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 $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ ...

26
votes

### Quantifier complexity of the definition of continuity of functions

It is truly a very nice question, one of those questions with an answer one feels must be right, but it is not so clear at first how to prove it.
Nevertheless, aiming at partial progress, I claim that ...

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 ...

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

### 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

### 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 (...

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 \...

15
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, ...

14
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 ...

14
votes

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

As for the second question, the equivalence relation on the class of countable ICC groups given by $G \sim \Gamma$ if and only if $L(G) \cong L(\Gamma)$ is very interesting and usually called $W^*$-...

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 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

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^\...

13
votes

Accepted

### Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$

For the first question (distinct regular cardinals $>\aleph_1$): Force ZFC + MA + $2^{\aleph_0}=\aleph_3$ over $L$ in the usual way (see Jech, Theorem 16.13; note the forcing is ccc and it forces ...

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

### Is $V=L$ equivalent to there being a $\Sigma_1$ well-ordering of the universe?

No, see the paper On $ Σ_1$ Well-Orderings of the Universe by Harrington and Jech.

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

### Homeomorphisms and "mod finite"

Define $f : C \to C$ by the formula
$$ f(x) = x_0 \cdot (x \oplus \sigma(x)) $$
where $\cdot$ is word concatenation, $\oplus : C \times C \to C$ is coordinatewise xor, and $\sigma(x)_i = x_{i+1}$ is ...

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