Selection properties of the split interval and the Continuum Hypothesis

Abstract

We prove that every usco multimap $\varPhi :X\rightarrow Y$ 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 $P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2$ 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.

1 Introduction

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

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

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

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

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

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

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

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

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

Let $\varPhi :X\multimap Y$ be a multimap between topological spaces. A function $f:X\rightarrow Y$ is called a selection of $\varPhi$ if $f(x)\in \varPhi (x)$ for every $x\in X$. The Axiom of Choice ensures that every multimap $\varPhi :X\multimap Y$ 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 [12] (see also [16, §5.2] or [14, 6.12]).

Theorem 1

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

We recall that a function $f:X\rightarrow Y$ between topological spaces is called Borel-measurable (resp. $F_\sigma$-measurable) if for every open set $U\subseteq Y$ 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 [4,5,6,7,8,9]. Positive results are known for two classes of compact spaces: fragmentable and linearly ordered.

Let us recall [3, 5.0.1] (see also [15, §6]) that a topological space K is fragmentable if K has a metric $\rho$ such that for every $\varepsilon >0$ each non-empty subset $A\subseteq K$ contains a non-empty relatively open set $U\subseteq A$ of $\rho$-diameter $<\varepsilon$. By [3, 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 [9].

Theorem 2

(Hansell, Jayne, Talagrand) Any usco map $\varPhi :X\rightarrow K$ 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 $\le$ 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 $x\le y$ in C the order interval $[x,y]:=\{z\in X:x\le z\le y\}$ is contained in X. We say that the topology of X is generated by the linear order $\le$ if the topology of X is generated by the subbase $\{(\leftarrow ,a),(a,\rightarrow ):a\in X\}$ consisting the the order-convex sets $(\leftarrow ,a):=\{x\in X:x<a\}$ and $(a,\rightarrow )=\{x\in X:a<x\}$.

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 metrizable space is $F_\sigma$-perfect.

The following selection theorem will be proved in Sect. 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 $\varPhi :X\multimap Y$ has a Borel ($F_\sigma$-measurable) selection.

Theorems 2, 3 suggest the following problem.

Problem 1

Is it true that any usco map $\varPhi :M\multimap K$ 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 ${{{\ddot{\mathbb I}}}^2}$, 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 $\langle x,i\rangle \le \langle y,j\rangle$ iff either $x<y$ or else $x=y$ and $i\le j$). The split interval plays a fundamental role in the theory of separable Rosenthal compacta [17]. 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 [17], 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 ${\varPhi :X\multimap {{\ddot{\mathbb I}}}}$ 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 ${p:{{\ddot{\mathbb I}}}\rightarrow \mathbb I}$, $p:\langle x,i\rangle \mapsto x$, be the natural projection of the split interval onto the unit interval $\mathbb I=[0,1]$, and

$$\begin{aligned} {P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2,\;\;P:\langle x,y\rangle \mapsto \langle p(x),p(y)\rangle ,} \end{aligned}$$

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

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

• 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 Theorem 4 in [17], 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 $\varPhi :\mathbb I^2\rightarrow K^2$ 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 [17], the space K contains a topological copy of the split interval ${{{\ddot{\mathbb I}}}}$. We lose no generality assuming that ${{{\ddot{\mathbb I}}}\subseteq K}$. By Theorem 4, under $\omega _1<\mathfrak c$, the usco multimap ${P^{-1}:\mathbb I^2\multimap {{\ddot{\mathbb I}}}^2\subseteq K^2}$ does not have Borel-measurable selections.$\square$

Now we pose some open problems suggested by Theorem 4.

