Uniquely Universal Sets

Arnold W. Miller

Abstract^{1}

We say that X×Y satisfies the Uniquely Universal property (UU) iff there exists an open set U ⊆ X × Y such that for every open set W ⊆ Y there is a unique cross section of U with U_{x} = W. Michael Hrus̆ák raised the question of when does X × Y satisfy UU and noted that if Y is compact, then X must have an isolated point. We consider the problem when the parameter space X is either the Cantor space 2^{ω} or the Baire space ω^{ω}. We prove the following:

1. If Y is a locally compact zero dimensional Polish space which is not compact, then 2^{ω} × Y has UU.

2. If Y is Polish, then ω^{ω} × Y has UU iff Y is not compact.

3. If Y is a σ-compact subset of a Polish space which is not compact, then ω^{ω} × Y has UU.

For any space Y with a countable basis there exists an open set U ⊆ 2^{ω}×Y which is universal for open subsets of Y, i.e., W ⊆ Y is open iff there exists x ∈ 2^{ω} with

U_{x} =^{def} {y ∈ Y : (x, y) ∈ U} = W.

To see this let {B_{n} : n < ω} be a basis for Y. Define

(x, y) ∈ U iff ∃n (x(n) = 1 and y ∈ B_{n}).

More generally if X contains a homeomorphic copy of 2^{ω} then X × Y will have a universal open set.

In 1995 Michael Hrušák mentioned the following problem to us. Most of the results in this note were proved in June and July of 2001.

1MSC2010: 03E15 Keywords: Universal set, Unique parameterization, Polish spaces, Cantor space, Baire space.

Last revised January 12, 2012.

Hrušák ’s problem.

Let X, Y be topological spaces, call X the parameter space, and Y the base space. When does there exists U ⊆ X × Y which is uniquely universal for the open subsets of Y? This means the U is open and for every open set W ⊆ Y there is a unique x ∈ X such that U_{x} = W.

Let us say that X × Y satisfies UU (uniquely universal property) if there exists such an open set U ⊆ X × Y which uniquely parameterizes the open subsets of Y. Note that the complement of U is a closed set which uniquely parameterizes the closed subsets of Y.

Proposition 1 (Hrušák ) 2^{ω} × 2^{ω} does not satisfy UU.

proof:

The problem is the empty set. Suppose U is uniquely universal for the closed subsets of 2^{ω}. Then there is an x_{0} such that U_{x_{0}} = ∅ but all other cross sections are nonempty. Take x_{n} → x_{0} but distinct from it. Since all other cross sections are non-empty we can choose y_{n} ∈ U_{x_{n}}. But then y_{n} has a convergent subsequence, say to y_{0}, but then y_{0} ∈ U_{x_{0}}. QED

More generally:

Proposition 2 (Hrušák ) Suppose X × Y has UU and Y is compact. Then X must have an isolated point.

proof:

Suppose U ⊆ X × Y witnesses UU for closed subsets of Y and U_{x_{0}} = ∅. For every y ∈ Y there exists U_{y} × V_{y} open containing (x_{0}, y) and missing U. By compactness of Y finitely many V_{y} cover Y. The intersection of the corresponding U_{y} isolates x_{0}. QED

Hence, for example, 2^{ω} × (ω + 1), ω^{ω} × (ω + 1), and ω^{ω} × 2^{ω} cannot have UU.

Proposition 3 Let 2^{ω} ⊕ 1 be obtained by attaching an isolated point to 2^{ω}. Then (2^{ω} ⊕ 1) × 2^{ω} has UU.

proof:

Define T ⊆ 2^{<ω} to be a nice tree iff

(a) s ⊆ t ∈ T implies s ∈ T and

(b) if s ∈ T, then either sˆ ⟨0⟩ or sˆ ⟨1⟩ in T.

Let NT ⊆ P(2^{<ω}) be the set of nice trees. Define the universal set U by

U = {(T, x) ∈ NT × 2^{ω} : ∀n x↾n ∈ T }.

Note that the empty tree T is nice and parameterizes the empty set. Also NT is a closed subset of P(2^{<ω}) with exactly one isolated point (the empty tree), and hence it is homeomorphic to 2^{ω} ⊕ 1. QED

Question 4 Does (2^{ω} ⊕ 1) × [0, 1] have UU?

Remark 5 2^{ω} × ω has the UU property. Just let (x, n) ∈ U iff x(n) = 1.

Question 6 Does either ℝ × ω or [0, 1] × ω have UU? Or more generally, is there any example of UU for a connected parameter space?

