Selection properties of the split interval and the Continuum Hypothesis

TL;DR

This paper investigates the selection properties of upper semi-continuous multivalued maps, proving the existence of measurable selections in some cases and showing the dependence on the Continuum Hypothesis for others, highlighting limitations in extending known results.

Contribution

It establishes the existence of $F_\sigma$-measurable selections for usco multimap into GO-spaces and demonstrates the CH-dependence of Borel selections for the split interval example.

Findings

Every usco multimap from a metrizable separable space to a GO-space has an $F_\sigma$-measurable selection.

The existence of Borel selections for a specific usco multimap on the split interval depends on the Continuum Hypothesis.

Results on Borel selections into fragmentable compact spaces cannot be generalized to all compact spaces.

Abstract

We prove that every usco multimap $\Phi:X\to Y$ from a metrizable separable space $X$ to a GO-space $Y$ has an $F_\sigma$-measurable selection. On the other hand, for the split interval $\ddot{\mathbb I}$ and the projection $P:\ddot{\mathbb I}^2\to{\mathbb I}^2$ of its square onto the unit square ${\mathbb I}^2$, the usco multimap $P^{-1}:{\mathbb I}^2\multimap\ddot{\mathbb I}^2$ has a Borel ($F_\sigma$-measurable) selection if and only if the Continuum Hypothesis holds. This CH-example shows that know results on Borel selections of usco maps into fragmentable compact spaces cannot be extended to a wider class of compact spaces.

Equations

$$P:{\ddot{\mathbb{I}}}^2\to {\mathbb{I}}^2,P:⟨x,y⟩\mapsto ⟨p(x),p(y)⟩,$$

$$Z_{ij}=\{z\in {\mathbb{I}}^2:s(z)\in {\ddot{\mathbb{I}}}_i\times {\ddot{\mathbb{I}}}_j\}$$

$$p\big{(}[0_{0},x_{0})\big{)}=[0,x),\;\;p\big{(}[0_{0},x_{1})\big{)}=[0,x],\;\;p\big{(}(x_{0},1_{1}]\big{)}=[x,1],\;\;p\big{(}(x_{1},1_{1}]\big{)}=(x,1].$$

$${\mathrm{L}}_a=\{⟨x,y⟩\in {\mathbb{R}}^2:x+y=a\}\text{mbox}and{\Gamma }^a=\{⟨x,y⟩\in {\mathbb{R}}^2:y-x=a\}$$

$$U=\bigcup _{a\in \Omega }{\mathrm{L}}_a\cap (Z_{00}\cup Z_{11})$$

$$\mathrm{L}$$

$$\Gamma$$

$$s(\langle x,y\rangle)=\begin{cases}\langle x_{1},y_{1}\rangle&\mbox{if $\langle x,y\rangle\in\mathrm{L}$},\\ \langle x_{1},y_{0}\rangle&\mbox{if $\langle x,y\rangle\in\Gamma$},\end{cases}$$

$$\big{[}x,a(x,y)\big{)}\times\big{[}y,b(x,y)\big{)}=s^{-1}\big{[}[x_{1},a(x,y)_{0})\times[y_{1},b(x,y)_{0})\big{]}\subset s^{-1}[V].$$

$$\big{[}x,a(x,y)\big{)}\times\big{(}b(x,y),y\big{]}=s^{-1}\big{[}[x_{1},a(x,y)_{0})\times(b(x,y)_{1},y_{0}]\big{)}\subset s^{-1}[V].$$

$$s^{-1}[V]=\Big{(}\bigcup_{\langle x,y\rangle\in\mathrm{L}_{V}}\big{[}x,a(x,y)\big{)}\times\big{[}y,b(x,y)\big{)}\Big{)}\cup\Big{(}\bigcup_{\langle x,y\rangle\in\Gamma_{V}}\big{[}x,a(x,y)\big{)}\times\big{(}b(x,y),y\big{]}\Big{)}.$$

$$\bigcup_{\langle x,y\rangle\in\mathrm{L}_{V}}\big{[}x,a(x,y))\times[y,b(x,y)\big{)}=\bigcup_{\langle x,y\rangle\in\mathrm{L}^{\prime}}\big{[}x,a(x,y)\big{)}\times\big{[}y,b(x,y)\big{)}$$

$$\bigcup_{\langle x,y\rangle\in\Gamma_{V}}\big{[}x,a(x,y)\big{)}\times\big{(}b(x,y),y\big{]}=\bigcup_{\langle x,y\rangle\in\Gamma^{\prime}}\big{[}x,a(x,y)\big{)}\times\big{(}b(x,y),y\big{]}.$$

$${\mathrm{L}}_{r,q}=\{⟨x,y⟩\in {\mathrm{L}}_V:a(x,y)=r,b(x,y)=q\}$$

$$\bigcup _{⟨x,y⟩\in {\mathrm{L}}_{r,q}^{\prime }}[x,r)\times [y,q)=\bigcup _{⟨x,y⟩\in {\mathrm{L}}_{r,q}}[x,r)\times [y,q).$$

$$\underline{y}(r^{\prime }):=\inf \{y:⟨x,y⟩\in {\mathrm{L}}_{r,q},x<r^{\prime }\}\text{mbox}and\underline{x}(q^{\prime }):=\inf \{x:⟨x,y⟩\in {\mathrm{L}}_{r,q},y<q^{\prime }\}.$$

$$\underline{y}(r^{\prime })=\inf \{y:⟨x,y⟩\in {\mathrm{L}}_{r,q}^{r^{\prime },0},x<r^{\prime }\}\text{mbox}and\underline{x}(q^{\prime })=\inf \{x:⟨x,y⟩\in {\mathrm{L}}_{r,q}^{0,q^{\prime }},y<q^{\prime }\}$$

$$\underline{y}(r^{\prime })=\min \{y:⟨x,y⟩\in {\mathrm{L}}_{r,q}^{r^{\prime },0},x<r^{\prime }\}\text{mbox}if\underline{y}(r^{\prime })=\min \{y:⟨x,y⟩\in {\mathrm{L}}_{r,q},x<r^{\prime }\}.$$

$$\underline{x}(q^{\prime })=\min \{x:⟨x,y⟩\in {\mathrm{L}}_{r,q}^{0,q^{\prime }},y<q^{\prime }\}\text{mbox}if\underline{x}(q^{\prime })=\min \{x:⟨x,y⟩\in {\mathrm{L}}_{r,q},y<q^{\prime }\}.$$

