Selection properties of the split interval and the Continuum Hypothesis

We prove that every usco multimap Φ:X→Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varPhi :X\rightarrow Y$$\end{document} from a metrizable separable space X to a GO-space Y has an Fσ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_\sigma $$\end{document}-measurable selection. On the other hand, for the split interval I¨\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\ddot{\mathbb I}}$$\end{document} and the projection P:I¨2→I2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P:{{\ddot{\mathbb I}}}^2\rightarrow \mathbb I^2$$\end{document} of its square onto the unit square I2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb I^2$$\end{document}, the usco multimap P-1:I2⊸I¨2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${P^{-1}:\mathbb I^2\multimap {{\ddot{\mathbb I}}}^2}$$\end{document} has a Borel (Fσ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F_\sigma $$\end{document}-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.


Introduction
By a multimap Φ : X Y between topological spaces X , Y we understand any subset Φ ⊆ X × Y , which can be thought as a function assigning to every point x ∈ X the subset Φ(x) := {y ∈ Y : x, y ∈ Φ} of Y . For a subset A ⊆ X we put Φ[A] = x∈A Φ(x). Each function f : X → Y can be thought as a single-valued For a multimap Φ : X Y , its inverse multimap Φ −1 : Y X is defined by Ivan Franko National University in Lviv, Lviv, Ukraine -Borel-measurable if for any Borel set B ⊆ Y the set Φ −1 [B] is Borel in X ; compact-valued if for every x ∈ X the subspace Φ(x) of Y is compact and nonempty; usco if Φ is upper semicontinuous and compact-valued.
It is well-known that for any surjective continuous function f : X → Y between compact Hausdorff spaces, the inverse multimap f −1 : Y X is usco. Let Φ : X Y be a multimap between topological spaces. A function f : X → Y is called a selection of Φ if f (x) ∈ Φ(x) for every x ∈ X . The Axiom of Choice ensures that every multimap Φ : X Y with non-empty values has a selection. The problem is to find selections possessing some additional properties like the continuity or measurability.

Theorem 1 Let X , Y be Polish spaces. Any Borel-measurable multimap Φ : X Y with non-empty values has a Borel-measurable selection.
We recall that a function f : X → Y between topological spaces is called Borelmeasurable (resp. F σ -measurable) if for every open set U ⊆ Y the preimage f −1 [U ] is Borel (or type F σ ) in X .
Let us recall [3, 5.0.1] (see also [15, §6]) that a topological space K is fragmentable if K has a metric ρ such that for every ε > 0 each non-empty subset A ⊆ K contains a non-empty relatively open set U ⊆ A of ρ-diameter < ε. By [3, 5.1.12], each fragmentable compact Hausdorff space contains a metrizable dense G δ -subspace.
The following selection theorem can be deduced from Theorem 1' and Lemma 6 in [9]. Theorem 2 (Hansell, Jayne, Talagrand) Any usco map Φ : X → K from a perfectly paracompact space X to a fragmentable compact space Y has an F σ -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 ≤ 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 ≤ y in C the order interval [x, y] := {z ∈ X : x ≤ z ≤ y} is contained in X . We say that the topology of X is generated by the linear order ≤ if the topology of X is generated by the subbase {(←, a), (a, →) : a ∈ X } consisting the the order-convex sets (←, a) := {x ∈ X : x < a} and (a, →) = {x ∈ 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 σ -perfect if every open set in X is of type F σ in X (i.e., can be represented as the countable union of closed sets). For example, every metrizable space is F σ -perfect.
The following selection theorem will be proved in Sect. 2.
Theorem 3 Let Y be a GO-space and X be an (F σ -perfect) topological space. If X or Y is countably cellular, then any usco map Φ : X Y has a Borel (F σ -measurable) selection.
Problem 1 Is it true that any usco map Φ : M K from a compact metrizable space M to a compact Hausdorff space K has a Borel (F σ -measurable) selection?
In this paper we prove that this problem has negative answer under the negation of the Continuum Hypothesis (i.e., under ω 1 < c). A suitable counterexample will be constructed using the split squareÏ 2 , which is the square of the split intervalÏ.
The split interval is the linearly ordered spaceÏ = [0, 1] × {0, 1} whose topology is generated by the lexicographic order (defined by x, i ≤ y, j 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 [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 Φ : X Ï from an F σ -perfect topological space X has an F σ -measurable selection. In contrast, the split squareÏ 2 has dramatically different selections properties. Let p :Ï → I, p : x, i → x, be the natural projection of the split interval onto the unit interval I = [0, 1], and P :Ï 2 → I 2 , P : x, y → p(x), p(y) , be the projection of the split squareÏ 2 onto the unit square I 2 .
Combining Theorem 4 with Theorem 4 in [17], we obtain the following consistent characterization of metrizable compacta.

