\big{(}\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}}\times(\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}}{\uparrow})\big{)}\>\subseteq\>{}^{\omega}\omega\times{}^{\omega}\omega\,.

Then, for all ${n}\neq{m}\in\omega,$

• (D)

${R}_{{c},{n}}\cap{R}_{{c},{m}}=\varnothing.$

To prove (D), suppose on the contrary that ${n}<{m}$ and there is ${p}=\langle{p}^{0},{p}^{1}\rangle\in{R}_{{c},{n}}\cap{R}_{{c},{m}}.$ We may assume without loss of generality that ${p}\in\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}}{\uparrow}\times\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}},$ so that ${p}^{0}\in\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{n}}{\uparrow}$ and ${p}^{1}\in\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}}.$ Since $\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}}\cap\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}}=\varnothing$ (because ${n}\neq{m}$), we have ${p}^{1}\nin\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}},$ so ${p}\nin\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{m}}{\uparrow}\times\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}}.$ Then ${p}\in\mathbf{S}_{{a}_{{c}}^{{0}{\frown}}{m}}\times(\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}}{\uparrow})$ because ${p}\in{R}_{{c},{m}},$ therefore ${p}^{1}\in\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}}{\uparrow}.$ This contradicts Remark 8, which says that $\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{n}}\cap(\mathbf{S}_{{a}_{{c}}^{{1}{\frown}}{m}}{\uparrow})=\varnothing,$ so (D) is proved.

Now, for each ${k}<\operatorname{\mathsf{length}}({b}_{{c}}),$ there is at most one ${n}\in\omega$ such that $\operatorname{\mathsf{shoot}}_{{L}}({b}_{{c}}{\upharpoonright}{k})\mathop{\dot{\in}}{R}_{{c},{n}}$ — this follows from (D) and (4) of Remark 11. Then there is some ${n}_{{c}}\in\omega$ such that, for all ${k}<\operatorname{\mathsf{length}}({b}_{{c}}),$

[TABLE]

Also we have

[TABLE]

so $\sigma^{0}({c})\coloneq{a}_{{c}}^{{0}{\frown}}{n}_{{c}}$ and $\sigma^{1}({c})\coloneq{a}_{{c}}^{{1}{\frown}}{n}_{{c}}$ define a strategy $\sigma$ that we search. ∎

Proof of Lemma 22.

This lemma says that for each game $\Gamma,$ there are sets ${M}^{0},{M}^{1}\in[\omega]^{\omega}$ and functions ${g}^{0}\colon{M}^{0}\to\omega,$ ${g}^{1}\colon{M}^{1}\to\omega$ that possess the following property:

• (◆)

For each candidate ${L}$ and each ${a}\in{}^{{<}\omega}\omega,$

• ➢ 

if game $\Gamma$ follows strategy $\sigma_{\scriptscriptstyle{L}}$ and $\,{L}_{{a}}\ni\langle\operatorname{\mathsf{res}}^{0}(\Gamma),\operatorname{\mathsf{res}}^{1}(\Gamma)\rangle,$

• ➢ 

then $\ \operatorname{\mathsf{shoot}}_{{L}}({a})\>\not\mathop{\dot{\in}}\>\widehat{\mathbf{S}}_{\operatorname{\mathsf{res}}^{0}(\Gamma)}({M}^{0},{g}^{0})\times\widehat{\mathbf{S}}_{\operatorname{\mathsf{res}}^{1}(\Gamma)}({M}^{1},{g}^{1}).$

Suppose that \Gamma=\big{\langle}\langle{c}_{{n}}\rangle_{{n}\in\omega},\langle{d}_{{n}}\rangle_{{n}\in\omega}\big{\rangle} is a game. Put ${r}^{0}\coloneq\operatorname{\mathsf{res}}^{0}(\Gamma)\in{}^{\omega}\omega$ and ${r}^{1}\coloneq\operatorname{\mathsf{res}}^{1}(\Gamma)\in{}^{\omega}\omega.$ Here are sets and functions that we must find:

