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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
Continuously selecting elements from unordered pairs
This is an expansion on the comment I made to the effect that the symmetric square of $\mathbb{Q}^2$ has a continuous selector, which is surprising since the symmetric square of $\mathbb{R}^2$ does no …
8
votes
Is it inconsistent for a model of set theory to contain its own first order theory?
No, this is not a problem. If $U$ is a transitive set with $\mathcal{P}(\omega) \subseteq U$ then $U$ contains the real $\{\ulcorner\sigma\urcorner \mid U \vDash \sigma\}$. So, for example, $V_\alpha$ …
4
votes
Accepted
Ensuring nonempty lightface Borel sets have elements via theories of second-order arithmetic
Given a tree $T \subseteq \omega^{\lt\omega}$, the statement "$B$ is an infinite path through $T$" is $\Pi^0_2$. Therefore, if $\mathcal{M}$ is such that every nonempty $\Pi^0_2$-class coded in $\math …
5
votes
disjoint translates of a dense uncountable set
Use of the Axiom of Choice is unnecessary here. One can write down an explicit $\mathbb{Q}$-linearly independent subset $\mathcal{T}$ of $\mathbb{R}$ with size $2^{\aleph_0}$ as I wrote in this answer …
2
votes
Set of perfect subsets of a Borel set
This appears not to be the case: there is a $F_\sigma$ set $B$ such that $S_B$ is not Borel. This is optimal since the bullets in the question explain how $S_B$ is Borel when $B$ is $G_\delta$.
There …
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 be p …
9
votes
Accepted
Large cardinal axioms and the perfect set property
Solovay showed that the following are equivalent:
$\boldsymbol{\Sigma}^1_2$ sets have the perfect set property
$\boldsymbol{\Pi}^1_1$ sets have the perfect set property
$\aleph_1^{L[a]} < \aleph_1$ …
14
votes
Accepted
2-colorings of the reals
Fred Galvin showed that if $c:[\mathbb{R}]^2\to\lbrace0,1\rbrace$ is such that $c^{-1}(0)$ and $c^{-1}(1)$ both have the Baire property, then there is a perfect set $P \subseteq \mathbb{R}$ which is $ …
4
votes
Accepted
A model of Krivine
Yes, that is correct. Krivine's observation is that by collapsing the continuum (or any larger cardinal) to $\aleph_0$ using finite conditions then any set of reals definable from ground model parame …
14
votes
About the axiom of choice, the fundamental theorem of algebra, and real numbers
Regarding Cauchy and Dedekind reals. The fact that every Dedekind real has a Cauchy representation is provable in very weak systems of intuitionistic analysis. The converse fact that every Cauchy real …
6
votes
Accepted
Comparing bornologies for domination/escaping
Note that $\mathfrak{b}=\mathfrak{d}$ is equivalent to the existence of a $<^\ast$-increasing sequence $(f_\alpha)_{\alpha<\mathfrak{d}}$ which is cofinal in $(\mathcal{N},{<^\ast})$, where $f <^\ast …
10
votes
Accepted
Continuity on a measure one set versus measure one set of points of continuity
As Jason and Gerald predicted, the answer is yes for Polish $X, Y$. (Indeed, it is sufficient for $X$ to be merely separable and metrizable and for $Y$ to be merely complete and metrizable.)
As Nate …
4
votes
When is $A$ "$L$-ish" whenever $B$ is "$L$-ish"?
Here is an extension of Barwise's theorem which may be of some use.
Theorem. Fix a real $a \subseteq \omega$ in $W$. Suppose the preorder $\preceq$ is first-order definable with parameter $a$ and that …
3
votes
Accepted
Are Cohen Generics Minimal Covers?
Indeed, this has been answered very negatively in the literature:
Abraham, Uri; Shore, Richard A., The degrees of constructibility of Cohen reals, Proc. Lond. Math. Soc., III. Ser. 53, 193-208 (1986). …