Corollary 1
Under ω 1 < c a Rosenthal compact space K is metrizable if and only if K has countable spread and each usco multimap Φ : I 2 → 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 intervalÏ. We lose no generality assuming thatÏ ⊆ K . By Theorem 4, under ω 1 < c, the usco multimap P −1 : I 2 Ï 2 ⊆ 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 Φ : X →Ï 2 from a metrizable (separable) space X has a Borel-measurable selection?
Observe that the map p :Ï → I is 2-to-1 and its square P : 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 squareÏ 2 contains a discrete subspace of cardinality continuum and hence has uncountable spread.
Remark 1 hris Heunen asked on Mathoverflow [10] whether for any continuous surjective function π : X → Y between compact Hausdorff spaces and any continuous map γ : I → Y there exists a Borel-measurable function g : I → X such that γ = π •g. Theorem 4 provides a consistent counterexample to this problem of Heunen. Indeed, consider the projection P :Ï 2 → I 2 and take any continuous surjective map γ : I → I 2 (whose existence was proved by Giuseppe Peano in 1890). By Theorem 2, the inverse multimap γ −1 : I 2 I has a Borel-measurable selection s : I 2 → I. Assuming that there exists a Borel-measurable function g : I →Ï 2 with γ = P •g, we conclude that g • s : I 2 →Ï 2 is a Borel-measurable selection of the multimap P −1 , with contradicts Theorem 4 under CH.

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 ≤ of any GO-space X is a closed subset of the square X × X.
Proof Given two elements x, y ∈ X with x ≤ y, use the Hausdorff property of X and find two disjoint order-convex neighborhoods O x , O y ⊆ X of the points x, y, respectively. We claim that the product O x × O y is disjoint with the linear order ≤. Assuming that this is not true, find elements x ∈ O x and y ∈ O y such that x ≤ 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 ≤ y that y < x. Then x < y < x < y and hence [y, x] ⊆ O x ∩ O y = ∅, which is a desired contradiction. This contradiction shows that the neighborhood O x × O y of the pair x, y is disjoint with ≤ and hence ≤ is a closed subset of X × X .
Proof Being a GO-space, Y has a base of the topology consisting of open orderconvex subsets with respect to some linear order ≤ on Y . By Lemma 1, the linear order ≤ is a closed subset of Y × Y . Then for every a ∈ Y the order-convex set (←, a] = {y ∈ Y : y ≤ a} is closed in Y , which implies that each non-empty compact subset of Y has the smallest element.

Claim 3 For any lower
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 ⊆ L (which means that for every x ∈ L there exists y ∈ C with x ≤ y). By Lemma 2, for

Proof
Claim 5 completes the proof of Lemma 2.

Lemma 3 Every usco multimap
Proof The Kuratowski-Zorn Lemma implies that the usco map Φ contains a minimal usco map : X Y . We claim that the image [X ] ⊆ Y is a countably cellular subspace of Y . Assuming the opposite, we can find an uncountable disjoint family (U α ) α∈ω 1 of non-empty open subsets in [X ]. For every α ∈ ω 1 , find x α ∈ X such that Φ(x α ) ∩ U α = ∅. By Lemma 3.1.2 [3], the minimality of the usco map implies that [V α ] ⊆ U α for some non-empty open set V α ⊆ X . Taking into account that the family (U α ) α∈ω 1 is disjoint, we conclude that the family (V α ) α∈ω 1 is disjoint, witnessing that the space X is not countably cellular. But this contradicts our assumption. This contradiction shows that the GO-subspace [X ] of Y is countably cellular. By Lemma 2, the usco map : X → [X ] has a Borel (F σ -measurable) selection, which is also a selection of the usco map Φ.
Finally, let us prove one selection property of the split interval, which will be used in the proof of Lemma 8.