$${\mathrm{L}}_{r,q}^{\prime \prime }:=\bigcup \{{\mathrm{L}}_{r,q}^{r^{\prime },0}\cup {\mathrm{L}}_{r,1}^{0,q^{\prime }}:r^{\prime },q^{\prime }\in \mathbb{Q},r^{\prime }<r,q^{\prime }<q\}$$

$$D_{r,q}=\Big{(}\bigcup_{\langle x,y\rangle\in\mathrm{L}_{r,q}}[x,r)\times[y,q)\Big{)}\setminus\Big{(}\bigcup_{\langle x,y\rangle\in\mathrm{L}_{r,q}^{\prime\prime}}[x,r)\times[y,q)\Big{)}$$

$$\langle x,y^{\prime}\rangle\in\big{(}[x,r)\times[y,q)\big{)}\setminus\{\langle x,y\rangle\}\subset\bigcup_{\langle x^{\prime\prime},y^{\prime\prime}\rangle\in L_{r,q}^{\prime\prime}}[x^{\prime\prime},r)\times[y^{\prime\prime},q)$$

$$\langle x^{\prime},y^{\prime}\rangle\in\big{(}[x,r)\times[y,q)\big{)}\setminus\{\langle x,y\rangle\}\subset\bigcup_{\langle x^{\prime\prime},y^{\prime\prime}\rangle\in L_{r,q}^{\prime\prime}}[x^{\prime\prime},r)\times[y^{\prime\prime},q)$$

$$D_{r,q}^{\prime }=\bigcup \{D_{r,q}\cap f_{\beta }:\beta \le \alpha ,f_{\beta }\text{mbox}isstrictlyincreasing\}$$

$$\displaystyle\bigcup\limits_{\langle x,y\rangle\in L_{r,q}}\big{(}[x,r)\times[y,q)\big{)}\subset\bigcup\limits_{\langle x,y\rangle\in\mathrm{L}_{r,q}^{\prime\prime}}[x,r)\times[y,q).$$

$$\bigcup_{\langle x,y\rangle\in\mathrm{L}^{\prime}}\big{(}[x,a(x,y))\times[y,b(x,y)\big{)}=\bigcup_{\langle x,y\rangle\in\mathrm{L}_{V}}\big{(}[x,a(x,y))\times[y,b(x,y)\big{)}.$$

$$\bigcup_{\langle x,y\rangle\in\mathrm{L}^{\prime}_{r,q}}\big{(}[x,a(x,y))\times[y,b(a,y))\big{)}=\bigcup_{\langle x,y\rangle\in\mathrm{L}^{\prime}_{r,q}}\big{(}[x,r)\times[y,q)\big{)}=\\ =\bigcup\limits_{\langle x,y\rangle\in\mathrm{L}_{r,q}}\big{(}[x,r)\times[y,q)\big{)}=\bigcup\limits_{\langle x,y\rangle\in\mathrm{L}_{r,q}}\big{(}[x,a(x,y))\times[y,b(x,y))\big{)}.$$

$$\bigcup_{\langle x,y\rangle\in\Gamma^{\prime}}\big{(}[x,a(x,y))\times(b(x,y),y]\big{)}=\bigcup_{\langle x,y\rangle\in\Gamma_{V}}\big{(}[x,a(x,y))\times(b(x,y),y]\big{)}.$$

Full text

1. Introduction

By a multimap $\Phi:X\multimap Y$ between topological spaces $X,Y$ we understand any subset $\Phi\subset X\times Y$, which can be thought as a function assigning to every point $x\in X$ the subset $\Phi(x):=\{y\in Y:\langle x,y\rangle\in\Phi\}$ of $Y$. For a subset $A\subset X$ we put $\Phi[A]=\bigcup_{x\in A}\Phi(x)$. Each function $f:X\to Y$ can be thought as a single-valued multimap $\{\langle x,f(x)\rangle:x\in X\}\subset X\times Y$.

For a multimap $\Phi:X\multimap Y$, its inverse multimap $\Phi^{-1}:Y\multimap X$ is defined by $\Phi^{-1}:=\{\langle y,x\rangle:\langle x,y\rangle\in\Phi\}$.

A multimap $\Phi:X\multimap Y$ is called

• •

lower semicontinuous if for any open set $U\subset Y$ the set $\Phi^{-1}[U]$ is open in $X$;

• •

upper semicontinuous if for any closed set $F\subset Y$ the set $\Phi^{-1}[F]$ is closed in $X$;

• •

Borel-measurable if for any Borel set $B\subset Y$ the set $\Phi^{-1}[B]$ is Borel in $X$;

• •

compact-valued if for every $x\in X$ the subspace $\Phi(x)$ of $Y$ is compact and non-empty;

• •

usco if $\Phi$ is upper semicontinuous and compact-valued.

It is well-known that for any surjective continuous function $f:X\to Y$ between compact Hausdorff spaces, the inverse multimap $f^{-1}:Y\multimap X$ is usco.

Let $\Phi:X\multimap Y$ be a multimap between topological spaces. A function $f:X\to Y$ is called a selection of $\Phi$ if $f(x)\in\Phi(x)$ for every $x\in X$. The Axiom of Choice ensures that every multimap $\Phi:X\multimap Y$ with non-empty values has a selection. The problem is to find selections possessing some additional properties like the continuity or measurability.

One of classical results in this direction is the following theorem of Kuratowski and Ryll-Nardzewski [11] (see also [15, §5.2] or [13, 6.12]).

Theorem 1**.**

Let $X,Y$ be Polish spaces. Any Borel-measurable multimap $\Phi:X\multimap Y$ with non-empty values has a Borel-measurable selection.

We recall that a function $f:X\to Y$ between topological spaces is called Borel-measurable (resp. $F_{\sigma}$-measurable) if for every open set $U\subset Y$ the preimage $f^{-1}[U]$ is Borel (or type $F_{\sigma}$) in $X$.

$F_{\sigma}$-Measurable selections of usco multimaps with values in non-metrizable compact spaces were studied by many mathematicians [3], [4], [5], [7], [8], [6]. Positive results are known for two classes of compact spaces: fragmentable and linearly ordered.

Let us recall [2, 5.0.1] (see also [14, §6]) that a topological space $K$ is fragmentable if $K$ has a metric $\rho$ such that for every $\varepsilon>0$ each non-empty subset $A\subset K$ contains a non-empty relatively open set $U\subset A$ of $\rho$-diameter $<\varepsilon$. By [2, 5.1.12], each fragmentable compact Hausdorff space contains a metrizable dense $G_{\delta}$-subspace.