Recall that a topological space is Polish iff it is completely metrizable and has a countable dense subset. A set is G_{δ} iff it is the countable intersection of open sets. The countable product of Polish spaces is Polish. A G_{δ} subset of a Polish space is Polish (Alexandrov). A space is zero-dimensional iff it has a basis of clopen sets. All compact zero-dimensional Polish spaces without isolated points are homeomorphic to 2^{ω} (Brouwer). A zero-dimensional Polish space is homeomorphic to ω^{ω} iff compact subsets have no interior (Alexandrov-Urysohn). For proofs of these facts see Kechris [5] p.13-39.

Proposition 7 Suppose Y is a zero dimensional Polish space. If Y is locally compact but not compact, then 2^{ω} ×Y has UU. So, for example, 2^{ω} ×(ω ×2^{ω}) has UU. proof:

Let B be a countable base for Y consisting of clopen compact sets. Define G ⊆ B is good iff

G = {b ∈ B : b ⊆ ⋃ G}

Let G ⊆ P(B) be the family of good subsets of B. We give the P(B) the topology from identifying it with 2^{B}. Since B is an infinite countable set P(B) is homeomorphic to 2^{ω}. A sequence G_{n} for n < ω converges to G iff for each finite F ⊆ B we have that G_{n} ∩ F = G ∩ F for all but finitely many n. Hence G is a closed subset of P(B) since by compactness b ⊆ ⋃ G iff b ⊆ ⋃ F for some finite F ⊆ G.

There is a one-to-one correspondence between good families and open subsets of Y: Given any open set U ⊆ Y define

G_{U} = {b ∈ B : b ⊆ U}

and given any good G define U_{G} = ⋃ G. (Note that the empty set is good.)

We claim that no G_{0} ∈ G is an isolated point. Suppose for contradiction it is. Then there must be a basic open set N with {G_{0}} = G ∩ N. A basic neighborhood has the following form

N = N(F_{0}, F_{1}) = {G ⊆ B : F_{0} ⊆ G and F_{1} ∩ G = ∅}

where F_{0}, F_{1} ⊆ B are finite.

For each b ∈ F_{1} since G_{0} is good, b is not a subset of ⋃ G_{0}, and since ⋃ F_{0} ⊆ ⋃ G_{0}, we can choose a point z_{b} ∈ b \ ⋃ F_{0}. Since Y is not compact, Y \ (⋃(F_{0} ∪ F_{1})) is nonempty. Fix z ∈ Y \ (⋃(F_{0} ∪ F_{1})).

Now let U_{1} = Y \ {z_{b} : b ∈ F_{1}} and let U_{2} = U_{1} \ {z}. Then G_{U_{1}}, G_{U_{2}} are distinct elements of N ∩ G.

Hence G is a compact zero-dimensional metric space without isolated points and therefore it is homeomorphic to 2^{ω}.

To get a uniquely universal open set U ⊆ G × Y define:

(G, y) ∈ U iff ∃b ∈ G y ∈ b.

QED

Example 26 is a countable Polish space Z such that 2^{ω} × Z has UU, but Z is not locally compact.

Lemma 8 Suppose f : X → Y is a continuous bijection and Y × Z has UU. Then X × Z has UU.

proof:

Given V ⊆ Y × Z witnessing UU, let

U = {(x, y) : (f(x), y) ∈ V }.

QED

Many uncountable standard Borel sets^{2} are the bijective continuous image of the Baire space ω^{ω}. According to the footnote on page 447 of Kuratowski [6] Sierpinski proved in a 1929 paper that any standard Borel set in which every point is a condensation point is the bijective continuous image of ω^{ω}. We weren’t able to find Sierpinski’s paper but we give a proof of his result in Lemma 21.

We first need a special case for which we give a proof.

Lemma 9 There is a continuous bijection f : ω^{ω} → 2^{ω}.

proof:

Let π : ω → ω + 1 be a bijection. It is automatically continuous. It induces a continuous bijection π : ω^{ω} → (ω + 1)^{ω}. But (ω + 1)^{ω} is a compact Polish space without isolated points, hence it is homeomorphic to 2^{ω}. QED

Remark. If C ⊆ 2^{ω} × ω^{ω} is the graph of f^{−1}, then C is a closed set which uniquely parameterizes the family of singletons of ω^{ω}.

Corollary 10 If 2^{ω} × Y has UU, then ω^{ω} × Y has UU.