Problem 2

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

Observe that the map ${p:{{\ddot{\mathbb I}}}\rightarrow \mathbb I}$ is 2-to-1 and its square ${P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2}$ is 4-to-1. A function $f:X\rightarrow Y$ is called n-to-1 for some $n\in \mathbb N$ if $|f^{-1}(y)|\le n$ for any $y\in Y$. By Theorem 3 of Todorčević [17], 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 $f:K\rightarrow M$ 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 Chris Heunen asked on Mathoverflow [10] whether for any continuous surjective function $\pi :X\rightarrow Y$ between compact Hausdorff spaces and any continuous map $\gamma :\mathbb I\rightarrow Y$ there exists a Borel-measurable function $g:\mathbb I\rightarrow X$ such that $\gamma =\pi \circ g$. Theorem 4 provides a consistent counterexample to this problem of Heunen. Indeed, consider the projection ${P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2}$ and take any continuous surjective map $\gamma :\mathbb I\rightarrow \mathbb I^2$ (whose existence was proved by Giuseppe Peano in 1890). By Theorem 2, the inverse multimap $\gamma ^{-1}:\mathbb I^2\multimap \mathbb I$ has a Borel-measurable selection $s:\mathbb I^2\rightarrow \mathbb I$. Assuming that there exists a Borel-measurable function ${g:\mathbb I\rightarrow {{\ddot{\mathbb I}}}^2}$ with $\gamma =P\circ g$, we conclude that ${g\circ s:\mathbb I^2\rightarrow {{\ddot{\mathbb I}}}^2}$ is a Borel-measurable selection of the multimap $P^{-1}$, with contradicts Theorem 4 under CH.

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 [13]. Probably this lemma is known but we could not find the precise reference in the literature.

Lemma 1

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

Proof

Given two elements $x,y\in X$ with $x\not \le y$, use the Hausdorff property of X and find two disjoint order-convex neighborhoods $O_x,O_y\subseteq X$ of the points x, y, respectively. We claim that the product $O_x\times O_y$ is disjoint with the linear order $\le$. Assuming that this is not true, find elements $x'\in O_x$ and $y'\in O_y$ such that $x'\le y'$. Taking into account that the sets $O_x,O_y$ are disjoint and order-convex, we conclude that $x'<y$ and $x<y'$. It follows from $x\not \le y$ that $y<x$. Then $x'<y<x<y'$ and hence $[y,x]\subseteq O_x\cap O_y=\emptyset$, which is a desired contradiction. This contradiction shows that the neighborhood $O_x\times O_y$ of the pair $\langle x,y\rangle$ is disjoint with $\le$ and hence $\le$ is a closed subset of $X\times X$. $\square$

Lemma 2

Any usco multimap $\varPhi :X\multimap Y$ 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 $\le$ on Y. By Lemma 1, the linear order $\le$ is a closed subset of $Y\times Y$. Then for every $a\in Y$ the order-convex set $(\leftarrow , a]=\{y\in Y:y\le a\}$ is closed in Y, which implies that each non-empty compact subset of Y has the smallest element.

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

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

Claim 1

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

Proof

For any $x\in f^{-1}[U]$ we get $\varPhi (x)\subseteq [f(x),\rightarrow )\subseteq U$. The upper semicontinuity of $\varPhi$ yields a neighborhood $O_x\subseteq X$ such that $\varPhi [O_x]\subseteq U$. Consequently, $f(O_x)\subseteq \varPhi [O_x]\subseteq U$, witnessing that the set $f^{-1}[U]$ is open in X. $\square$

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

Claim 2

For any closed lower set $L\subseteq Y$ 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. $\square$

Claim 3

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

Proof

If L has a 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 a largest element. Then the countable cellularity of Y implies that L has a countable cofinal subset $C\subseteq L$ (which means that for every $x\in L$ there exists $y\in C$ with $x\le y$). 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. $\square$

Claim 4

For any open order-convex subset $U\subseteq Y$ 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 $\overrightarrow{U}=\bigcup _{u\in U}[u,\rightarrow )$. 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). $\square$

Claim 5

For every open set $U\subseteq Y$ 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 $O_x\subseteq U$. 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 (of 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. $\square$

