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An important and fundamental axiom in set theory sometimes called Zermelo's axiom of choice. It was formulated by Zermelo in 1904 and states that, given any set of mutually disjoint nonempty sets, there exists at least one set that contains exactly one element in common with each of the nonempty sets. The axiom of choice is related to the first of Hilbert's problems.
19
votes
2
answers
834
views
Do choice principles in all generic extensions imply AC in $V$?
It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a …
15
votes
Accepted
Is Global Choice conservative over Zermelo with Choice?
Edit: Here's my preprint filling in the details of the construction in this answer: https://arxiv.org/pdf/2312.11902.pdf
Global Choice is not conservative over ZC. We'll build a model of ZC which sati …
13
votes
Accepted
Consistency of a strange (choice-wise) set of reals
The existence of such a set follows from $``\mathbb{R}$ is a countable union of countable sets.$"$ Let $\mathbb{R} = \bigcup_{n<\omega} S_n,$ each $S_n$ countable. Let $T_n = \{x \in \mathbb{R}: \exis …
12
votes
Accepted
Is every set being cardinal definable consistent with ZF + negation of Choice?
This is consistent. Kanovei constructed a model $M$ with an infinite Dedekind finite set of reals which is lightface projectively definable. By descending to $L(R),$ we can further assume it satisfies …
11
votes
1
answer
491
views
Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$
The choice principle $\text{AC}_{\text{WO}}$ proves a large amount of cardinal arithmetic. It's well-known to imply DC, that successor cardinals are regular, and that for all $X$, there is $\lambda$ s …
11
votes
Accepted
How much choice is necessary to prove this statement?
Your statement is equivalent to the assertion that there is a function choosing an enumeration of every countable ordinal. From an enumeration of $\alpha,$ you can easily inject it into a countable se …
10
votes
Do the surreal numbers enjoy the transfer principle in ZFC?
A partial answer to the focused question: it's not provable in ZFC that there is an OD class $\mathbb{Z}^*$ such that $(\mathbb{R}, +, \cdot, \mathbb{Z}) \equiv (\mathrm{No}, +, \cdot, \mathbb{Z}^*).$ …
9
votes
Accepted
Must strange sequences wear Russellian socks?
For $n \ge 2$ and $\langle A_i \rangle$ such that $|A_i| = n$ for all $i,$ TFAE:
$|\prod A_i| \neq |\mathbb{R}|.$
$\bigsqcup A_i$ is uncountable.
$\bigsqcup A_i$ is un-orderable.
There are $k \in [2, …
9
votes
Accepted
Building the real from Dedekind finite sets
Q1: There is no such partition. Let $\langle X_n \rangle$ be a countable partition of $\mathbb{R}.$ We will construct $n,$ an open interval $I,$ and an injection $g: \omega \rightarrow I \cap X_n$ wit …
7
votes
Accepted
The Parity Principle and $\mathbf{C}_2$ (choice for $2$-sets)
Over ZFA, the Parity Principle is strictly weaker. We'll show it follows from Multiple Choice, the assertion that for any family of nonempty sets $\mathcal{F},$ there is $g: \mathcal{F} \rightarrow [\ …
7
votes
Accepted
How hard is it to get "absolutely" no amorphous sets?
Turning my comment into an answer, an $X$ which is the universe of any finitely axiomatized theory with an infinite model must be orderable, and there must be a bijection between between $X$ and $X^2. …
6
votes
Accepted
Stronger negation of AC given by rejecting "infinite hat" puzzles
Naturally, the more generalizations of the infinite hats puzzle we consider, the stronger it is to assert that none of them have a paradoxical solution. One of the variants you linked in your question …
6
votes
Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?
I’ll prove the following: over Z set theory, AC is equivalent to cardinal trichotomy holding for the sets between $X$ and $\mathcal{P}^3(X)$ for all infinite $X.$ I suspect the superscript 3 is unnece …
5
votes
Accepted
Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$
($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$
Pf: Let $\lambda$ be such …
4
votes
Sequential continuity and the Axiom of Choice
If the graph of $f$ has a Borel code $c,$ then ZF proves equivalence of these two continuity concepts. Discontinuity of $f$ at $x$ is downward absolute to $L[c, x],$ in which we can find a sequence $x …