Question 11 Is the converse of Corollary 10 false? That is: Does there exist Y such that ω^{ω} × Y has UU but 2^{ω} × Y does not have UU?

Lemma 12 Suppose X is a zero-dimensional Polish space without isolated points. Then there exists a continuous bijection f : ω^{ω} → X. proof:

2A standard Borel set is a Borel subset of a Polish space.

Construct a subtree T ⊆ ω^{<ω} and (C_{s} ⊆ X : s ∈ T) nonempty clopen sets such that:

1. C_{⟨⟩} = X,

2. if s ∈ T is a terminal node, then C_{s} is compact, and

3. if s ∈ T is not terminal, then sˆ ⟨n⟩ ∈ T for every n ∈ ω and C_{s} is partitioned by (C_{sˆ ⟨n⟩} : n < ω) into nonempty clopen sets each of diameter^{3} less than \frac{1}{|s|+1}.

For each terminal node s ∈ T choose a continuous bijection f_{s} : [s] → C_{s} given by Lemma 9. Define f : ω^{ω} → X by f(x) = f_{x↾n}(x) if there exists n such that x↾n is a terminal node of T and otherwise determine f(x) by the formula:

{f(x)} = ⋂_{n<ω} C_{x↾n}

Checking that f is a continuous bijection is left to the reader. QED

Remark. An easy modification of the above argument shows that any zero-dimensional Polish space is homeomorphic to a closed subspace of ω^{ω}. It also gives the classical result that if no clopen sets are compact, then X is homeomorphic to ω^{ω}. A different proof of Lemma 12 is given in Moschovakis [8] p. 12.

Definition 13 We use cl(X) to denote the closure of X.

Proposition 14 Suppose Y is Polish but not compact. Then ω^{ω} × Y has the UU. So for example, ω^{ω} × ω^{ω} and ω^{ω} × ℝ both have UU.

proof:

We assume that the metric on Y is complete and bounded. Let B be a countable basis for Y of nonempty open sets which has the property that no finite subset of B covers Y.

For s, t ∈ B define t ⊲ s iff cl(t) ⊆ s and diam(t) ≤ \frac{1}{2} diam(s).

3This is with respect to a fixed complete metric on X.

Lemma 15 Suppose G ⊆ B has the following properties:

(1) for all t, s ∈ B if t ⊆ s ∈ G, then t ∈ G and

(2) ∀s ∈ B if (∀t ⊲ s t ∈ G), then s ∈ G,

then for any s ∈ B if s ⊆ ⋃ G, then s ∈ G.

proof:

Suppose (1) and (2) hold but for some s ⊆ ⋃ G we have s ∉ G.

Note that there cannot be a sequence (s_{n} : n ∈ ω) starting with s_{0} = s, and with s_{n+1} ⊲ s_{n} and s_{n} ∉ G for each n. This is because if {x} = ⋂_{n∈ω} s_{n}, then x ∈ s ⊆ ⋃ G and so for some t ∈ G we have x ∈ t. But then for some sufficiently large n we have that s_{n} ⊆ t putting s_{n} ∈ G by (1).

Hence there must be some t⊴s with t ∉ G but for all r⊲t we have r ∈ G. This is a contradiction to (2). QED

Let G ⊆ P(B) be the set of all G which satisfy the hypothesis of Lemma 15. Then we have a G_{δ} subset G of 2^{B}, which uniquely parameterizes the open sets. Hence the set U witnesses the unique universal property:

U = {(G, x) ∈ G × Y : x ∈ ⋃ G}.

To finish the proof it is enough to see that no point in G is isolated. If G has an isolated point, there must be s_{1}, …, s_{n} and t_{1}, …, t_{m} from B such that

N = {G ∈ G : s_{1} ∈ G, …, s_{n} ∈ G, t_{1} ∉ G, …, t_{m} ∉ G}

is a singleton. Let W = s_{1} ∪ ⋯ ∪ s_{n}. Since N contains the point

G = {s ∈ B : s ⊆ W }

it must be that N = {G}. For each j choose x_{j} ∈ t_{j} \ W. Since the union of the s_{i} and t_{j} does not cover Y, we can choose

z ∈ Y \ (s_{1} ∪ ⋯ ∪ s_{n} ∪ t_{1} ∪ ⋯ ∪ t_{m}).

Take r ∈ B with z ∈ r but x_{j} ∉ r for each j = 1, …, m. Let

G′ = {s ∈ B : s ⊆ r ∪ W }.

Then G′ ∈ N but G′ ≠ G which shows that N is not a singleton.