Lemma 3

Every usco multimap $\varPhi :X\multimap Y$ 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 $\varPhi$ contains a minimal usco map $\Psi :X\multimap Y$. We claim that the image $\Psi [X]\subseteq Y$ 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 $\varPhi (x_\alpha )\cap U_\alpha \ne \emptyset$. By Lemma 3.1.2 [3], the minimality of the usco map $\Psi$ implies that $\Psi [V_\alpha ]\subseteq U_\alpha$ for some non-empty open set $V_\alpha \subseteq X$. 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 $\Psi :X\rightarrow \Psi [X]$ has a Borel ($F_\sigma$-measurable) selection, which is also a selection of the usco map $\varPhi$. $\square$

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

Proof

Given any open subset ${U\subseteq {{\ddot{\mathbb I}}}}$, 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 $I_x\subseteq U$ containing s(x). It is well-known (see e.g. [2, 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 $I_x\subseteq {{\ddot{\mathbb I}}}$ 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 $\mathbb I$. $\square$

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 the ordered pair of points x, y. In this way we distinguish ordered pairs from the order intervals $(x,y):=\{z:x<z<y\}$ in linearly ordered spaces.

The split interval ${{\ddot{\mathbb I}}}=\mathbb I\times \{0,1\}$ carries the lexicographic order defined by $\langle x,i\rangle \le \langle y,j\rangle$ iff either $x<y$ or ($x=y$ and $i\le j$). It is well-known that the topology generated by the lexicographic order on ${{\ddot{\mathbb I}}}$ is compact and Hausdorff, see [2, 3.10.C(b)]. By $p:{{\ddot{\mathbb I}}}\rightarrow \mathbb I$, $p:\langle x,i\rangle \mapsto x$, we denote the coordinate projection and by $P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2$, $P:\langle x,y\rangle \mapsto \langle p(x),p(y)\rangle$ 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 ${P^{-1}:\mathbb I^2\multimap {{\ddot{\mathbb I}}}^2}$ has no Borel selections.

Proof

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

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

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

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

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

$$\begin{aligned}&p\big ([0_0,x_0)\big )=[0,x),\;\;p\big ([0_0,x_1)\big )=[0,x],\;\;\\&p\big ((x_0,1_1]\big )=[x,1],\;\;p\big ((x_1,1_1]\big )=(x,1].\end{aligned}$$

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

$$\begin{aligned}\mathrm {L}_a=\{\langle x,y\rangle \in \mathbb R^2:x+y=a\}\quad \text{ and }\quad {\Gamma }^a=\{\langle x,y\rangle \in \mathbb R^2:y-x=a\} \end{aligned}$$

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 $B\subseteq \mathrm L_a\cap Z_{00}$ of cardinality $|B|=\omega _1$. For every point $\langle x,y\rangle \in B\subseteq Z_{00}$, 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 $\langle x,y\rangle \in s^{-1}(U_{\langle x,y\rangle })\subseteq p(U_{\langle x,y\rangle })=[0,x]\times [0,y]$ 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. $\square$

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 $\Omega \subseteq [\frac{1}{2},\frac{3}{2}]$ 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

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

has cardinality $|U|\le \omega _1$. Since $|U|\le \omega _1<\mathfrak c$, 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 $\{b\}\cup \Omega \subseteq [\frac{1}{2},\frac{3}{2}]$ 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 $X\subseteq \Gamma ^b\setminus U\subseteq \Gamma ^b\cap (Z_{01}\cup Z_{10})$. 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 ${P^{-1}:\mathbb I^2\multimap {{\ddot{\mathbb I}}}^2}$ has an $F_\sigma$-measurable selection.

First we introduce some terminology related to monotone functions.