Selection properties of the split squareÏ 2 under the negation of CH
In this section we study the selection properties of the split squareÏ 2 under the negation of the Continuum Hypothesis.
By x, y 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 intervalÏ = I × {0, 1} carries the lexicographic order defined by x, i ≤ y, j iff either x < y or (x = y and i ≤ j). It is well-known that the topology generated by the lexicographic order onÏ is compact and Hausdorff, see [2, 3.10.C(b)]. By p :Ï → I, p : x, i → x, we denote the coordinate projection and by P :Ï 2 → I 2 , P : x, y → p(x), p(y) the square of the map p.
Proof To derive a contradiction, assume that the multimap P −1 has a Borel-measurable selection s : For a real number x ∈ I by x 0 and x 1 we denote the points x, 0 and x, 1 of the split intervalÏ.
For any numbers i, j ∈ {0, 1} consider the set For a point a ∈Ï, let [0 0 , a) and (a, 1 1 ] be the order intervals inÏ with respect to the lexicographic order. Observe that for any x ∈ I we have For every a ∈ (0, 2) ⊆ R consider the lines L a = { x, y ∈ R 2 : x + y = a} and a = { x, y ∈ R 2 : y − x = a} on the plane.

Claim 6
For every a ∈ R the intersection L a ∩ Z 00 is at most countable.
Proof If for some a ∈ R the intersection L a ∩ Z 00 is uncountable, then we can choose a non-Borel subset B ⊆ L a ∩ Z 00 of cardinality |B| = ω 1 . For every point x, y ∈ B ⊆ Z 00 , the definition of the set Z 00 ensures that s( x, y ) = x 0 , y 0 and hence the set By analogy we can prove the following claims.

Claim 7
For every a ∈ R the intersection L a ∩ Z 11 is at most countable.

Claim 8 For every b ∈ R the intersection b ∩ (Z 01 ∪ Z 10 ) is at most countable.
Now fix any subset set ⊆ [ 1 2 , 3 2 ] of cardinality | | = ω 1 . By Claims 6, 7, for every a ∈ the intersection L a ∩ (Z 00 ∪ Z 11 ) is at most countable. Consequently the union has cardinality |U | ≤ ω 1 . Since |U | ≤ ω 1 < c, there exists a real number b ∈ [ 1 2 , 3 2 ] such that the line b does not intersect the set U . Since {b} ∪ ⊆ [ 1 2 , 3 2 ] for every a ∈ the intersection b ∩L a ∩I 2 is not empty. Then the set X = a∈ L a ∩ b ⊂ I 2 is uncountable and X ⊆ b \ U ⊆ b ∩ (Z 01 ∪ Z 10 ). But this contradicts Claim 8.

Selection properties of the split squareÏ 2 under the Continuum Hypothesis
In this section we shall prove that under the continuum hypothesis the usco multimap P −1 : I 2 Ï 2 has an F σ -measurable selection.
First we introduce some terminology related to monotone functions. A subset f ⊆ I 2 is called -a function if for any x, y , x , y ∈ f the equality x = x implies y = y ; strictly increasing if for any x, y , x , y ∈ f the strict inequality x < x implies y < y ; strictly decreasing if for any x, y , x , y ∈ 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 ⊂ I 2 is a subset of a Borel strictly increasing functionf ⊂ I 2 .
Proof It follows that the strictly increasing function f is a strictly increasing bijective function between the sets pr 1 [ f ] = {x ∈ I : ∃y ∈ I x, y ∈ f } and pr 2 [ f ] = {y ∈ I : ∃x ∈ I x, y ∈ 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 ⊆ f such that the set f \ D f coincides with the graph of some increasing homeomorphism between subsets of I. Replacing D f by a larger countable set, we can assume that , where pr 1 , pr 2 : I 2 → I are coordinate projections. By the Lavrentiev Theorem [11, 3.9], the homeomorphism f \ D f extends to a (strictly increasing) homeomorphism h ⊆ I 2 between G δ -subsets of I 2 such that f \ D f is dense in h. It is easy to check that the Borel subsetf = (h \ (pr 1 [D f ] × pr 2 [D f ]) ∪ D f is a strictly increasing function extending f .