The following selection theorem can be deduced from Theorem 1’ and Lemma 6 in [8].

Theorem 2** (Hansell, Jayne, Talagrand).**

Any usco map $\Phi:X\to K$ from a perfectly paracompact space $X$ to a fragmentable compact space $Y$ has an $F_{\sigma}$-measurable selection.

A similar selection theorem holds for usco maps into countably cellular GO-spaces. A Hausdorff topological space $X$ is called a generalized ordered space (briefly, a GO-space) if $X$ admits a linear order $\leq$ such that the topology of $X$ is generated by a base consisting of open order-convex subsets of $X$. A subset $C$ of a linearly ordered space $X$ is called order-convex if for any points $x\leq y$ in $C$ the order interval $[x,y]:=\{z\in X:x\leq z\leq y\}$ is contained in $X$. We say that the topology of $X$ is generated by the linear order $\leq$ if the topology of $X$ is generated by the subbase $\{(\leftarrow,a),(a,\to):a\in X\}$ consisting the the order-convex sets $(\leftarrow,a):=\{x\in X:x<a\}$ and $(a,\to)=\{x\in X:a<x\}$.

A topological space $X$ is countably cellular if every disjoint family of open sets in $X$ is at most countable. It is easy to see that each separable topological space is countably cellular. A topological space is called $F_{\sigma}$-perfect if every open set in $X$ is of type $F_{\sigma}$ in $X$ (i.e., can be represented as the countable union of closed sets). For example, every metruzable space is $F_{\sigma}$-perfect.

The following selection theorem will be proved in Section 2.

Theorem 3**.**

Let $Y$ be a GO-space and $X$ be an ($F_{\sigma}$-perfect) topological space. If $X$ or $Y$ is countably cellular, then any usco map $\Phi:X\multimap Y$ has a Borel ($F_{\sigma}$-measurable) selection.

Theorems 2, 3 suggest the following problem.

Problem 1**.**

Is it true that any usco map $\Phi:M\multimap K$ from a compact metrizable space $M$ to a compact Hausdorff space $K$ has a Borel ($F_{\sigma}$-measurable) selection?

In this paper we prove that this problem has negative answer under the negation of the Continuum Hypothesis (i.e., under $\omega_{1}<\mathfrak{c}$). A suitable counterexample will be constructed using the split square $\ddot{\mathbb{I}}$, which is the square of the split interval $\ddot{\mathbb{I}}$.

The split interval is the linearly ordered space $\ddot{\mathbb{I}}=[0,1]\times\{0,1\}$ whose topology is generated by the lexicographic order (defined by $\langle x,i\rangle\leq\langle y,j\rangle$ iff either $x<y$ or else $x=y$ and $i<j$). The split interval plays a fundamental role in the theory of separable Rosenthal compacta [16]. Let us recall that a topological space is called Rosenthal compact if it is homeomorphic to a compact subspace of the space $B_{1}(P)$ of functions of the first Baire class on a Polish space $P$. It is well-known (and easy to see) that the split interval is Rosenthal compact and so is its square. By Theorem 4 of Todorčević [16], each non-metrizable Rosenthal compact space of countable spread contains a topological copy of the split interval. A topological space has countable spread if it contains no uncountable discrete subspaces.

By Theorem 3, any usco map $\Phi:X\multimap\ddot{\mathbb{I}}$ from an $F_{\sigma}$-perfect topological space $X$ has an $F_{\sigma}$-measurable selection. In contrast, the split square $\ddot{\mathbb{I}}^{2}$ has dramatically different selections properties. Let $p:\ddot{\mathbb{I}}\to\mathbb{I}$, $p:\langle x,i\rangle\mapsto x$, be the natural projection of the split interval onto the unit interval $\mathbb{I}=[0,1]$, and

[TABLE]

be the projection of the split square $\ddot{\mathbb{I}}^{2}$ onto the unit square $\mathbb{I}^{2}$.

Theorem 4**.**

The following conditions are equivalent:

1. (1)

the usco multimap $P^{-1}:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ has a Borel-measurable selection; 2. (2)

the usco multimap $P^{-1}:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ has an $F_{\sigma}$-measurable selection; 3. (3)

$\omega_{1}=\mathfrak{c}$.

The implication $(2)\Rightarrow(1)$ of Theorem 4 is trivial and the implications $(1)\Rightarrow(3)\Rightarrow(2)$ are proved in Lemmas 2 and 8, respectively.

Combining Theorem 4 with the Todorčević dichotomy for Rosenthal compact spaces, we obtain the following consistent characterization of metrizable compacta.

Corollary 1**.**

Under $\omega_{1}<\mathfrak{c}$ a Rosenthal compact space $K$ is metrizable if and only if $K$ has countable spread and each usco multimap $\Phi:\mathbb{I}^{2}\to K^{2}$ has a Borel-measurable selection.

Proof.

The “only if” part follows from Theorem 2. To prove the “if” part, assume that a Rosenthal compact $K$ is not metrizable but has countable spread. By Theorem 4 of [16], the space $K$ contains a topological copy of the split interval $\ddot{\mathbb{I}}$. We lose no generality assuming that $\ddot{\mathbb{I}}\subset K$. By Theorem 4, under $\omega_{1}<\mathfrak{c}$, the usco multimap $P^{-1}:\mathbb{I}^{2}\multimap\ddot{\mathbb{I}}^{2}\subset K^{2}$ does not have Borel-measurable selections. ∎

Now we pose some open problems suggested by Theorem 4.

Problem 2**.**

Assume CH. Is it true that each usco map $\Phi:X\to\ddot{\mathbb{I}}^{2}$ from a metrizable (separable) space $X$ has a Borel-measurable selection?

Observe that the map $p:\ddot{\mathbb{I}}\to\mathbb{I}$ is 2-to-1 and its square $P:\ddot{\mathbb{I}}^{2}\to\mathbb{I}^{2}$ is 4-to-1. A function $f:X\to Y$ is called $n$-to-1 for some $n\in\mathbb{N}$ if $|f^{-1}(y)|\leq n$ for any $y\in Y$. By Theorem 3 of Todorčević [16], every Rosenthal compact space of countable spread admits a 2-to-1 map onto a metrizable compact space. Let us observe that the splitted square $\ddot{\mathbb{I}}^{2}$ contains a discrete subspace of cardinality continuum and hence has uncountable spread.

Problem 3**.**

Let $n\in\{2,3\}$. Is there an $n$-to-1 map $f:K\to M$ from a (Rosenthal) compact space $K$ to a metrizable compact space $M$ such that the inverse multimap $f^{-1}:M\multimap K$ has no Borel selections?

