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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.

2 votes
1 answer
136 views

Complexity of deciding if an incomplete first-order theory has a stable completion

I'm curious about the problem of deciding if a given incomplete first-order theory has a stable completion from a descriptive set theory point of view. It seems likely that this problem is $\Pi_1^1$-c …
James E Hanson's user avatar
4 votes
0 answers
64 views

Borel rank collapse in Hilbert cube modulo $\sigma$-ideal generated by zero-dimensional sets

Both of the commonly studied $\sigma$-ideals (meager sets and null sets) in Polish spaces with a natural measure (i.e. $\mathbb{R}$, $[0,1]$, $[0,1]^\omega$, $2^{\omega}$, etc.) have the nice property …
James E Hanson's user avatar
4 votes
0 answers
82 views

When do Borel propositional theories have topologically tame truth assignments?

Let $(P_r)_{r \in \mathbb{R}}$ be an $\mathbb{R}$-indexed family of propositional variables. Let $\mathcal{L}$ be the collection of all propositional sentences formed from the variables $(P_r)_{r \in …
James E Hanson's user avatar
10 votes
0 answers
164 views

How nice can sets of reals be under $\mathsf{ZF} + \mathsf{BPI}$?

It's well known that the full axiom of choice is not needed to prove the existence of non-measurable subsets of $\mathbb{R}$. In particular, the Boolean prime ideal theorem ($\mathsf{BPI}$) is suffici …
James E Hanson's user avatar
10 votes
1 answer
242 views

How much of second-order arithmetic do you need for $\mathbf{\Sigma}^1_1$-determinacy to giv...

This is in some sense a follow-up to this question. The answer there says that over $\mathsf{Z}_2$ (second-order arithmetic), (boldface) $\mathbf{\Sigma}^1_1$-determinacy is enough to entail the exist …
James E Hanson's user avatar
5 votes
0 answers
138 views

If $\omega_1$ is not inaccessible in $L$, how hard can it be to find a non-measurable $\Sigm...

In his wonderfully titled paper Can you take Solovay's inaccessible away? Shelah showed that if every $\mathbf{\Sigma}^1_3$ set of reals is Lebesgue measurable, then $\omega_1$ is an inaccessible card …
James E Hanson's user avatar