It follows from Lemma 8 and 12 that ω^{ω} × Y has the UU. QED

Proposition 2 and 14 show that:

Corollary 16 For Y Polish:

ω^{ω} × Y has UU iff Y is not compact.

Question 17 Does 2^{ω} × ω^{ω} have UU?

Question 18 Does 2^{ω} × ℝ have UU?

Our next result follows from Proposition 22 but has a simpler proof so we give it first.

Proposition 19 If X is a countable metric space which is not compact, then ω^{ω} × X has UU. So, for example, ω^{ω} × ℚ has UU.

proof:

We produce a uniquely universal set for the open subsets of X.

First note that there exists a countable basis B for X with the property that it is closed under finite unions and X \ B is infinite for every B ∈ B. To see this fix {x_{n} : n < ω} ⊆ X an infinite set without a limit point, i.e., an infinite closed discrete set. Given a countable basis B replace it with finite unions of sets from

{B \ {x_{m} : m > n} : n ∈ ω and B ∈ B}.

We may assume also that B includes the empty set.

Next let

ℙ = {(B, F) : B ∈ B, F ∈ [X]^{<ω}, and B ∩ F = ∅}.

Then ℙ is a partial order determined by p ≤ q iff B_{q} ⊆ B_{p} and F_{q} ⊆ F_{p}. For p ∈ ℙ we write p = (B_{p}, F_{p}). For p, q ∈ ℙ we write p ⊥ q to stand for p and q are incompatible, i.e., there does not exist r ∈ ℙ with r ≤ p and r ≤ q.

We will code open subsets of X by good filters on ℙ. Define the family G of good filters on ℙ to be the set of all G ⊆ ℙ such that 1. p ≤ q and p ∈ G implies q ∈ G,

2. ∀p, q ∈ G exists r ∈ G with r ≤ p and r ≤ q,

3. ∀x ∈ X ∃p ∈ G x ∈ B_{p} ∪ F_{p}, and

4. ∀p ∈ ℙ either p ∈ G or ∃q ∈ G p ⊥ q.

Since the poset ℙ is countable we can identify P(ℙ) with P(ω) and hence 2^{ω}. We give G ⊆ P(ℙ) the subspace topology. Note that G is G_{δ} in this topology. Note also that the sets

[p] = {G ∈ G : p ∈ G}

form a basis for G (use conditions (2) and (4) to deal with finitely many p_{i} in G and finitely many q_{j} not in G).

Note that since X \B_{p} is always infinite, for any p ∈ ℙ there exists r, q ≤ p with r ⊥ q. Namely, for some x ∈ X \(B_{p} ∪F_{p}) put x into B_{r} ∩F_{q}. It follows that no element of G is isolated. So G is a zero-dimensional Polish space without isolated points. Hence by Lemma 12 there is a continuous bijection f : ω^{ω} → G.

For G ∈ G, let

U_{G} = ⋃{B_{p} : p ∈ G}.

For any U ⊆ X open, define

G_{U} = {p ∈ ℙ : B_{p} ⊆ U and F_{p} ∩ U = ∅}.

The maps G → U_{G} and U → G_{U} show that there is a one-to-one correspondence between G and the open subsets of X.

Finally define U ⊆ G × X by

(G, x) ∈ U iff ∃p ∈ G x ∈ B_{p}.

This witnesses the UU property for G × X and so by Lemma 8, we have UU for ω^{ω} × X. QED

Question 20 Does 2^{ω} × ℚ have UU?

Our next result Proposition 22 implies Proposition 19 but needs the following Lemma:

Lemma 21 (Sierpinski) Suppose B is a Borel set in a Polish space for which every point is a condensation point. Then there exists a continuous bijection from ω^{ω} to B.

proof:

We use that every Borel set is the bijective image of a closed subset of ω^{ω}. This is due to Lusin-Souslin see Kechris [5] p.83 or Kuratowski-Mostowski [7] p.426.

Using the fact that every uncountable Borel set contains a perfect subset it is easy to construct K_{n} for n < ω satisfying:

1. K_{n} ⊆ B are pairwise disjoint,

2. K_{n} are homeomorphic copies of 2^{ω} which are nowhere dense in B, and

3. every nonempty open subset of B contains infinitely many K_{n}.