Remark 1**.**

Theorem 4 provides a consistent counterexample to the problem [9] of Chris Heunen, posed on Mathoverflow.

2. Proof of Theorem 3

Theorem 3 follows from Lemmas 2 and 3, proved in this section.

First we prove one lemma, showing that our definition of a GO-space agrees with the original definition of Lutzer [12]. Probably this lemma is known but we could not find the precise reference in the literature.

Lemma 1**.**

The linear order $\leq$ of any GO-space $X$ is a closed subset of the square $X\times X$.

Proof.

Given two elements $x,y\in X$ with $x\not\leq y$, use the Hausdorff property of $X$ and find two disjoint order-convex neighborhoods $O_{x},O_{y}\subset X$ of the points $x,y$, respectively. We claim that the product $O_{x}\times O_{y}$ is disjoint with the linear order $\leq$. Assuming that this is not true, find elements $x^{\prime}\in O_{x}$ and $y^{\prime}\in O_{y}$ such that $x^{\prime}\leq y^{\prime}$. Taking into account that the sets $O_{x},O_{y}$ are disjoint and order-convex, we conclude that $x^{\prime}<y$ and $x<y^{\prime}$. It follows from $x\not\leq y$ that $y<x$. Then $x^{\prime}<y<x<y^{\prime}$, which contradicts the assumption. This contradiction shows that the neighborhood $O_{x}\times O_{y}$ of the pair $\langle x,y\rangle$ is disjoint with $\leq$ and hence $\leq$ is a closed subset of $X\times X$. ∎

Lemma 2**.**

Any usco multimap $\Phi:X\multimap Y$ from an ($F_{\sigma}$-perfect) topological space $X$ to a countably cellular GO-space space $Y$ has a Borel ($F_{\sigma}$-measurable) selection.

Proof.

Being a GO-space, $Y$ has a base of the topology consisting of open order-convex subsets with respect to some linear order $\leq$ on $Y$. By Lemma 1, the linear order $\leq$ is a closed subset of $Y\times Y$. Then for every $a\in Y$ the order-convex set $(\leftarrow,a]=\{y\in Y:y\leq a\}$ is closed in $Y$, which implies that each non-empty compact subset of $Y$ has the smallest element.

Then for any usco multmap $\Phi:X\multimap Y$ we can define a selection $f:X\to Y$ of $\Phi$ assigning to each point $x\in X$ the smallest element $f(x)$ of the non-empty compact set $\Phi(x)\subset Y$. We claim that this selection is $F_{\sigma}$-measurable.

A subset $U\subset Y$ is called upper if for any $u\in U$ the order-convex set $[u,\to)=\{y\in Y:u\leq y\}$ is contained in $U$.

Claim 1**.**

For any upper open set $U\subset Y$ the preimage $f^{-1}[U]$ is open in $X$.

Proof.

For any $x\in f^{-1}[U]$ we get $\Phi(x)\subset[f(x),\to)\subset U$. The upper semicontinuity of $\Phi$ yields a neighborhood $O_{x}\subset X$ such that $\Phi[O_{x}]\subset U$. Consequently, $f(O_{x})\subset\Phi[O_{x}]\subset U$, witnessing that the set $f^{-1}[U]$ is open in $X$. ∎

A subset $L\subset Y$ is lower if for every $a\in L$ the order-convex set $(\leftarrow,a]=\{y\in Y:y\leq a\}$ is contained in $L$.

Claim 2**.**

For any closed lower set $L\subset Y$ the preimage $f^{-1}[L]$ is closed in $X$.

Proof.

Observe the the complement $X\setminus L$ is an open upper set in $Y$. By Claim 1, the preimage $f^{-1}[X\setminus L]$ is open in $X$ and hence its complement $X\setminus f^{-1}[X\setminus L]=f^{-1}[L]$ is closed in $X$. ∎

Claim 3**.**

For any lower set $L\subset Y$ the preimage $f^{-1}(L)$ is of type $F_{\sigma}$ in $X$.

Proof.

If $L$ has the largest element $\lambda$, then $L=(\leftarrow,\lambda]$ and $f^{-1}[L]=f^{-1}[(\leftarrow,\lambda]]$ is closed by Claim 2. So, we assume that $L$ does not have the largest element. Then the countable cellularity of $Y$ implies that $L$ has a countable cofinal subset $C\subset L$ (which means that for every $x\in L$ there exists $y\in C$ with $x\leq y$). By Lemma 2, for every $c\in C$ the preimage $f^{-1}[(\leftarrow,c]]$ is closed in $X$. Since $L=\bigcup_{c\in C}(\leftarrow,c]$, the preimage $f^{-1}[L]=\bigcup_{c\in C}f^{-1}[(\leftarrow,c]]$ is of type $F_{\sigma}$ in $X$. ∎

Claim 4**.**

For any open order-convex subset $U\subset Y$ the preimage $f^{-1}[U]$ is a Borel subset of $X$ (of type $F_{\sigma}$ if the space $X$ is $F_{\sigma}$-perfect).

Proof.

The order-convexity of $U$ implies that $U=\overleftarrow{U}\cap\overrightarrow{U}$ where $\overleftarrow{U}=\bigcup_{u\in U}(\leftarrow,u]$ and $\overrightarrow{U}=\bigcup_{u\in U}[u,\to)$. Taking into account that $Y$ has a base of order-convex sets, one can show that the upper set $\overrightarrow{U}$ is open in $X$. By Claim 1, the preimage $f^{-1}[\overrightarrow{U}]$ is open in $X$ (of type $F_{\sigma}$ if the space $X$ is $F_{\sigma}$-perfect). By Claim 3, the preimage $f^{-1}[\overleftarrow{U}]$ is of type $F_{\sigma}$ in $X$. Then $f^{-1}(U)=f^{-1}[\overleftarrow{U}]\cap f^{-1}[\overrightarrow{U}]$ is Borel (of type $F_{\sigma}$ if $X$ is $F_{\sigma}$-perfect). ∎

Claim 5**.**

For every open set $U\subset Y$ the preimage $f^{-1}[B]$ is Borel subset of $X$ (of type $F_{\sigma}$ if $X$ is $F_{\sigma}$-perfect).

Proof.

