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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. …
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5 votes
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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 …
Will Brian's user avatar
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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]^ …
Will Brian's user avatar
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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 …
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4 votes
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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} > …
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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 …
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15 votes
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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 …
Will Brian's user avatar
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3 votes
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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 …
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5 votes
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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 …
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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 …
Will Brian's user avatar
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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 …
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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 …
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12 votes
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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 …
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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 …
Will Brian's user avatar
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