Let B_{0} = B \ ⋃_{n<ω} K_{n}. We may assume B_{0} is nonempty, otherwise just split K_{0} into two pieces. Since it is a Borel set, there exists C ⊆ ω^{ω} closed and a continuous bijection f : C → B_{0}. Define

Γ = {s ∈ ω^{<ω} : [s] ∩ C = ∅ and [s^{∗}] ∩ C ≠ ∅}

where s^{∗} is the unique t ⊆ s with |t| = |s| − 1. Without loss we may assume that C is nowhere dense and hence Γ infinite. Let Γ = {s_{n} : n < ω} be a one-one enumeration. Note that {C} ∪ {[s_{n}] : n < ω} is a partition of ω^{ω}.

Inductively choose l_{n} > l_{n−1} with K_{l_{n}} a subset of the ball of radius \frac{1}{n+1} around f(x_{n}) for some x_{n} ∈ C ∩ [s_{n}^{∗}]. For each n < ω let f_{n} : [s_{n}] → K_{l_{n}} be a continuous bijection.

Then g = f ∪⋃_{n<ω} f_{n} is a continuous bijection from ω^{ω} to B_{0} ∪⋃_{n<ω} K_{l_{n}}.

To see that it is continuous suppose for contradiction that u_{n} → u as n → ∞ and |g(u_{n})−g(u)| > ϵ > 0 all n. Since C is closed if infinitely many u_{n} are in C, so is u and we contradict continuity of f. If u ∈ [s_{n}], then we contradict the continuity of f_{n}, So, we may assume that all u_{n} are not in C but u is in C. By the continuity of f we may find s ⊆ u with f([s] ∩ C) inside a ball of radius \frac{ϵ}{3} around f(u). Find n with \frac{1}{n+1} < \frac{ϵ}{3} for which there is m such that u_{m} ∈ [s_{n}] and s_{n} ⊇ s. But then g(u_{m}) = f_{n}(u_{m}) ∈ K_{l_{n}} and K_{l_{n}} is in a ball of radius \frac{1}{n+1} around some f(x_{n}) with x_{n} ∈ [s_{n}^{∗}] ∩ C. This is a contradiction:

d(g(u), g(u_{m})) ≤ d(f(u), f(x_{n})) + d(f(x_{n}), f_{n}(u_{m})) ≤ \frac{2}{3}ϵ.

Next let I = ω \ {l_{n} : n < ω}. Then there exists continuous bijection

h : I × ω^{ω} → ⋃_{i∈I} K_{i}.

Finally g∪h is a continuous bijection from ω^{ω}⊕(I×ω^{ω}) to B_{0}∪⋃_{n<ω} K_{n} = B. Since ω^{ω} ⊕ (I × ω^{ω}) is a homeomorphic copy of ω^{ω} we are done. QED

Proposition 22 ω^{ω}×Y has UU for any σ-compact but not compact subspace Y of a Polish space. So for example, ω^{ω} × (ℚ × 2^{ω}) has UU.

proof:

Let Y = ⋃_{n<ω} K_{n} where each K_{n} is compact. Since Y is not compact it contains an infinite closed discrete set D. Choose a countable basis B for Y such that for any b ∈ B the closure of b contains at most finitely many points of D. Define G ⊆ B to be good iff for every b ∈ B if cl(b) ⊆ ⋃ G then b ∈ G. Let G ⊆ P(B) be the family of good sets.

Note that G is a Π_{3}^{0} set:

G ∈ G iff ∀b ∈ B (∀n cl(b) ∩ K_{n} ⊆ ⋃ G) → b ∈ G

Note that cl(b)∩K_{n} ⊆ ⋃ G iff there is a finite F ⊆ G with cl(b)∩K_{n} ⊆ ⋃ F.

To finish the proof it is necessary to see that basic open sets in G are uncountable. Given b_{i}, c_{j} ∈ B for i < n and j < m suppose that

N = {G ∈ G : b_{0} ∈ G, …, b_{n−1} ∈ G, c_{0} ∉ G, …, c_{m−1} ∉ G}

is nonempty. Since it is nonempty we can choose points

u_{j} ∈ cl(c_{j}) \ ⋃_{i<n} cl(b_{i})

for j < m. Note that the set

Z = D \ cl( ⋃_{i<n,j<m} b_{i} ∪ c_{j})

is an infinite discrete closed set. But then given any Q ⊆ Z we can find an open set U_{Q} with ⋃_{i<n} cl(b_{i}) ⊆ U_{Q}, u_{j} ∉ U_{Q} for j < m, and U_{Q} ∩ Z = Q. Let

G_{Q} = {b ∈ B : cl(b) ⊆ U_{Q}}

Since each G_{Q} ∈ N we have that N is uncountable. By Lemma 21 and 8, we are done. QED

Question 23 For what Borel spaces Y does ω^{ω} × Y have UU?