• ✎ 

${M}^{0}\>\coloneq\>\{\operatorname{\mathsf{length}}({c}^{0}_{2{k}+1})-1:{k}\in\omega\};$

• ✎ 

${M}^{1}\>\coloneq\>\{\operatorname{\mathsf{length}}({c}^{1}_{2{k}+2})-1:{k}\in\omega\};$

• ✎ 

${g}^{0}\,$ is the function from ${M}^{0}$ to $\omega$ such that ${g}^{0}({m})={r}^{0}({m})+1$ for all ${m}\in{M}^{0};$

• ✎ 

${g}^{1}\,$ is the function from ${M}^{1}$ to $\omega$ such that ${g}^{1}({m})={r}^{1}({m})+1$ for all ${m}\in{M}^{1}.$

We must show that (◆) holds. Suppose that ${L}$ is a candidate, ${a}\in{}^{{<}\omega}\omega,$ game $\Gamma$ follows strategy $\sigma_{\scriptscriptstyle{L}},$ and ${L}_{{a}}\ni\langle{r}^{0},{r}^{1}\rangle;$ we must prove that $\operatorname{\mathsf{shoot}}_{{L}}({a})\>\not\mathop{\dot{\in}}\>\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})\times\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1}).$ Suppose on the contrary that

• (E1)

$\operatorname{\mathsf{shoot}}_{{L}}({a})\>\mathop{\dot{\in}}\>\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})\times\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1}).$

Let ${b}_{{-}1}\coloneq\langle\rangle\in{}^{{<}\omega}\omega.$ Recall that ${d}^{{i}}_{{n}}=\sigma_{\scriptscriptstyle{L}}^{i}({c}_{{n}})$ for all ${n}\in\omega$ and ${i}\in{2}$ because game $\Gamma$ follows strategy $\sigma_{\scriptscriptstyle{L}}.$ If ${n}\in\omega,$ then, by definition of $\sigma_{\scriptscriptstyle{L}}$ (see Notaiton 21), for ${c}_{{n}}=\langle{c}^{0}_{{n}},{c}^{1}_{{n}}\rangle\in{}^{{<}\omega}\omega\times{}^{{<}\omega}\omega,$ there is ${b}_{{n}}\in{}^{{<}\omega}\omega$ such that

• (E2)

$\mathbf{S}_{{c}^{0}_{{n}}}\times\mathbf{S}_{{c}^{1}_{{n}}}\>\supseteq\>{L}_{{b}_{{n}}}\>\supseteq\>\mathbf{S}_{{d}^{0}_{{n}}}\times\mathbf{S}_{{d}^{1}_{{n}}}\quad$ and

• (E3)

\operatorname{\mathsf{shoot}}_{{L}}({b}_{{n}}{\upharpoonright}{k})\>\not\mathop{\dot{\in}}\>\big{(}\mathbf{S}_{{d}^{0}_{{n}}}{\uparrow}\times\mathbf{S}_{{d}^{1}_{{n}}}\big{)}\cup\big{(}\mathbf{S}_{{d}^{0}_{{n}}}\times(\mathbf{S}_{{d}^{1}_{{n}}}{\uparrow})\big{)}\quad for all ${k}<\operatorname{\mathsf{length}}({b}_{{n}}).$

We have ${L}_{{b}_{{n}}}\supset{L}_{{b}_{{n}+1}}$ and ${L}_{{b}_{{n}}}\ni\langle{r}^{0},{r}^{1}\rangle$ for all ${n}\in\omega$ by (E2) and (1) of Remark 16. Then ${b}_{{n}}\sqsubset{b}_{{n}+1}$ for all ${n}\in\omega$ by (2) of Remark 3. Also we have ${L}_{{a}}\ni\langle{r}^{0},{r}^{1}\rangle$ by the choice of ${a},$ so ${L}_{{a}}\cap{L}_{{b}_{{n}}}\neq\varnothing,$ and hence ${a}\sqsupseteq{b}_{{n}}$ or ${a}\sqsubseteq{b}_{{n}}$ for all ${n}\in\omega$ by (3) of Remark 3. Therefore there is $\dot{n}\in\omega$ such that ${b}_{\dot{n}-1}\sqsubseteq{a}\sqsubset{b}_{\dot{n}}.$