By the definition of the topology of $Y$, each point $x\in U$ has an open order-convex neighborhood $O_{x}\subset U$. By the Kuratowski-Zorn Lemma, each open order-convex subset of $U$ is contained in a maximal open order convex subset of $U$. Let $\mathcal{C}\subset\mathcal{B}$ be the family of maximal open order-convex subsets of $U$. Observe that $U=\bigcup\mathcal{C}$ and any distinct sets $C,D\in\mathcal{C}$ are disjoint: otherwise the union $C\cup D$ would be an open order convex subset of $U$ and by the maximality of $C$ and $D$, $C=C\cup D=D$. Since the space $Y$ is countably cellular, the family $\mathcal{C}$ is at most countable. By Claim 4, for every $C\in\mathcal{C}$ the preimage $f^{-1}(C)$ is Borel (an type $F_{\sigma}$-set if $X$ is $F_{\sigma}$-perfect) and so is the countable union $f^{-1}[U]=\bigcup_{C\in\mathcal{C}}f^{-1}$. ∎

Claim 5 completes the proof of Lemma 2. ∎

Lemma 3**.**

Every usco multimap $\Phi:X\multimap Y$ from a countably cellular ($F_{\sigma}$-perfect) topological space $X$ into a GO-space $Y$ has a Borel ($F_{\sigma}$-measurable) selection.

Proof.

The Kuratowski-Zorn Lemma implies that the usco map $\Phi$ contains a minimal usco map $\Psi:X\multimap Y$. We claim that the image $\Psi[X]\subset Y$ is a countably cellular subspace of $Y$. Assuming the opposite, we can find an uncountable disjoint family $(U_{\alpha})_{\alpha\in\omega_{1}}$ of non-empty open subsets in $\Psi[X]$. For every $\alpha\in\omega_{1}$, find $x_{\alpha}\in X$ such that $\Phi(x_{\alpha})\cap U_{\alpha}\neq\emptyset$. By Lemma 3.1.2 [2], the minimality of the usco map $\Psi$ implies that $\Psi[V_{\alpha}]\subset U_{\alpha}$ for some non-empty open set $V_{\alpha}\subset X$. Taking into account that the family $(U_{\alpha})_{\alpha\in\omega_{1}}$ is disjoint, we conclude that the family $(V_{\alpha})_{\alpha\in\omega_{1}}$ is disjoint, witnessing that the space $X$ is not countably cellular. But this contradicts our assumption. This contradiction shows that the GO-subspace $\Psi[X]$ of $Y$ is countably cellular. By Lemma 2, the usco map $\Psi:X\to\Psi[X]$ has a Borel ($F_{\sigma}$-measurable) selection, which is also a selection of the usco map $\Phi$. ∎

Finally, let us prove one selection property of the split interval, which will be used in the proof of Lemma 8.

Lemma 4**.**

Any selection of the multimap $p^{-1}:\mathbb{I}\to\ddot{\mathbb{I}}$ is $F_{\sigma}$-measurable.

Proof.

Given any open subset $U\subset\ddot{\mathbb{I}}$, we need to show that $s^{-1}[U]$ is of type $F_{\sigma}$ in $\mathbb{I}$. For every $x\in s^{-1}[U]$, find an open order-convex set $I_{x}\subset U$ containing $s(x)$. It is well-known (see e.g. [1, 3.10.C(a)]) that the split interval $\ddot{\mathbb{I}}$ is hereditarily Lindelöf. Consequently, there exists a countable set $C\in s^{-1}[U]$ such that $\bigcup_{x\in s^{-1}[U]}I_{x}=\bigcup_{x\in C}I_{x}$ and hence $s^{-1}[U]=\bigcup_{x\in C}s^{-1}[I_{x}]$. For every $x\in C$ the order-convexity of the interval $I_{x}\subset\ddot{\mathbb{I}}$ implies that its preimage $s^{-1}[I_{x}]$ is a convex subset of $\mathbb{I}$, containing $x$. Since convex subsets of $\mathbb{I}$ are of type $F_{\sigma}$, the countable union $s^{-1}[U]=\bigcup_{x\in C}I_{x}$ is an $F_{\sigma}$-set in $\mathbb{I}$. ∎

3. Selection properties of the split square $\ddot{\mathbb{I}}^{2}$ under the negation of CH

In this section we study the selection properties of the split square $\ddot{\mathbb{I}}^{2}$ under the negation of the Continuum Hypothesis.

By $\langle x,y\rangle$ we denote ordered pairs of elements $x,y$. In this way we distinguish ordered pairs from the order intervals $(x,y):=\{z\in x\leq z\leq y\}$ in linearly ordered spaces.

The split interval $\ddot{\mathbb{I}}=\mathbb{I}\times\{0,1\}$ carries the lexicographic order defined by $\langle x,i\rangle\leq\langle y,j\rangle$ iff either $x<y$ or ($x=y$ and $i<j$). It is well-known that the topology generated by the lexicographic order on $\ddot{\mathbb{I}}$ is compact and Hausdorff, see [1, 3.10.C(b)]. By $p:\ddot{\mathbb{I}}\to\mathbb{I}$, $p:\langle x,i\rangle\mapsto x$, we denote the coordinate projection and by $P:\ddot{\mathbb{I}}^{2}\to\mathbb{I}^{2}$, $P:\langle x,y\rangle\mapsto\langle p(x),p(y)\rangle$ the square of the map $p$.

The following lemma proves the implication $(1)\Rightarrow(3)$ of Theorem 4.

Lemma 5**.**

If $\omega_{1}<\mathfrak{c}$, then the multimap $P^{-1}:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ has no Borel selections.

Proof.

To derive a contradiction, assume that the multimap $P^{-1}$ has a Borel-measurable selection $s:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$.

For a real number $x\in\mathbb{I}$ by $x_{0}$ and $x_{1}$ we denote the points $\langle x,0\rangle$ and $\langle x,1\rangle$ of the split interval $\ddot{\mathbb{I}}$. Then $\ddot{\mathbb{I}}=\ddot{\mathbb{I}}_{0}\cup\ddot{\mathbb{I}}_{1}$ where $\ddot{\mathbb{I}}_{i}=\{x_{i}:x\in\mathbb{I}\}$ for $i\in\{0,1\}$.

For any numbers $i,j\in\{0,1\}$ consider the set

[TABLE]

and observe that $\mathbb{I}^{2}=\bigcup_{i,j=0}^{1}Z_{ij}$.

For a point $a\in\ddot{\mathbb{I}}$, let $[0_{0},a)$ and $(a,1_{1}]$ be the order intervals in $\ddot{\mathbb{I}}$ with respect to the lexicographic order. Observe that for any $x\in\mathbb{I}$ we have

[TABLE]

For every $a\in(0,2)\subset\mathbb{R}$ consider the lines

[TABLE]

on the plane.