For example, does ω^{ω} × ℚ^{ω} have UU?

Proposition 24 For every Y a Σ_{1}^{1} set there exists a Σ_{1}^{1} set X such that X × Y has UU.

proof:

Suppose Y ⊆ Z where Z is Polish and B is a countable base for Z. Define G ⊆ P(B) by

G ∈ G iff ∀b ∈ B [(b ∩ Y ⊆ ⋃ G) → b ∈ G]

Note that b ∩ Y ⊆ ⋃ G is Π_{1}^{1} and so G is Σ_{1}^{1}. QED

We use the next Lemma for Example 26.

Lemma 25 For any space Y (ω × 2^{ω}) × Y has UU iff 2^{ω} × Y has UU.

proof:

Suppose C ⊆ (ω ×2^{ω})×Y is a closed set uniquely universal for the closed subsets of Y.

Since the whole space Y occurs as a cross section of C without loss we may assume that Y = C_{(0,0)} where 0 is the constant zero function.

For each n > 0 let

K_{n} = {(0^{n}ˆ ⟨⋆, x_{0}, x_{1}, …⟩, y) ∈ 2^{ω} × Y : ((n, x), y) ∈ C}

By 0^{n}ˆ ⟨⋆, x_{0}, x_{1}, …⟩ we mean a sequence of n zeros followed by the special symbol ⋆ and then the (binary) digits of x. Note that the K_{n} converge to 0. Let

K_{0} = {(x, y) : ((0, x), y) ∈ C}

Let B = ⋃_{n<ω} K_{n} ⊆ S × Y where

S = 2^{ω} ∪ ⋃_{n>0}{0^{n}ˆ ⟨⋆, x_{0}, x_{1}, …⟩ : x ∈ 2^{ω}}.

Note that S is homeomorphic to 2^{ω} and there is a one-to-one correspondence between the cross sections of B and cross sections of C. Note that B is closed in S ×Y: If (x_{n}, y_{n}) ∈ B for n < ω is a sequence converging to (x, y) ∈ S ×Y and x is not the zero vector it is easy to see that (x, y) ∈ B. On the other hand if x is the zero vector, then since B0 = Y, it is automatically true that (x, y) ∈ B.

Hence UU holds for 2^{ω} × Y.

For the opposite direction just note that (ω + 1) × 2^{ω} is homeomorphic to 2^{ω} and there is a continuous bijection from ω × 2^{ω} onto (ω + 1) × 2^{ω}. QED

Next we describe our counterexample to a converse of Proposition 7. Let Z = (ω × ω) ∪ {∞}. Let each D_{n} = {n} × ω be an infinite closed discrete set and let the sequence of D_{n} “converge” to ∞, i.e., each neighborhood of ∞ contains all but finitely many D_{n}. Equivalently Z is homeomorphic to:

Z′ = {x ∈ ω^{ω} : |{n : x(n) ≠ 0}| ≤ 1}.

The point ∞ is the constant zero map, while D_{n} are the points in Z′ with x(n) ≠ 0. Note that Z′ is a closed subset of ω^{ω}, hence it is Polish. This seems to be the simplest nonlocally compact Polish space.

Example 26 Z is a countable nonlocally compact Polish space such that 2^{ω} × Z has the UU. proof:

Z = ⋃_{n<ω} D_{n} ∪ {∞}. Note that X ⊆ Z is closed iff ∞ ∈ X or X ⊆ ⋃_{i≤k} D_{i} for some k < ω. By Lemma 25 it is enough to see that (ω × 2^{ω}) × Z has the UU.

Let P_{n} = {n} × 2^{ω}.

Use P_{0} to uniquely parameterize all subsets of Z which contain the point at infinity, see Remark 5.

Use P_{1} to uniquely parameterize all X ⊆ D_{0}, including the empty set. For n = 1 + 2^{k−1}(2l − 1) with k, l > 0 use P_{n} to uniquely parameterize all

X ⊆ ⋃_{i≤k} D_{i} such that D_{k} meets X and the minimal element of D_{k} ∩ X is the l-th element of D_{k}. QED

Our next two results have to do with Question 17.