Let ${c}^{0}_{{-}1}\coloneq{c}^{1}_{{-}1}\coloneq\langle\rangle\in{}^{{<}\omega}\omega.$ We have $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}{L}_{{a}}$ by (1) of Remark 11, ${L}_{{a}}\subseteq{L}_{{b}_{\dot{n}-1}}$ by the choice of $\dot{n}$ and (2) of Remark 3, and ${L}_{{b}_{\dot{n}-1}}\subseteq\,\mathbf{S}_{{c}^{0}_{\dot{n}-1}}\!\!\times\mathbf{S}_{{c}^{1}_{\dot{n}-1}}$ by (E2) (and because ${L}_{{b}_{-1}}={}^{\omega}\omega\times{}^{\omega}\omega=\mathbf{S}_{{c}^{0}_{-1}}\times\mathbf{S}_{{c}^{1}_{-1}}$ by (L2) of Definition 2), so $\operatorname{\mathsf{shoot}}_{{L}}({a})\mathop{\dot{\in}}\mathbf{S}_{{c}^{0}_{\dot{n}-1}}\!\!\times\mathbf{S}_{{c}^{1}_{\dot{n}-1}}$ by (2) of Remark 11. Therefore

[TABLE]

by (E1) and (3) of Remark 11, and then using (2) of Remark 11 we get

[TABLE]

Let

• ✎ 

${T}^{0}\>\coloneq\>\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})\cap\mathbf{S}_{{c}^{0}_{\dot{n}-1}}$ and

• ✎ 

${T}^{1}\>\coloneq\>\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1})\cap\mathbf{S}_{{c}^{1}_{\dot{n}-1}},$

so that

• (E4)

$\operatorname{\mathsf{shoot}}_{{L}}({a})\>\mathop{\dot{\in}}\>{T}^{0}\times{T}^{1}.$

Next we show the following:

• ($\text{E5}^{0}$)

if $\dot{n}$ is even, then ${T}^{0}\subseteq\mathbf{S}_{{d}^{0}_{\dot{n}}}\quad$ and

• ($\text{E5}^{1}$)

if $\dot{n}$ is odd, then ${T}^{1}\subseteq\mathbf{S}_{{d}^{1}_{\dot{n}}}.$

To prove ($\text{E5}^{0}$), assume that $\dot{n}$ is even and ${q}\in{T}^{0}.$ Recall that $\widehat{\mathbf{S}}_{{r}^{0}}({M}^{0},{g}^{0})=\{{r}^{0}\}\cup\bigcup_{{m}\in{M}^{0}}\widetilde{\mathbf{S}}^{{g}^{0}({m})}_{{p}^{0}{\upharpoonright}{m}}.$ If ${q}={r}^{0},$ then ${q}\in\mathbf{S}_{{d}^{0}_{\dot{n}}}$ by (1) of Remark 16. If ${q}\neq{r}^{0},$ then there is $\dot{m}\in{M}^{0}$ such that ${q}\in\widetilde{\mathbf{S}}^{{g}^{0}(\dot{m})}_{{r}^{0}{\upharpoonright}\dot{m}},$ so

• ($\text{E6}^{0}$)

${q}\in\widetilde{\mathbf{S}}^{{r}^{0}(\dot{m})+1}_{{r}^{0}{\upharpoonright}\dot{m}}$

by the choice of ${g}^{0}.$ Since $\dot{m}\in{M}^{0},$ there is $\dot{k}\in\omega$ such that

• ($\text{E7}^{0}$)