Lemma 7
Each strictly decreasing function f ⊂ I 2 is a subset of a Borel strictly decreasing functionf ⊂ I 2 .
Now we are ready to prove the main result of this section.
Proof Let M be the set of infinite strictly monotone Borel functions f ⊂ I 2 .
Since ω 1 = c, the set M can be written as So, for any point z ∈ I 2 we can find the smallest ordinal α z < ω 1 such that z ∈ f α z . Consider the sets L := {z ∈ I 2 : f α z is strictly increasing} and Define a selection s : I 2 →Ï 2 of the multimap P −1 : We claim that the function s : Let Q := { n m : n, m ∈ N, n < m} be the set of rational numbers in the interval (0, 1).
Consider the subsets L V := L ∩ s −1 (V ) and V := ∩ s −1 (V ). For every x, y ∈ L V we have s( x, y ) = x 1 , y 1 ∈ V and by the definition of the topology of the split interval, we can find rational numbers a(x, y), b(x, y) ∈ Q such that x < a(x, y), y < b(x, y) and s( x, y ) = x 1 , On the other hand, for every x, y ∈ V there are rational numbers a(x, y), b(x, y) ∈ Q such that x < a(x, y), b(x, y) < y and s( x, y ) = x 1 , It follows that This equality and the following claim imply that the set s −1 [V ] is of type F σ in I 2 .
Claim 9 There are countable subsets L ⊆ L V and ⊆ V such that We shall show how to find the countable set L ⊆ L V . The countable set ⊆ V can be found by analogy.
For rational numbers r , q ∈ Q, consider the set and observe that L V = r ,q∈Q L p,q .

Claim 10
For any rational numbers r , q ∈ Q there exists a countable subset L r ,q ⊆ L r ,q such that Proof For every rational numbers r ≤ r and q ≤ q, consider the numbers y(r ) := inf{y : x, y ∈ L r ,q , x < r } and x(q ) := inf{x : x, y ∈ L r ,q , y < q }.

Claim 11
x,y ∈L r ,q Proof Fix any pairs x, y ∈ L r ,q and x , y ∈ [x, r ) × [y, q) \ { x, y }. Three cases are possible: 1.
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 x(q ) ensures that x(q ) ≤ x < x . By the choice of the family L 0,q r ,q , there exists x , y ∈ L 0,q r ,q ⊂ L r ,q such that x < x < r and y < q < y < q. Then x , y ∈ [x , r ) × [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 x(q ) ≤ x = x . If x(q ) < x , then by the definition of the family L 0,q So, we assume that x(q ) = x = x and hence x(q ) = x = min{x : x , y ∈ L p,q : y < q }. In this case x = x(q ) = x for some x , y ∈ L 0,q r ,q ⊂ L r ,q with y < q < y < q. Then x , y ∈ [x , r ) × [y , q).
By analogy, in the third case (x < x < r and y = y ) we can find a pair x , y ∈ L r ,q such that x , y ∈ [x , r ) × [y , q).
Claim 11 implies that the set

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 x, y , x, y ∈ D r ,q with y < y . Applying Claim 11, we conclude that and hence x, y / ∈ D r ,q , which contradicts the choice of the pair x, y . This contradiction shows that D r ,q is a function.

T. Banakh
Assuming that D r ,q is not strictly decreasing, we can find pairs x, y , x , y ∈ D r ,q such that x < x and y ≤ y . Applying Claim 11, we conclude that and hence x , y / ∈ D r ,q , which contradicts the choice of the pair x, y . 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 α for some ordinal α < ω 1 . Since the intersection of a strictly increasing function and a strictly decreasing function contains at most one point, the set D r ,q = {D r ,q ∩ f β : β ≤ α, f β is strictly increasing} is at most countable. We claim that D r ,q = D r ,q . To derive a contradiction, assume that D r ,q \ D r ,q contains some pair z = x, y . It follows from z ∈ D r ,q ⊆ f α that α z ≤ α. Since z / ∈ D r ,q , the strictly monotone function f α z z is not strictly increasing and hence f α z is strictly decreasing. Then the definition of the set L guarantees that z / ∈ L, which contradicts the inclusion z ∈ D r ,q ⊆ L r ,q ⊆ L. Now consider the countable subset L r ,q := L r ,q ∪ D r ,q of L r ,q and observe that Proof By Claim 10, for any rational numbers r , q ∈ Q there exists a countable subset L r ,q ⊆ L r ,q such that  (x, y)) .
Since L V = r ,q∈Q L r ,q , the countable set L := r ,q∈Q L r ,q has the required property.
Claims 14 and 15 complete the proof of Claim 9 and the proof of Lemma 8.