Claim 6**.**

For every $a\in\mathbb{R}$ the intersection ${\mathrm{L}}_{a}\cap Z_{00}$ is at most countable.

Proof.

If for some $a\in\mathbb{R}$ the intersection ${\mathrm{L}}_{a}\cap Z_{00}$ is uncountable, then we can choose a non-Borel subset $B\subset\mathrm{L}_{a}\cap Z_{00}$ of cardinality $|B|=\omega_{1}$. For every point $\langle x,y\rangle\in B\subset Z_{00}$, the definition of the set $Z_{00}$ ensures that $s(\langle x,y\rangle)=\langle x_{0},y_{0}\rangle$ and hence the set $U_{\langle x,y\rangle}=[0_{0},x_{1})\times[0_{0},y_{1})=[0_{0},x_{0}]\times[0_{0},y_{0}]$ is an open neighborhood of $s(\langle x,y\rangle)$ in $\ddot{\mathbb{I}}^{2}$. Observe that $\langle x,y\rangle\in s^{-1}(U_{\langle x,y\rangle})\subset p(U_{\langle x,y\rangle})=[0,x]\times[0,y]$ and hence $\mathrm{L}_{a}\cap s^{-1}(U_{\langle x,y\rangle})=\{\langle x,y\rangle\}$. Then for the open set $U=\bigcup_{\langle x,y\rangle\in B}U_{\langle x,y\rangle}$ the preimage $s^{-1}[U]$ is not Borel in $\mathbb{I}^{2}$ because the intersection $s^{-1}[U]\cap\mathrm{L}_{a}=B$ is not Borel. But this contradicts the Borel measurability of $s$. ∎

By analogy we can prove the following claims.

Claim 7**.**

For every $a\in\mathbb{R}$ the intersection $\mathrm{L}_{a}\cap Z_{11}$ is at most countable.

Claim 8**.**

For every $b\in\mathbb{R}$ the intersection $\Gamma^{b}\cap(Z_{01}\cup Z_{10})$ is at most countable.

Now fix any subset set $\Omega\subset[\frac{1}{2},\frac{3}{2}]$ of cardinality $|\Omega|=\omega_{1}$. By Claims 6, 7, for every $a\in\Omega$ the intersection $\mathrm{L}_{a}\cap(Z_{00}\cup Z_{11})$ is at most countable. Consequently the union

[TABLE]

has cardinality $|U|\leq\omega_{1}$. Since $|U|\leq\omega_{1}<\mathfrak{c}$, there exists a real number $b\in[\frac{1}{2},\frac{3}{2}]$ such that the line $\Gamma^{b}$ does not intersect the set $U$. Since $\{b\}\cup\Omega\subset[\frac{1}{2},\frac{3}{2}]$ for every $a\in\Omega$ the intersection $\Gamma^{b}\cap\mathrm{L}_{a}\cap\mathbb{I}^{2}$ is not empty. Then the set $X=\bigcup_{a\in\Omega}\mathrm{L}_{a}\cap\Gamma^{b}\subset\mathbb{I}^{2}$ is uncountable and $X\subset\Gamma^{b}\setminus U\subset\Gamma^{b}\cap(Z_{01}\cup Z_{10})$. But this contradicts Claim 8. ∎

4. Selection properties of the split square $\ddot{\mathbb{I}}^{2}$ under the Continuum Hypothesis

In this section we shall prove that under the continuum hypothesis the usco multimap $P^{-1}:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ has an $F_{\sigma}$-measurable selection.

First we introduce some terminology related to monotone functions.

A subset $f\subset\mathbb{I}^{2}$ is called a

1. (1)

a function if for any $\langle x_{1},y_{1}\rangle,\langle x_{2},y_{2}\rangle\in f$ the equality $x_{1}=x_{2}$ implies $y_{1}=y_{2}$; 2. (2)

strictly increasing if for any $\langle x_{1},y_{1}\rangle,\langle x_{2},y_{2}\rangle\in f$ the strict inequality $x_{1}<x_{2}$ implies $y_{1}<y_{2}$; 3. (3)

strictly decreasing if for any $\langle x_{1},y_{1}\rangle,\langle x_{2},y_{2}\rangle\in f$ the inequality $x_{1}<x_{2}$ implies $y_{1}>y_{2}$; 4. (4)

strictly monotone if $f$ is strictly increasing or strictly decreasing.

Lemma 6**.**

Each strictly increasing function $f\subset\mathbb{I}^{2}$ is a subset of a Borel strictly increasing function $\bar{f}\subset\mathbb{I}^{2}$.

Proof.

It follows that the strictly increasing function $f$ is a strictly increasing bijective function between the sets $\mathrm{pr}_{1}[f]=\{x\in\mathbb{I}:\exists y\in\mathbb{I}\;\;\langle x,y\rangle\in f\}$ and $\mathrm{pr}_{2}[f]=\{y\in\mathbb{I}:\exists x\in\mathbb{I}\;\;\langle x,y\rangle\in f\}$. It is well-known that monotone functions of one real variable have at most countably many discontinuity points. Consequently, the sets of discontinuity points of the strictly monotone functions $f$ and $f^{-1}$ are at most countable. This allows us to find a countable set $D_{f}\subset f$ such that the set $f\setminus D_{f}$ coincides with the graph of some increasing homeomorphism between subsets of $\mathbb{I}$. Replacing $D_{f}$ by a larger countable set, we can assume that $D_{f}=f\cap(\mathrm{pr}_{1}[D_{f}]\times\mathrm{pr}_{2}[D_{f}])$, where $\mathrm{pr}_{1},\mathrm{pr}_{2}:\mathbb{I}^{2}\to\mathbb{I}$ are coordinate projections. By the Lavrentiev Theorem [10, 3.9], the homeomorphism $f\setminus D_{f}$ extends to a (strictly increasing) homeomorphism $h\subset\mathbb{I}^{2}$ between $G_{\delta}$-subsets of $\mathbb{I}^{2}$ such that $f\setminus D_{f}$ is dense in $h$. It is easy to check that the Borel subset $\bar{f}=(h\setminus(\mathrm{pr}_{1}[D_{f}]\times\mathrm{pr}_{2}[D_{f}])\cup D_{f}$ is a strictly increasing function extending $f$. ∎

By analogy we can prove

Lemma 7**.**

Each strictly decreasing function $f\subset\mathbb{I}^{2}$ is a subset of a Borel strictly decreasing function $\bar{f}\subset\mathbb{I}^{2}$.