$\dot{m}=\operatorname{\mathsf{length}}({c}^{0}_{2\dot{k}+1})-1.$

We have ${q}\nin{\mathbf{S}}_{{r}^{0}{\upharpoonright}(\dot{m}+1)}$ by ($\text{E6}^{0}$) and (1) of Remark 13, so ${q}\nin{\mathbf{S}}_{{r}^{0}{\upharpoonright}\operatorname{\mathsf{length}}({c}^{0}_{2\dot{k}+1})}={\mathbf{S}}_{{c}^{0}_{2\dot{k}+1}}$ by ($\text{E7}^{0}$) and (2) of Remark 16. On the other hand, ${q}\in{\mathbf{S}}_{{c}^{0}_{\dot{n}-1}}$ by definition of ${T}^{0},$ therefore ${\mathbf{S}}_{{c}^{0}_{2\dot{k}+1}}\nsupseteq{\mathbf{S}}_{{c}^{0}_{\dot{n}-1}},$ hence ${c}^{0}_{2\dot{k}+1}\nsqsubseteq{c}^{0}_{\dot{n}-1}$ by (2) of Remark 3, so $2\dot{k}+1\nleqslant\dot{n}-1$ by (5) of Remark 16, that is, $2\dot{k}+1>\dot{n}-1.$ Then $2\dot{k}\geqslant\dot{n}-1$ and also $2\dot{k}\neq\dot{n}-1$ because $\dot{n}$ is even, so we get

• ($\text{E8}^{0}$)

$2\dot{k}\geqslant\dot{n}.$

Now, ${q}\in\widetilde{\mathbf{S}}^{{r}^{0}(\dot{m})+1}_{{r}^{0}{\upharpoonright}\dot{m}}$ by ($\text{E6}^{0}$), $\widetilde{\mathbf{S}}^{{r}^{0}(\dot{m})+1}_{{r}^{0}{\upharpoonright}\dot{m}}\subseteq\widetilde{\mathbf{S}}_{{r}^{0}{\upharpoonright}\dot{m}}$ by (1) of Remark 13, $\widetilde{\mathbf{S}}_{{r}^{0}{\upharpoonright}\dot{m}}={\mathbf{S}}_{{r}^{0}{\upharpoonright}(\operatorname{\mathsf{length}}({c}^{0}_{2\dot{k}+1})-1)}$ by ($\text{E7}^{0}$), ${\mathbf{S}}_{{r}^{0}{\upharpoonright}(\operatorname{\mathsf{length}}({c}^{0}_{2\dot{k}+1})-1)}\subseteq{\mathbf{S}}_{{r}^{0}{\upharpoonright}\operatorname{\mathsf{length}}({d}^{0}_{2\dot{k}})}$ by (4) of Remark 16 and (2) of Remark 3, ${\mathbf{S}}_{{r}^{0}{\upharpoonright}\operatorname{\mathsf{length}}({d}^{0}_{2\dot{k}})}={\mathbf{S}}_{{d}^{0}_{2\dot{k}}}$ by (3) of Remark 16, and ${\mathbf{S}}_{{d}^{0}_{2\dot{k}}}\subseteq{\mathbf{S}}_{{d}^{0}_{\dot{n}}}$ by ($\text{E8}^{0}$), (6) of Remark 16, and (2) of Remark 3. Therefore ${q}\in\mathbf{S}_{{d}^{0}_{\dot{n}}},$ which proves ($\text{E5}^{0}$).

The proof of ($\text{E5}^{1}$) is similar: Assume that $\dot{n}$ is odd and ${q}\in{T}^{1}.$ $\widehat{\mathbf{S}}_{{r}^{1}}({M}^{1},{g}^{1})=\{{r}^{1}\}\cup\bigcup_{{m}\in{M}^{1}}\widetilde{\mathbf{S}}^{{g}^{1}({m})}_{{p}^{1}{\upharpoonright}{m}}.$ If ${q}={r}^{1},$ then ${q}\in\mathbf{S}_{{d}^{1}_{\dot{n}}}.$ If ${q}\neq{r}^{1},$ then there is $\dot{m}\in{M}^{1}$ such that ${q}\in\widetilde{\mathbf{S}}^{{g}^{1}(\dot{m})}_{{r}^{1}{\upharpoonright}\dot{m}},$ so

• ($\text{E6}^{1}$)

${q}\in\widetilde{\mathbf{S}}^{{r}^{1}(\dot{m})+1}_{{r}^{1}{\upharpoonright}\dot{m}}\>.$