Proposition 27 Existence of UU for 2^{ω} × ω^{ω} is equivalent to:

There exist a C ⊆ P(ω^{<ω}) homeomorphic to 2^{ω} such that every T ∈ C is a subtree of ω^{<ω} (possibly with terminal nodes) and such that for every closed C ⊆ ω^{ω} there exists a unique T ∈ C with C = [T].

proof:

Given C ⊆ 2^{ω} × ω^{ω} witnessing UU for closed subsets of ω^{ω}. Let

[T] = {(s, t) : ([s] × [t]) ∩ C ≠ ∅}.

Define f : 2^{ω} → P(ω^{<ω}) = 2^{ω^{<ω}} by f(x)(s) = 1 iff (x↾n, s) ∈ T where n = |s|. Then f is continuous, since f(x)(s) depends only on x↾n where n = |s|.

If T_{x} = f(x), then [T_{x}] = C_{x}. Hence f is one-to-one, so its image C is as described. QED

Proposition 28 Suppose ω^{ω} × Y has UU where Y is any topological space in which open sets are F_{σ}. Then there exists U ⊆ 2^{ω} × Y an F_{σ} set such that all cross sections U_{x} are open and for every open W ⊆ Y there is a unique x ∈ 2^{ω} with U_{x} = W. proof:

Let ω ⊕ 1 denote the discrete space with one isolated point adjoined and let ω+1 denote the compact space consisting of a single convergent sequence. Then (ω ⊕ 1)^{ω} is homeomorphic to ω^{ω} and (ω + 1)^{ω} is homeomorphic to 2^{ω}.

Assume that U ⊆ (ω ⊕ 1)^{ω} × Y is an open set witnessing UU. Then U is an F_{σ} set in (ω + 1)^{ω} × Y.

To see this note that a basic clopen set in (ω ⊕ 1)^{ω} could be defined by some s ∈ (ω ⊕ 1)^{<ω} by

[s] = {x ∈ (ω ⊕ 1)^{ω} : s ⊆ x}.

But it is easy to check that [s] ⊆ (ω+1)^{ω} is closed in the topology of (ω+1)^{ω}. Since U is open in (ω ⊕ 1)^{ω} × Y there exists s_{n} and open sets W_{n} ⊆ Y such that

U = ⋃_{n<ω}[s_{n}] × W_{n}

Hence U is the countable union of F_{σ} sets and so it is F_{σ} in (ω + 1)^{<ω}. QED

Here ⊕ refers to the topological sum of disjoint copies or equivalently a clopen separated union.

Proposition 29 Suppose X_{i} × Y_{i} has UU for i ∈ I. Then

(∏_{i∈I} X_{i}) × (⨁_{i∈I} Y_{i}) has UU.

So, for example, if 2^{ω} × Y has UU, then 2^{ω} × (ω × Y) has UU.

proof:

Define

((x_{i})_{i∈I}, y) ∈ U iff ∃i ∈ I (x_{i}, y) ∈ U_{i}

where the U_{i} ⊆ X_{i} × Y_{i} witness UU. QED

Except for Proposition 2 we have given no negative results. The following two propositions are the best we could do in that direction.

Proposition 30 Suppose U ⊆ X × Y is an open set universal for the open subsets of Y. If X is second countable, then so is Y.

proof:

U is the union of open rectangles of the form B × C with B open in X and C open in Y. Clearly we may assume that B is from a fixed countable basis for X. Since ⋃_{i} B × C_{i} = B × ⋃_{i} C_{i} we may write U as a countable union:

U = ⋃_{n<ω} B_{n} × C_{n}

where the B_{n} are basic open sets in X and the C_{n} are open subsets of Y. But this implies that {C_{n} : n < ω} is a basis for Y since for each x ∈ X

U_{x} = ⋃{C_{n} : x ∈ B_{n}}

QED

Proposition 31 There exists a partition X ∪ Y = 2^{ω} into Bernstein sets X and Y such that for every Polish space Z neither Z × X nor Z × Y has UU.

proof:

Note that up to homeomorphism there are only continuum many Polish spaces. If there is a UU set for Z ×X, then there is an open U ⊆ Z ×2^{ω} such that U ∩ (Z × X) is UU. Note that the cross sections of U must be distinct open subsets of 2^{ω}. Hence it suffices to prove the following:

Given U_{α} for α < 𝔠 such that each U_{α} is a family of open subsets of 2^{ω} either

(a) there exists distinct U, V ∈ U_{α} with U ∩ X = V ∩ X or

(b) there exists U ⊆ 2^{ω} open such that U ∩X ≠ V ∩X for all V ∈ U_{α}.

And the same for Y in place of X.

