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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.
31
votes
About the axiom of choice, the fundamental theorem of algebra, and real numbers
The fundamental theorem of algebra is, unless I miscounted quantifiers, a $\Pi^1_2$ sentence of second-order arithmetic and therefore absolute between the full universe and Gödel's constructible unive …
23
votes
Accepted
Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets
It suffices to express $\mathbb R$ as the union of $\aleph_1$ (not necessarily disjoint) Borel sets such that no countably many of them cover $\mathbb R$, because then you can list them in an $\omega_ …
21
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 …
17
votes
Accepted
Does there exist an uncountable separable metric space $X$ such that every subset of $X$ is ...
Under Martin's Axiom plus the negation of CH, every set $X$ of reals of size $<\mathfrak c$ is a Q-set, which means that every subset of $X$ is an $F_\sigma$-set with respect to the subspace topology …
14
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^ …
14
votes
Accepted
Indeterminacy of long games
Your statement of AD is incorrect. $X$ should be at most $\omega$ (and at least 2), not an arbitrary set. Specifically, for $X=\aleph_1$, determinacy of games in $X^\omega$ is inconsistent with ZF. …
13
votes
Continuously selecting elements from unordered pairs
This should have been a comment, but it got a bit too long.
A possibly useful necessary condition, in regular spaces, for the existence of a continuous selector is that there should not exist three …
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 unde …
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, an …
9
votes
Accepted
Measurably-isomorphic subsets of polish spaces and the continuum hypothesis
Every Borel subset (and in fact every analytic subset) of a Polish space either is countable or has a perfect subset. In particular, an uncountable Borel subset in a Polish space has the cardinality …
9
votes
Follow up question: Shelah's "Can you take Solovay's inaccessible away?"
Gabe Goldberg has already given a couple of proofs, but here's another one, just because I like it. Being a model of ZFC, $L[x]$ sees a one-to-one map of its $\omega_1$ into its reals. That map, in t …
7
votes
Accepted
"Compactness length" of Baire space
Baire space is the union of $\mathfrak d$ (the dominating number) compact subsets. So, using equivalence relations that collapse those sets one at a time (i.e., one equivalence class is the set to be …
7
votes
Accepted
Subset of the reals with zero inner measure and "full" outer measure in $\mathsf{ZF}+\mathsf...
I'll start as in Monroe Eskew's answer: Assume $X$ isn't measurable and get Borel sets $A,B$, with $A\subseteq X\subseteq B$, such that the measure of $A$ (resp. $B$) is the inner (resp. outer) measur …
7
votes
Accepted
Iteration of random reals
Since $x$ is random over $V$, the fact that $C_x^{V[x]}$ is null is (like any fact about $x$ in $V[x]$) forced by some condition in $V$. This condition is the equivalence class, modulo the null ideal …
6
votes
How badly does compactness fail in $\mathcal{L}_{\omega_1\omega}$?
This is really a comment, but (a) it's too long and (b) is ought to be attached to both of the answers. Not only does the characterization of the $\kappa$-compactness property not depend on the availa …