Since $\dot{m}\in{M}^{1},$ there is $\dot{k}\in\omega$ such that

• ($\text{E7}^{1}$)

$\dot{m}=\operatorname{\mathsf{length}}({c}^{1}_{2\dot{k}+2})-1.$

We have ${q}\nin{\mathbf{S}}_{{r}^{1}{\upharpoonright}(\dot{m}+1)}={\mathbf{S}}_{{r}^{1}{\upharpoonright}\operatorname{\mathsf{length}}({c}^{1}_{2\dot{k}+2})}={\mathbf{S}}_{{c}^{1}_{2\dot{k}+2}}.$ On the other hand, ${q}\in{\mathbf{S}}_{{c}^{1}_{\dot{n}-1}},$ therefore ${\mathbf{S}}_{{c}^{1}_{2\dot{k}+2}}\nsupseteq{\mathbf{S}}_{{c}^{1}_{\dot{n}-1}},$ hence ${c}^{1}_{2\dot{k}+2}\nsqsubseteq{c}^{1}_{\dot{n}-1},$ so $2\dot{k}+2\nleqslant\dot{n}-1,$ that is, $2\dot{k}+2>\dot{n}-1.$ Then $2\dot{k}+1\geqslant\dot{n}-1$ and also $2\dot{k}+1\neq\dot{n}-1$ because $\dot{n}$ is odd, so we get

• ($\text{E8}^{1}$)

$2\dot{k}+1\geqslant\dot{n}.$

By the same argument as above, we can write

[TABLE]

This proves ($\text{E5}^{1}$).

Now, ($\text{E5}^{0}$) and ($\text{E5}^{1}$) imply

• (E5)

${T}^{0}\subseteq\mathbf{S}_{{d}^{0}_{\dot{n}}}\quad\mathsf{or}\quad{T}^{1}\subseteq\mathbf{S}_{{d}^{1}_{\dot{n}}}.$

Also, for each ${i}\in{2},$ we have ${T}^{{i}}\subseteq\widehat{\mathbf{S}}_{{r}^{{i}}}({M}^{{i}},{g}^{{i}})$ by definition of ${T}^{{i}},$ $\widehat{\mathbf{S}}_{{r}^{i}}({M}^{i},{g}^{i})\subseteq{r}^{{i}}{\downfilledspoon}$ by (2) of Remark 13, and ${r}^{{{i}}}{\downfilledspoon}\,\subseteq\,\mathbf{S}_{{d}^{{{i}}}_{\dot{n}}}{\uparrow}$ (because ${r}^{{{i}}}\in\mathbf{S}_{{d}^{{{i}}}_{\dot{n}}}$ by (1) of Remark 16), so

• (E9)

${T}^{0}\,\subseteq\,\mathbf{S}_{{d}^{0}_{\dot{n}}}{\uparrow}\quad\mathsf{and}\quad{T}^{1}\,\subseteq\,\mathbf{S}_{{d}^{1}_{\dot{n}}}{\uparrow}.$

It follows from (E5) and (E9) that ${T}^{0}\times{T}^{1}\subseteq\mathbf{S}_{{d}^{0}_{\dot{n}}}\times(\mathbf{S}_{{d}^{1}_{\dot{n}}}{\uparrow})$ or ${T}^{0}\times{T}^{1}\subseteq\mathbf{S}_{{d}^{0}_{\dot{n}}}{\uparrow}\times\mathbf{S}_{{d}^{1}_{\dot{n}}},$ so

[TABLE]

Then, using (E4) and (2) of Remark 11, we get

[TABLE]

But also, since ${a}\sqsubset{b}_{\dot{n}}$ by the choice of $\dot{n},$ there is ${k}<\operatorname{\mathsf{length}}({b}_{\dot{n}})$ such that ${a}={b}_{\dot{n}}{\upharpoonright}{k}.$ Then $\operatorname{\mathsf{shoot}}_{{L}}({a})=\operatorname{\mathsf{shoot}}_{{L}}({b}_{\dot{n}}{\upharpoonright}{k})$ and we have a contradiction with (E3), which says

[TABLE]

∎