A subset $f\subseteq \mathbb I^2$ is called

• a function if for any $\langle x,y\rangle ,\langle x',y'\rangle \in f$ the equality $x=x'$ implies $y=y'$;

• strictly increasing if for any $\langle x,y\rangle ,\langle x',y'\rangle \in f$ the strict inequality $x<x'$ implies $y<y'$;

• strictly decreasing if for any $\langle x,y\rangle ,\langle x',y'\rangle \in f$ the inequality $x< x'$ implies $y>y'$;

• 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 $D_f\subseteq f$ 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 $\mathrm {pr}_1,\mathrm {pr}_2:\mathbb I^2\rightarrow \mathbb I$ are coordinate projections. By the Lavrentiev Theorem [11, 3.9], the homeomorphism $f\setminus D_f$ extends to a (strictly increasing) homeomorphism $h\subseteq \mathbb I^2$ 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. $\square$

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

$$\begin{aligned} \begin{aligned} \mathrm {L}&:=\{z\in \mathbb I^2:f_{\alpha _z}\hbox { is strictly increasing}\}\hbox { and }\\ \Gamma&:=\{z\in \mathbb I^2:f_{\alpha _z} \,\hbox {is strictly decresing}\}=\mathbb I^2 \setminus \mathrm L. \end{aligned}\end{aligned}$$

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

$$\begin{aligned}s(\langle x,y\rangle )={\left\{ \begin{array}{ll}\langle x_1,y_1\rangle &{}\quad \hbox {if}\,\, \langle x,y\rangle \in \mathrm L,\\ \langle x_1,y_0\rangle &{}\quad \text{ if } \,\,\langle x,y\rangle \in \Gamma , \end{array}\right. } \end{aligned}$$

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

We claim that the function ${s:\mathbb I^2\rightarrow {{\ddot{\mathbb I}}}^2}$ is $F_\sigma$-measurable. Given any open set ${U\subseteq {{\ddot{\mathbb I}}}^2}$, we should prove that its preimage $s^{-1}[U]$ of type $F_\sigma$ in $\mathbb I^2$. Consider the open subset ${V:=U\cap (0_1,1_0)^2\subseteq {{\ddot{\mathbb I}}}^2}$ of U. Using Lemma 4, it can be shown that the set $s^{-1}[U\setminus V]\subseteq \mathbb I^2\setminus (0,1)^2$ 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 $s(\langle x,y\rangle )=\langle x_1,y_1\rangle \in \big [x_1,a(x,y)_0\big )\times \big [y_1,b(x,y)_0\big )\subseteq V$. Then

$$\begin{aligned}\big [x,a(x,y)\big )\times \big [y,b(x,y)\big )=s^{-1}\big [[x_1,a(x,y)_0)\times [y_1,b(x,y)_0)\big ]\subseteq s^{-1}[V].\end{aligned}$$

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 $s(\langle x,y\rangle )=\langle x_1,y_0\rangle =\big [x_1,a(x,y)_0\big )\times \big (b(x,y)_1,y_0\big ]\subseteq V$. In this case

$$\begin{aligned}\big [x,a(x,y)\big )\times \big (b(x,y),y\big ]=s^{-1}\big [[x_1,a(x,y)_0)\times (b(x,y)_1,y_0]\big )\subseteq s^{-1}[V].\end{aligned}$$

It follows that

$$\begin{aligned} s^{-1}[V]&=\Big (\bigcup _{\langle x,y\rangle \in \mathrm L_V}\big [x,a(x,y)\big )\times \big [y,b(x,y)\big )\Big )\\&\cup \Big (\bigcup _{\langle x,y\rangle \in \Gamma _V}\big [x,a(x,y)\big )\times \big (b(x,y),y\big ]\Big ). \end{aligned}$$

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 $\mathrm L'\subseteq \mathrm L_V$ and $\Gamma '\subseteq \Gamma _V$ such that

$$\begin{aligned}\bigcup _{\langle x,y\rangle \in \mathrm L_V} \big [x,a(x,y))\times [y,b(x,y)\big )=\bigcup _{\langle x,y\rangle \in \mathrm L'} \big [x,a(x,y)\big )\times \big [y,b(x,y)\big )\end{aligned}$$

and

$$\begin{aligned}\bigcup _{\langle x,y\rangle \in \Gamma _V} \big [x,a(x,y)\big )\times \big (b(x,y),y\big ]=\bigcup _{\langle x,y\rangle \in \Gamma '} \big [x,a(x,y)\big )\times \big (b(x,y),y\big ].\end{aligned}$$

We shall show how to find the countable set $\mathrm L'\subseteq \mathrm L_V$. The countable set $\Gamma '\subseteq \Gamma _V$ can be found by analogy.

