Selection properties of the split interval and the Continuum Hypothesis

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.


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 multimap { x, f (x) : For a multimap Φ : X ⊸ Y , its inverse multimap Φ −1 : Y ⊸ X is defined by Φ −1 := { y, x : x, y ∈ Φ}.
A multimap Φ : X ⊸ Y is called • compact-valued if for every x ∈ X the subspace Φ(x) of Y is compact and non-empty; • 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.
We recall that a function f : X → Y between topological spaces is called Borel-measurable (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 [2, 5.0.1] (see also [14, §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 [2, 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 [8].
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 metruzable space is F σ -perfect.
The following selection theorem will be proved in Section 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, 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 [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 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 the Todorčević dichotomy for Rosenthal compact spaces, 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 [16], 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 [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 squareÏ 2 contains a discrete subspace of cardinality continuum and hence has uncountable spread.

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 ≤ 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 ′ , which contradicts the assumption. 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 order-convex 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. Then for any usco multmap Φ : X ⊸ Y we can define a selection f : X → Y of Φ assigning to each point x ∈ X the smallest element f (x) of the non-empty compact set Φ(x) ⊂ Y . We claim that this selection is F σ -measurable.
Proof. If L has the largest element λ, then L = (←, λ] and f −1 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 ⊂ L (which means that for every x ∈ L there exists y ∈ C with x ≤ y). By Lemma 2, for every c ∈ C the preimage Since the space Y is countably cellular, the family C is at most countable. By Claim 4, for every C ∈ C the preimage f −1 (C) is Borel (an type F σ -set if X is F σ -perfect) and so is the countable union f −1 [U ] = C∈C f −1 .

Claim 4. For any open order
Claim 5 completes the proof of Lemma 2.
Lemma 3. Every usco multimap Φ : X ⊸ Y from a countably cellular (F σ -perfect) topological space X into a GO-space Y has a Borel (F σ -measurable) selection.
Proof. The Kuratowski-Zorn Lemma implies that the usco map Φ contains a minimal usco map Ψ : X ⊸ Y . We claim that the image Ψ[X] ⊂ Y is a countably cellular subspace of Y . Assuming the opposite, we can find an uncountable disjoint family (U α ) α∈ω1 of non-empty open subsets in Ψ[X]. For every α ∈ ω 1 , find x α ∈ X such that Φ(x α )∩U α = ∅. By Lemma 3.1.2 [2], the minimality of the usco map Ψ implies that Ψ[V α ] ⊂ U α for some non-empty open set V α ⊂ X. Taking into account that the family (U α ) α∈ω1 is disjoint, we conclude that the family (V α ) α∈ω1 is disjoint, witnessing that the space X is not countably cellular. But this contradicts our assumption. This contradiction shows that the GO-subspace Ψ[X] of Y is countably cellular. By Lemma 2, the usco map Ψ : X → Ψ[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. x ∈ C the order-convexity of the interval I x ⊂Ï implies that its preimage s −1 [I x ] is a convex subset of I, containing x. Since convex subsets of I are of type F σ , the countable union s −1 [U ] = x∈C I x is an F σ -set in I.

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 ordered pairs of elements 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 [1, 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 every a ∈ (0, 2) ⊂ R consider the lines L a = { x, y ∈ R 2 : x + y = a} and Γ 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.

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 : First we introduce some terminology related to monotone functions. A subset f ⊂ I 2 is called a (1) a function if for any x 1 , y 1 , x 2 , y 2 ∈ f the equality x 1 = x 2 implies y 1 = y 2 ; (2) strictly increasing if for any x 1 , y 1 , x 2 , y 2 ∈ f the strict inequality x 1 < x 2 implies y 1 < y 2 ; (3) strictly decreasing if for any x 1 , y 1 , x 2 , y 2 ∈ f the inequality x 1 < x 2 implies y 1 > y 2 ; (4) 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 D f = f ∩(pr 1 [D f ]×pr 2 [D f ]), where pr 1 , pr 2 : I 2 → I are coordinate projections. By the Lavrentiev Theorem [10, 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 . By analogy we can prove 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 M = {f α } α<ω1 . It is clear α<ω1 f α = I 2 . So, for any point z ∈ I 2 we can find the smallest ordinal α z < ω 1 such that z ∈ f αz . Consider the sets L := {z ∈ I 2 : f αz is strictly increasing} and Γ := {z ∈ I 2 : f αz is strictly decresing} = I 2 \ L.
Define a selection s : I 2 →Ï 2 of the multimap P −1 : I 2 ⊸Ï 2 letting s( x, y ) = x 1 , y 1 if x, y ∈ L, We claim that the function s : I 2 →Ï 2 is F σ -measurable. Given any open set U ⊂Ï 2 , we should prove that its preimage s −1 [U ] of type F σ in I 2 . Consider the open subset V := U ∩ (0 1 , 1 0 ) 2 ⊂Ï 2 of U . Using Lemma 4, it can be shown that the set s −1 [U \ V ] ⊂ I 2 \ (0, 1) 2 is of type F σ in I 2 . Therefore, it remains to show that the preimage s −1 [V ] is of type F σ in I 2 .
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 Γ V := Γ ∩ 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 ∈ Γ 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 x, a(x, y) × b(x, y), y .
This equality and the following claim imply that the set s −1 [V ] is of type F σ in I 2 .
We shall show how to find the countable set L ′ ⊂ L V . The countable set Γ ′ ⊂ Γ V can be found by analogy.
For rational numbers r, q ∈ Q, consider the set L r,q = { x, y ∈ L V : a(x, y) = r, b(x, y) = q} 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 implies that the set
is contained in 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 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.
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 αz ∋ z is not strictly increasing and hence f αz 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 x,y ∈Lr,q [x, r) × [y, q) ⊂ x,y ∈L ′′ r,q [x, r) × [y, q).
This completes the proof of Claim 10.
Proof. By Claim 10, for any rational numbers r, q ∈ Q there exists a countable subset L ′ r,q ⊂ L r,q such that Since L V = r,q∈Q L r,q , the countable set L ′ := r,q∈Q L ′ r,q has the required property. By analogy with Claim 14 we can prove Claim 15. There exists a countable subset Γ ′ ⊂ Γ V such that

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.