Now we are ready to prove the main result of this section.

Lemma 8**.**

Under $\omega_{1}=\mathfrak{c}$ the multifunction $P^{-1}:\mathbb{I}^{2}\multimap\ddot{\mathbb{I}}^{2}$ has an $F_{\sigma}$-measurable selection.

Proof.

Let $\mathcal{M}$ be the set of infinite strictly monotone Borel functions $f\subset\mathbb{I}^{2}$. Since $\omega_{1}=\mathfrak{c}$, the set $\mathcal{M}$ can be written as $\mathcal{M}=\{f_{\alpha}\}_{\alpha<\omega_{1}}$. It is clear $\bigcup_{\alpha<\omega_{1}}f_{\alpha}=\mathbb{I}^{2}$. So, for any point $z\in\mathbb{I}^{2}$ we can find the smallest ordinal $\alpha_{z}<\omega_{1}$ such that $z\in f_{\alpha_{z}}$. Consider the sets

[TABLE]

Define a selection $s:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ of the multimap $P^{-1}:\mathbb{I}^{2}\multimap\ddot{\mathbb{I}}^{2}$ letting

[TABLE]

for $\langle x,y\rangle\in\mathbb{I}^{2}$.

We claim that the function $s:\mathbb{I}^{2}\to\ddot{\mathbb{I}}^{2}$ is $F_{\sigma}$-measurable. Given any open set $U\subset\ddot{\mathbb{I}}^{2}$, we should prove that its preimage $s^{-1}[U]$ of type $F_{\sigma}$ in $\mathbb{I}^{2}$. Consider the open subset $V:=U\cap(0_{1},1_{0})^{2}\subset\ddot{\mathbb{I}}^{2}$ of $U$. Using Lemma 4, it can be shown that the set $s^{-1}[U\setminus V]\subset\mathbb{I}^{2}\setminus(0,1)^{2}$ is of type $F_{\sigma}$ in $\mathbb{I}^{2}$. Therefore, it remains to show that the preimage $s^{-1}[V]$ is of type $F_{\sigma}$ in $\mathbb{I}^{2}$.

Let $\mathbb{Q}:=\{\frac{n}{m}:n,m\in\mathbb{N},\;n<m\}$ be the set of rational numbers in the interval $(0,1)$.

Consider the subsets $\mathrm{L}_{V}:=\mathrm{L}\cap s^{-1}(V)$ and $\Gamma_{V}:=\Gamma\cap s^{-1}(V)$. For every $\langle x,y\rangle\in\mathrm{L}_{V}$ we have $s(\langle x,y\rangle)=\langle x_{1},y_{1}\rangle\in V$ and by the definition of the topology of the split interval, we can find rational numbers $a(x,y),b(x,y)\in\mathbb{Q}$ such that $x<a(x,y)$, $y<b(x,y)$ and s(\langle x,y\rangle)=\langle x_{1},y_{1}\rangle\in\big{[}x_{1},a(x,y)_{0}\big{)}\times\big{[}y_{1},b(x,y)_{0}\big{)}\subset V. Then

[TABLE]

On the other hand, for every $\langle x,y\rangle\in\Gamma_{V}$ there are rational numbers $a(x,y),b(x,y)\in\mathbb{Q}$ such that $x<a(x,y)$, $b(x,y)<y$ and s(\langle x,y\rangle)=\langle x_{1},y_{0}\rangle=\big{[}x_{1},a(x,y)_{0}\big{)}\times\big{(}b(x,y)_{1},y_{0}\big{]}\subset V. In this case

[TABLE]

It follows that

[TABLE]

This equality and the following claim imply that the set $s^{-1}[V]$ is of type $F_{\sigma}$ in $\mathbb{I}^{2}$.

Claim 9**.**

There are countable subsets $\mathrm{L}^{\prime}\subset\mathrm{L}_{V}$ and $\Gamma^{\prime}\subset\Gamma_{V}$ such that

[TABLE]

and

[TABLE]

We shall show how to find the countable set $\mathrm{L}^{\prime}\subset\mathrm{L}_{V}$. The countable set $\Gamma^{\prime}\subset\Gamma_{V}$ can be found by analogy.

For rational numbers $r,q\in\mathbb{Q}$, consider the set

[TABLE]

and observe that $\mathrm{L}_{V}=\bigcup_{r,q\in\mathbb{Q}}\mathrm{L}_{p,q}.$

Claim 10**.**

For any rational numbers $r,q\in\mathbb{Q}$ there exists a countable subset $\mathrm{L}_{r,q}^{\prime}\subset\mathrm{L}_{r,q}$ such that

[TABLE]

Proof.

For every rational numbers $r^{\prime}\leq r$ and $q^{\prime}\leq q$, consider the numbers

[TABLE]

Choose countable subsets $\mathrm{L}_{r,q}^{r^{\prime}\!,0},L_{r,q}^{0,q^{\prime}}\subset\mathrm{L}_{r,q}$ such that

[TABLE]

and moreover,

[TABLE]

and

[TABLE]

Consider the countable subset

[TABLE]

of $\mathrm{L}_{r,q}$.

Claim 11**.**

\displaystyle\bigcup\limits_{\langle x,y\rangle\in L_{r,q}}\big{(}[x,r)\times[y,q)\big{)}\setminus\{\langle x,y\rangle\}\subset\bigcup\limits_{\langle x,y\rangle\in\mathrm{L}_{r,q}^{\prime\prime}}[x,r)\times[y,q).**

Proof.

Fix any pairs $\langle x,y\rangle\in\mathrm{L}_{r,q}$ and \langle x^{\prime},y^{\prime}\rangle\in\big{(}[x,r)\times[y,q)\big{)}\setminus\{\langle x,y\rangle\}. Three cases are possible:

1. (1)

$x<x^{\prime}<r$ and $y<y^{\prime}<q$; 2. (2)

$x=x^{\prime}$ and $y<y^{\prime}<q$; 3. (3)

$x<x^{\prime}<r$ and $y=y^{\prime}$.

In the first case there exist rational numbers $r^{\prime},q^{\prime}$ such that $x<r^{\prime}<x^{\prime}<r$ and $y<q^{\prime}<y^{\prime}<q$. The definition of $\underline{x}(q^{\prime})$ ensures that $\underline{x}(q^{\prime})\leq x<x^{\prime}$. By the choice of the family $\mathrm{L}_{r,q}^{0,q^{\prime}}$, there exists $\langle x^{\prime\prime},y^{\prime\prime}\rangle\in\mathrm{L}_{r,q}^{0,q^{\prime}}\subset\mathrm{L}_{r,q}^{\prime\prime}$ such that $x^{\prime\prime}<x^{\prime}<r$ and $y^{\prime\prime}<q^{\prime}<y^{\prime}<q$. Then $\langle x^{\prime},y^{\prime}\rangle\in[x^{\prime\prime},r)\times[y^{\prime\prime},q)$.