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

$$\begin{aligned}\mathrm L_{r,q}=\{\langle x,y\rangle \in \mathrm L_V:a(x,y)=r,\;b(x,y)=q\}\end{aligned}$$

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 $\mathrm L_{r,q}'\subseteq \mathrm L_{r,q}$ such that

$$\begin{aligned}\bigcup _{\langle x,y\rangle \in \mathrm L'_{r,q}}[x,r)\times [y,q)=\bigcup _{\langle x,y\rangle \in \mathrm L_{r,q}}[x,r)\times [y,q).\end{aligned}$$

Proof

For every rational numbers $r'\le r$ and $q'\le q$, consider the numbers

$$\begin{aligned} \underline{y}(r'):=\inf \{y:\langle x,y\rangle \in \mathrm L_{r,q},\;x<r'\} \text{ and } \underline{x}(q'):=\inf \{x:\langle x,y\rangle \in \mathrm L_{r,q},\;y<q'\}.\end{aligned}$$

Choose countable subsets $\mathrm L_{r,q}^{r'\!,0},L_{r,q}^{0,q'}\subseteq \mathrm L_{r,q}$ such that

$$\begin{aligned}\underline{y}(r')=\inf \{y:\langle x,y\rangle \in \mathrm L^{r'\!,0}_{r,q},\;x<r'\} \text{ and } \underline{x}(q')=\inf \{x:\langle x,y\rangle \in \mathrm L^{0,q'}_{r,q},\;y<q'\} \end{aligned}$$

and moreover,

$$\begin{aligned}\underline{y}(r')=\min \{y:\langle x,y\rangle \in \mathrm L^{r'\!,0}_{r,q},\;x<r'\} \text{ if } \underline{y}(r')=\min \{y:\langle x,y\rangle \in \mathrm L_{r,q},\;x<r'\}. \end{aligned}$$

and

$$\begin{aligned} \underline{x}(q')=\min \{x:\langle x,y\rangle \in \mathrm L^{0,q'}_{r,q},\;y<q'\} \text{ if } \underline{x}(q')=\min \{x:\langle x,y\rangle \in \mathrm L_{r,q},\;y<q'\}.\end{aligned}$$

Consider the countable subset

$$\begin{aligned}\mathrm L_{r,q}'':=\textstyle \bigcup \{\mathrm L^{r'\!,0}_{r,q}\cup \mathrm L_{r,1}^{0,q'}:r',q'\in \mathbb Q,\; r'<r,\;q'<q\}\end{aligned}$$

of $\mathrm L_{r,q}$.

Claim 11

$\displaystyle \bigcup \limits _{\langle x,y\rangle \in L_{r,q}}\big ([x,r)\times [y,q)\big )\setminus \{\langle x,y\rangle \}\subseteq \bigcup \limits _{\langle x,y\rangle \in \mathrm L_{r,q}''}[x,r)\times [y,q).$

Proof

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

• 1. $x<x'<r$ and $y<y'<q$;

• 2. $x=x'$ and $y<y'<q$;

• 3. $x<x'<r$ and $y=y'$.

