35
votes

Accepted

### A rare property of Hausdorff spaces

Yes, there is such a space. Let $X=2^{\omega_1}$ be the space of
binary sequences of length $\omega_1$, in the order topology
generated by the lexical order. So $X$ consists of the branches
through ...

29
votes

### A rare property of Hausdorff spaces

How about this example: $[0,1]^A$ with the product topology, where $A$ is uncountable. Then every nonempty $G_\delta$ set is uncountable. Since your sets $f^{-1}(x)$ are $G_\delta$ sets, this has ...

28
votes

Accepted

### Closed balls vs closure of open balls

The following theorem (or its corollary) implies negative answer to the original question.
Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ ...

25
votes

Accepted

### How (non-)computable is set theory?

The question is extremely interesting, and I have looked into this kind of thing with various colleagues (including Russell Miller and Kameryn Williams), although our investigation has not yet ...

22
votes

### Examples of statements with a high quantifier complexity

My favorite example is the 5-quantifier definition of an almost-periodic function: $f$ is almost-periodic iff
$$\forall \epsilon>0\, \exists t>0\ \forall a\ \exists s \in [a,a+t]\ \forall x\, |f(...

21
votes

Accepted

### Is there a Borel subset of $ \mathbb{R}^{2} $, with finite vertical cross-sections, whose projection onto the first component is non-Borel?

No, no such set exists. This is a special case of the Lusin–Novikov theorem; see e.g. Kechris, Classical Descriptive Set Theory, Theorem 18.10.
In general, let $X,Y$ be standard Borel spaces, ...

20
votes

Accepted

### Two strengthenings of "strong measure zero"

Strategically strong measure zero is equivalent to countable. To prove the nontrivial direction, suppose $X$ is strategically strong measure zero and $s$ is a winning strategy for player II. Consider ...

20
votes

Accepted

### Can an injective $f: \Bbb{R}^m \to \Bbb{R}^n$ have a closed graph for $m>n$?

There is no such function.
Suppose $f: \mathbb R^m \rightarrow \mathbb R^n$ is an injective function with $\Gamma_f$ closed in $\mathbb R^{m+n}$. For each $i \in \mathbb N$, let $K_i = f^{-1}([-i,i]^...

18
votes

Accepted

### Avoiding countable subgroups of general uncountable groups

Counterexample for Problem 1: According to this answer Saharon Shelah constructed a "Jónsson group" of order $\aleph_1,$ i.e., an uncountable group $G$ in which every proper subgroup is countable. ...

18
votes

Accepted

### Topological proof that a Vitali set is not Borel

Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $\sigma$-algebra containing the open ...

17
votes

Accepted

### Is every element of $\omega_1$ the rank of some Borel set?

In $\mathsf{ZFC}$ the Borel hierarchy of an uncountable Polish space $X$ has length exactly $\omega_1$, meaning that $\mathbf{\Sigma}^0_\xi(X)\neq\mathbf{\Pi}^0_\xi(X)$ for all $\xi<\omega_1$.
The ...

16
votes

Accepted

### Does there exist an uncountable partition of a Polish space so that the union of any collection of blocks is Borel?

Suppose $\{P_i : i < \kappa\}$ is such a partition. Let $f: R \to R$ be a function satisfying $|f[P_i]| = 1$ and for all $i < j < \kappa$, $f[P_i] \cap f[P_j] = \phi$. Then $f$ is Borel so ...

16
votes

Accepted

### Is there an uncountable Borel almost disjoint family?

Firstly, as commented by Asaf Karagila, it is routine to build a Borel uncountable family of almost disjoint subsets of $\omega$: fix a one-to-one map $f$ from the set $2^{<\omega}$ of finite ...

16
votes

Accepted

### Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ with injective restriction $f|\mathbb Q^\omega$?

It looks like no.
Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{\omega-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \...

14
votes

### Application of Fraïssé construction in set theory

A Fraïssé construction lies at the heart of the proof of my embedding theorems.
Theorem. (J. D. Hamkins, Every countable model of set theory
embeds into its own constructible universe, JML 13(2), ...

