All Questions
Tagged with axiom-of-choice descriptive-set-theory
18
questions
3
votes
0
answers
161
views
Periodicity in the cumulative hierarchy
Under Reinhardt cardinals in ZF, the cumulative hierarchy exhibits a periodicity in that for large enough $λ$, certain properties of $V_λ$ depend on whether $λ$ is even vs odd. See Periodicity in the ...
- 6,365
4
votes
0
answers
144
views
Consistency of definability beyond P(Ord) in ZF
Is it consistent with ZF that the satisfaction relation of $L(P(Ord))$ is $Δ^V_2$ definable? More generally, is it consistent with ZF that there is a $Δ^V_2$ formula (taking $α$ as a parameter) that ...
- 6,365
7
votes
3
answers
447
views
How much Dependent Choice is provable in $Z_2$? And what about Projective Determinacy?
So, second order arithmetic, $Z_2$, is capable of proving quite a few things. One thing which would be of use is dependent choice for $\mathbb{R}$.
Basically, dependent choice on $\mathbb{R}$ says ...
- 323
6
votes
0
answers
210
views
Failure of Baire's grand theorem when the hypothesis is weakened to separable metric space
The statement of Baire grand theorem gives a characterization of Baire class 1 functions between a completely metrizable separable space (aka Polish space) and a separable metrizable space. The ...
- 2,236
2
votes
1
answer
190
views
How much choice is needed to prove this statement?
Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):
There exists $(C_\alpha, x_\alpha)_{\alpha \in \omega_1}$ s.t. $C_\alpha \subseteq \mathbb{N}^\mathbb{N}$ is closed ...
- 2,236
9
votes
1
answer
692
views
How much choice is necessary to prove this statement?
Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):
There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ ...
- 2,236
6
votes
0
answers
395
views
General theory of the reals in Solovay-like models
Solovay's model is a famous model of $\sf ZF$ where we start in $L$ with $\kappa$ inaccessible, and we collapse all the ordinals below $\kappa$ to be countable, without collapsing $\kappa$ itself, and ...
- 38.5k
5
votes
0
answers
169
views
How much choice is required for a countably-infinite index subgroup of the real additive group?
The existence of such subgroups implies the existence of a non-measurable set; simply intersect each of the cosets with $[0,1]$. The results will all have equal outer measure, but their union will be ...
- 1,252
8
votes
1
answer
418
views
Undetermined games of "overdetermined" type
This is motivated by a previous question of mine, but I think it is ultimately more interesting (and hopefully easier to answer in the positive). In that question, a class of games (on $\omega$, of ...
- 20.7k
7
votes
0
answers
167
views
Choice and the Baire property in non-separable complete metric spaces
It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has ...
- 29.5k
8
votes
0
answers
535
views
A Banach-Tarski game
This is partially inspired by the question https://math.stackexchange.com/questions/1383397/cutting-a-banach-tarski-cake, which I find intriguing if unclearly written.
A paradoxical family of subsets ...
- 20.7k
5
votes
0
answers
424
views
Cardinal characteristics without choice
(I'm taking my definition of a cardinal characteristic from Blass' excellent article http://www.math.lsa.umich.edu/~ablass/need.pdf, which cites Vojtas/Fremlin/Miller; theirs is more general, but I'm ...
- 20.7k
18
votes
4
answers
1k
views
Pathological behavior of Borel sets?
Usually in set theory, Borel sets are much more nicely behaved than arbitrary sets of reals. One reason for this is Borel determinacy, which immediately yields measurability, Baireness, and the ...
- 20.7k
5
votes
0
answers
440
views
How many Dedekind-finite sets can $\mathbb{R}$ be partitioned into?
Building off Asaf Karagila's answer to my previous question (Can $\mathbb{R}$ be partitioned into dedekind-finite sets?) on partitioning $\mathbb{R}$ into strictly Dedekind-finite sets:
(1) What ...
- 20.7k
9
votes
1
answer
625
views
Can $\mathbb{R}$ be partitioned into dedekind-finite sets?
Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...
- 20.7k