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Descriptive Set Theory is the study of definable subsets of Polish spaces, where definable is taken to mean from the Borel or projective hierarchies. Other topics include infinite games and determinacy, definable equivalence relations and Borel reductions between them, Polish groups, and effective descriptive set theory.
4
votes
Accepted
Topological spaces with too many open sets
Consistently, $\omega^*$ has this property. In the paper
E. van Douwen, K. Kunen, and J. van Mill, "There can be $C^*$-embedded dense proper subspaces of $\beta\omega - \omega$," Proc. Amer. …
5
votes
Accepted
A property of the Frechet filter and every ultrafilter
The filters that you're looking for don't exist. Every filter with your property, other than the Frechet filter, maps to an ultrafilter via a finite-to-one map. Let's call the filters with your proper …
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]^ …
8
votes
What is the descriptive complexity of a set added by Cohen forcing?
Jensen proved in this paper that, beginning with $V = L$, it is possible to add a real $a$ by forcing such that $a$ is $\Delta^1_3$ in $L[a]$.
This result is the best possible, in the sense that one …
4
votes
Accepted
Are there such a complete metric space X of weight k (w(X)=k) and ....?
As K.P. and Ramiro both point out in the comments, it follows from $\mathsf{CH}$ that the answer is no. I claim that it is also consistent that the answer is yes.
It is consistent that $\mathfrak{c} > …
11
votes
Accepted
What is an example of a meager space X such that X is concentrated on countable dense set?
ADDED LATER
The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph_1$.
If $\mathfrak{b} = \aleph_1$, then there is such a space.
To see this, first note th …
15
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 …
3
votes
Accepted
Covering measure one sets by closed null sets
The answer to the first question is yes: it is always true that $\kappa_{\mathcal E} = \mathrm{cov}(\mathcal E)$. The answer to the second question is no.
As Piotr points out, a negative answer to th …
5
votes
Accepted
Is it consistent that the additivity of Lebesgue null sets is greater than $\frak h$?
Yes, this is consistent. In fact, it is consistent that $\mathrm{add}(\mathcal N) > \mathfrak{s}$. This was proved by Ihoda and Shelah in
Ihoda, Jaime I.; Shelah, Saharon, Souslin forcing, J. Symb. Lo …
12
votes
Accepted
Is there a Bernstein subset $X$ of $\mathbb{R}$ such that no continuous map $f : X → [0,1]$ ...
No, if $X$ is a Bernstein set then there is a continuous surjection $X \rightarrow [0,1]$.
To see this, note that there is a continuous surjection $f: \mathbb R \rightarrow [0,1]$ such that the preima …
3
votes
The "strong" measure number
Great question! This is not a complete answer, but hopefully it gets the ball rolling . . .
Theorem: $\mathfrak s_{-}\geq \mathrm{cov}(\mathcal M)$, where $\mathrm{cov}(\mathcal M)$ is the smalles …
5
votes
Small uncountable cardinals related to $\sigma$-continuity
Forcing to add uncountably many Cohen reals makes $\sigma = \aleph_1$.
To see this, suppose $X$ is an uncountable set of mutually generic Cohen reals over $V$. To be precise: ``mutually generic'' mea …
12
votes
Accepted
A strong Borel selection theorem for equivalence relations
Let $X$ be the Cantor space $2^\omega$, and let $E$ be the relation of "equivalence mod $\mathrm{Fin}$" -- i.e., $xEy$ if and only if $\{n \in \omega :\, x(n) \neq y(n) \}$ is finite. The equivalence …
5
votes
Can all uncountable (but small) families of sets with positive measure have an uncountable s...
The answer to both questions (1) and (2) is yes.
It is consistent, and in fact it follows from Martin's Axiom for $\aleph_1$, that:
For every uncountable family $\mathcal F$ of positive-measure subse …