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 ...
- 216k
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 ...
- 39.9k
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 $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ ...
- 39.7k
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 ...
- 216k
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 ...
- 216k
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, ...
- 28.6k
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 ...
- 70.8k
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]^...
- 17k
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 ...
- 26.2k
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 ...
- 2,620
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 ...
- 9,771
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 (...
- 2,156
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 ...
- 16.1k
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 \...
- 96.9k
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, ...
- 57.9k
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 ...
- 43.4k
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 $W^*$-...
- 3,241
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 ...
- 17k
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....
- 31.9k
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 $...
- 2,490
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 ...
- 3,738
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^\...
- 70.8k
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 ...
- 6,999
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 ...
- 7,337
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 ...
- 70.8k
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, ...
- 70.8k
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.
- 31.3k
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 ...
- 57.9k
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 ...
- 6,162
Only top scored, non community-wiki answers of a minimum length are eligible