Next, assume that $x=x^{\prime}$ and $y<y^{\prime}<q$. In this case we can choose a rational number $q^{\prime}$ such that $y<q^{\prime}<y^{\prime}$. It follows that $\underline{x}(q^{\prime})\leq x=x^{\prime}$. If $\underline{x}(q^{\prime})<x^{\prime}$, then by the definition of the family $\mathrm{L}_{r,q}^{0,q^{\prime}}$, there exists $\langle x^{\prime\prime},y^{\prime\prime}\rangle\in\mathrm{L}_{r,q}^{0,q^{\prime}}\subset\mathrm{L}_{r,q}$ such that $x^{\prime\prime}<x^{\prime}<r$ and $y^{\prime\prime}<q^{\prime}<y^{\prime}<q$. Then $\langle x^{\prime},y^{\prime}\rangle\in[x^{\prime\prime},r)\times[y^{\prime\prime},q)$.

So, we assume that $\underline{x}(q^{\prime})=x^{\prime}=x$ and hence $\underline{x}(q^{\prime})=x=\min\{x^{\prime\prime}:\langle x^{\prime\prime},y^{\prime\prime}\rangle\in\mathrm{L}_{p,q}:y^{\prime\prime}<q^{\prime}\}$. In this case $x^{\prime}=\underline{x}(q^{\prime})=x^{\prime\prime}$ for some $\langle x^{\prime\prime},y^{\prime\prime}\rangle\in\mathrm{L}_{r,q}^{0,q^{\prime}}\subset\mathrm{L}_{r,q}^{\prime\prime}$ with $y^{\prime\prime}<q^{\prime}<y^{\prime}<q$. Then $\langle x^{\prime},y^{\prime}\rangle\in[x^{\prime\prime},r)\times[y^{\prime\prime},q)$.

By analogy, in the third case ($x<x^{\prime}<r$ and $y=y^{\prime\prime}$) we can find a pair $\langle x^{\prime\prime},y^{\prime\prime}\rangle\in L_{r,q}^{\prime\prime}$ such that $\langle x^{\prime},y^{\prime}\rangle\in[x^{\prime\prime},r)\times[y^{\prime\prime},q)$. ∎

Claim 11 implies that the set

[TABLE]

is contained in $\mathrm{L}_{r,q}.$ ∎

Claim 12**.**

The set $D_{r,q}$ is a strictly decreasing function.

Proof.

First we show that $D_{r,q}$ is a function. Assuming that $D_{r,q}$ is not a function, we can find two pairs $\langle x,y\rangle,\langle x,y^{\prime}\rangle\in D_{r,q}$ with $y<y^{\prime}$. Applying Claim 11, we conclude that

[TABLE]

and hence $\langle x,y^{\prime}\rangle\notin D_{r,q}$, which contradicts the choice of the pair $\langle x,y^{\prime}\rangle$. This contradiction shows that $D_{r,q}$ is a function.

Assuming that $D_{r,q}$ is not strictly decreasing, we can find pairs $\langle x,y\rangle,\langle x^{\prime},y^{\prime}\rangle\in D_{r,q}$ such that $x<x^{\prime}$ and $y\leq y^{\prime}$. Applying Claim 11, we conclude that

[TABLE]

and hence $\langle x^{\prime},y^{\prime}\rangle\notin D_{r,q}$, which contradicts the choice of the pair $\langle x,y^{\prime}\rangle$. This contradiction shows that $D_{r,q}$ is strictly decreasing. ∎

Claim 13**.**

The set $D_{r,q}$ is at most countable.

Proof.

To derive a contradiction, assume that $D_{r,q}$ is uncountable. By Lemma 7, the strictly decreasing function $D_{r,q}$ is contained in some Borel strictly decreasing function, which is equal to $f_{\alpha}$ for some ordinal $\alpha<\omega_{1}$. Since the intersection of a strictly increasing function and a strictly decreasing function contains at most one point, the set

[TABLE]

is at most countable. We claim that $D_{r,q}=D^{\prime}_{r,q}$. To derive a contradiction, assume that $D_{r,q}\setminus D^{\prime}_{r,q}$ contains some pair $z=\langle x,y\rangle$. It follows from $z\in D_{r,q}\subset f_{\alpha}$ that $\alpha_{z}\leq\alpha$. Since $z\notin D_{r,q}^{\prime}$, the strictly monotone function $f_{\alpha_{z}}\ni z$ is not strictly increasing and hence $f_{\alpha_{z}}$ is strictly decreasing. Then the definition of the set $\mathrm{L}$ guarantees that $z\notin\mathrm{L}$, which contradicts the inclusion $z\in D_{r,q}\subset\mathrm{L}_{r,q}\subset\mathrm{L}$. ∎

Now consider the countable subset $\mathrm{L}^{\prime}_{r,q}:=\mathrm{L}_{r,q}^{\prime\prime}\cup D_{r,q}$ of $\mathrm{L}_{r,q}$ and observe that

[TABLE]

This completes the proof of Claim 10.

Claim 14**.**

There exists a countable subset $\mathrm{L}^{\prime}\subset\mathrm{L}_{V}$ such that

[TABLE]

Proof.

By Claim 10, for any rational numbers $r,q\in\mathbb{Q}$ there exists a countable subset $\mathrm{L}_{r,q}^{\prime}\subset\mathrm{L}_{r,q}$ such that

[TABLE]

Since $\mathrm{L}_{V}=\bigcup_{r,q\in\mathbb{Q}}\mathrm{L}_{r,q}$, the countable set $\mathrm{L}^{\prime}:=\bigcup_{r,q\in\mathbb{Q}}\mathrm{L}^{\prime}_{r,q}$ has the required property. ∎

By analogy with Claim 14 we can prove

Claim 15**.**

There exists a countable subset $\Gamma^{\prime}\subset\Gamma_{V}$ such that

[TABLE]

Claims 14 and 15 complete the proof of Claim 9 and the proof of Lemma 8. ∎

5. Acknowledgement

The author expresses his sincere thanks to the Mathoverflow user @YCor (Yves Cornulier) for the idea of the proof of Lemma 5 and to Dušan Repovš for valuable information on measurable selectors.