14
votes

Accepted

### Partitions of the real line into Borel subsets

The answer to both problems is no!
If the Cohen forcing is used to add lots of reals to a countable transitive model of GCH, then in the resulting extension, any partition of the real line into Borel ...

13
votes

Accepted

### Woodin on Posner-Robinson for the hyperjump and sharp

MR2449474 (2009j:03067) Woodin, W. Hugh. A tt version of the
Posner-Robinson theorem. Computational prospects of infinity. Part II.
Presented talks, 355–392, Lect. Notes Ser. Inst. Math. Sci. Natl....

13
votes

Accepted

### Can an ultrapower be undone by forcing?

For set-forcing, the answer is no, see the following article
Joel David Hamkins, Greg Kirmayer, and Norman Lewis Perlmutter, Generalizations of the Kunen inconsistency, Ann. Pure Appl. Logic 163 (...

13
votes

Accepted

### Can the Turing degrees be linearly ordered?

You can't linearly order the Vitali ($\mathcal{P}(\omega)/\mathrm{Fin}$) degrees if every set of reals has the property of Baire, since you can't even choose between complementary degrees. The set of $...

13
votes

Accepted

### The Axiom of Determinacy and the Banach-Mazur game

The claim is false. The Banach-Mazur game, also known as the $**$-game shows (and is equivalent to) that every set of reals has the Baire property. What is true, as you've noted, is that if one has a ...

13
votes

### Is the set of all ICC amenable groups countable?

The set of countable ICC groups doesn't exist, so I think you're asking about isomorphism classes.
There are continuum many non-isomorphic locally finite fields (e.g., take, for $S$ any set of primes, ...

13
votes

Accepted

### Do Borel subsets of the plane with null sections have Borel projections?

Notice that $\omega^\omega$ can be embedded to a null subset of itself by sending any sequence $(a_0,a_1,a_2,\dots)$ to $(a_0,0,a_1,0,a_2,0,\dots)$. So any Borel phenomenon that can happen in $\omega^\...

12
votes

Accepted

### Does Turing determinacy imply full determinacy?

This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) ...

12
votes

### Application of Fraïssé construction in set theory

If you are willing to allow uncountable generalizations of Fraisse's/Hrushovski construction then the conditions you need essentially are those that make up an Abstract Elementary Class. Abstract ...

12
votes

Accepted

### Descriptive Complexity of Knot Equivalence

A perfect timing for this question, since I just uploaded a paper on this topic to arXive (see below). Let us specify the definitions. A knot is a homeomorphic image of the circle in $\mathbb{R}^3$. ...

12
votes

### Can Sacks forcing add a Cohen generic real over $L$?

Regarding Question 2:
Zapletal's solution to the "Half-Cohen problem" gives an example of a 2-stage iteration $P*\dot Q$ such that $P$ does not add Cohen reals, and $P$ forces that $\dot Q$ does not ...

12
votes

Accepted

### Ordinal-definable witnesses to the perfect set property?

Let $A$ be the set of those reals that are not OD. Then $A$ is OD (since I've just defined it), and it's uncountable (since its complement, being a well-orderable set of reals, must be countable ...

12
votes

Accepted

### What is the descriptive complexity of a set added by Cohen forcing?

Since Cohen forcing is weakly homogeneous, all hereditarily ordinal definable sets in the Cohen extension are already in the ground model. That applies in particular to any ordinal definable real, ...

12
votes

Accepted

### Automorphisms of power set lattice mod finite

This is known as Rudin-Shelah problem. Note that, by Stone duality, this is equivalent to determine the self-homeomorphism group of the Stone-Cech boundary of $N$. Notably, consider the group induced ...

12
votes

Accepted

### Uncountable disjoint closed coverings of $[0,1]$

This is question has a long and interesting history, which is discussed in Arnie Miller's paper cited below. The first construction of a model of ZFC + $\aleph_1 < 2^{\aleph_0}$ where $[0,1]$ can ...

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