Let P_{α} for α < 𝔠 list all perfect subsets of 2^{ω} and let {z_{α} : α < 𝔠} = 2^{ω}. Construct X_{α}, Y_{α} ⊆ 2^{ω} with

1. X_{α} ∩ Y_{α} = ∅ 2. α < β implies X_{α} ⊆ X_{β} and Y_{α} ⊆ Y_{β}

3. |X_{α} ∪ Y_{α}| = |α| + ω

4. If there exists distinct U, V ∈ U_{α} such that UΔV is a countable set disjoint from X_{α}, then there exists such a pair with UΔV ⊆ Y_{α+1}

5. If there exists distinct U, V ∈ U_{α} such that UΔV is a countable set disjoint from Y_{α+1}, then there exists such a pair with UΔV ⊆ X_{α+1}

6. P_{α} meets both X_{α+1} and Y_{α+1}

7. z_{α} ∈ (X_{α+1} ∪ Y_{α+1})

First we do (4) then (5) and then take care of (6) and (7).

Let X = ⋃_{α<𝔠} X_{α} and Y = ⋃_{α<𝔠} Y_{α}.

Fix α and let us verify (a) or (b) holds. Take any point p ∈ X \ X_{α+1}. If (b) fails there must be U, V ∈ U_{α} with X = X ∩ U and X \ {p} = X ∩ V. Then (UΔV) ∩ X_{α+1} = ∅. Since X is Bernstein and (UΔV) ∩ X has only one point in it, it must be that UΔV is countable. Then by our construction we have chosen distinct U, V ∈ U_{α} with UΔV ⊆ Y therefore U ∩X = V ∩X, so (a) holds.

A similar argument goes through for Y. QED

Finally, and conveniently close to the bibliography, we note some papers in the literature which are related to the property UU. Friedberg [3] proved that there is one-to-one recursively enumerable listing of the recursively enumerable sets. This is the same as saying that there is a (light-face) Σ_{1}^{0} subset U ⊆ ω × ω which is uniquely universal for the Σ_{1}^{0} subsets of ω. Brodhead and Cenzer [1] prove the analogous result for (light-face) Σ_{1}^{0} subsets of 2^{ω}.

Becker [2] considers unique parameterizations of the family of countable sets by Borel or analytic sets. Gao, Jackson, Laczkovich, and Mauldin [4] consider several other problems of unique parameterization.

References

[1] Brodhead, Paul; Cenzer, Douglas; Effectively closed sets and enumerations. Arch. Math. Logic 46 (2008), no. 7-8, 565-582.

[2] Becker, Howard; Borel and analytic one-one parametrizations of the countable sets of reals. Proc. Amer. Math. Soc. 103 (1988), no. 3, 929-932.

[3] Friedberg, Richard M.; Three theorems on recursive enumeration. I. Decomposition. II. Maximal set. III. Enumeration without duplication. J. Symb. Logic 23 1958 309-316.

[4] Gao, Su; Jackson, Steve; Laczkovich, Mikls; Mauldin, R. Daniel; On the unique representation of families of sets. Trans. Amer. Math. Soc. 360 (2008), no. 2, 939-958.

[5] Kechris, Alexander S.; Classical descriptive set theory. Graduate Texts in Mathematics, 156. Springer-Verlag, New York, 1995. xviii+402 pp. ISBN: 0-387-94374-9

[6] Kuratowski, K.; Topology. Vol. I. New edition, revised and augmented. Translated from the French by J. Jaworowski Academic Press, New York-London; Pastwowe Wydawnictwo Naukowe, Warsaw 1966 xx+560 pp.

[7] Kuratowski, Kazimierz; Mostowski, Andrzej; Set theory. With an introduction to descriptive set theory. Translated from the 1966 Polish original. Second, completely revised edition. Studies in Logic and the Foundations of Mathematics, Vol. 86. North-Holland Publishing Co., Amsterdam-New York-Oxford; PWN—Polish Scientific Publishers, Warsaw, 1976. xiv+514 pp.

[8] Moschovakis, Yiannis N.; Descriptive set theory. Studies in Logic and the Foundations of Mathematics, 100. North-Holland Publishing Co., Amsterdam-New York, 1980. xii+637 pp. ISBN: 0-444-85305-7

Arnold W. Miller miller@math.wisc.edu http://www.math.wisc.edu/∼miller University of Wisconsin-Madison Department of Mathematics, Van Vleck Hall 480 Lincoln Drive Madison, Wisconsin 53706-1388