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

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

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

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

Claim 11 implies that the set

$$\begin{aligned}D_{r,q}=\left( \bigcup _{\langle x,y\rangle \in \mathrm L_{r,q}}[x,r)\times [y,q)\right) \setminus \left( \bigcup _{\langle x,y\rangle \in \mathrm L_{r,q}''}[x,r)\times [y,q)\right) \end{aligned}$$

is contained in $\mathrm L_{r,q}.$ $\square$

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 $\langle x,y\rangle ,\langle x,y'\rangle \in D_{r,q}$ with $y<y'$. Applying Claim 11, we conclude that

$$\begin{aligned}\langle x,y'\rangle \in \big ([x,r)\times [y,q)\big )\setminus \{\langle x,y\rangle \}\subseteq \bigcup _{\langle x'',y''\rangle \in L_{r,q}''}[x'',r)\times [y'',q)\end{aligned}$$

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

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

$$\begin{aligned}\langle x',y'\rangle \in \big ([x,r)\times [y,q)\big )\setminus \{\langle x,y\rangle \}\subseteq \bigcup _{\langle x'',y''\rangle \in L_{r,q}''}[x'',r)\times [y'',q)\end{aligned}$$

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

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

$$\begin{aligned}D_{r,q}'=\textstyle \bigcup \{D_{r,q}\cap f_\beta :\beta \le \alpha ,\;f_\beta \hbox { is strictly increasing}\}\end{aligned}$$

is at most countable. We claim that $D_{r,q}=D'_{r,q}$. To derive a contradiction, assume that $D_{r,q}\setminus D'_{r,q}$ contains some pair $z=\langle x,y\rangle$. It follows from $z\in D_{r,q}\subseteq f_\alpha$ that $\alpha _z\le \alpha$. Since $z\notin D_{r,q}'$, 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 $z\in D_{r,q}\subseteq \mathrm L_{r,q}\subseteq \mathrm L$. $\square$

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

$$\begin{aligned}\displaystyle \bigcup \limits _{\langle x,y\rangle \in L_{r,q}}\big ([x,r)\times [y,q)\big )\subseteq \bigcup \limits _{\langle x,y\rangle \in \mathrm L_{r,q}''}[x,r)\times [y,q).\end{aligned}$$

This completes the proof of Claim 10.

Claim 14

There exists a countable subset $\mathrm L'\subseteq \mathrm L_V$ such that

$$\begin{aligned}\bigcup _{\langle x,y\rangle \in \mathrm L'}\big ([x,a(x,y))\times [y,b(x,y)\big )= \bigcup _{\langle x,y\rangle \in \mathrm L_V}\big ([x,a(x,y))\times [y,b(x,y)\big ).\end{aligned}$$

Proof

By Claim 10, for any rational numbers $r,q\in \mathbb Q$ there exists a countable subset $\mathrm L_{r,q}'\subseteq \mathrm L_{r,q}$ such that

$$\begin{aligned} \bigcup _{\langle x,y\rangle \in \mathrm L'_{r,q}}\big ([x,a(x,y))\times [y,b(a,y))\big )=\bigcup _{\langle x,y\rangle \in \mathrm L'_{r,q}}\big ([x,r)\times [y,q)\big )=\\ =\bigcup \limits _{\langle x,y\rangle \in \mathrm L_{r,q}}\big ([x,r)\times [y,q)\big )= \bigcup \limits _{\langle x,y\rangle \in \mathrm L_{r,q}}\big ([x,a(x,y))\times [y,b(x,y))\big ). \end{aligned}$$

Since $\mathrm L_V=\bigcup _{r,q\in \mathbb Q}\mathrm L_{r,q}$, the countable set $\mathrm L':=\bigcup _{r,q\in \mathbb Q}\mathrm L'_{r,q}$ has the required property. $\square$

By analogy with Claim 14 we can prove

Claim 15

There exists a countable subset $\Gamma '\subseteq \Gamma _V$ such that

$$\begin{aligned} \bigcup _{\langle x,y\rangle \in \Gamma '}\big ([x,a(x,y))\times (b(x,y),y]\big )= \bigcup _{\langle x,y\rangle \in \Gamma _V}\big ([x,a(x,y))\times (b(x,y),y]\big ).\end{aligned}$$

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