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Results tagged with axiom-of-choice
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user 109573
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.
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 [\ …
- 4,316
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 …
- 4,316
3
votes
Can a Vitali set be Lebesgue measurable? (ZF)
The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds …
- 4,316
2
votes
Can a Vitali set be Lebesgue measurable? (ZF)
The answer to your first question is no. We will show there is a null Vitali set in Cohen's first model $M.$
Recall that in $M,$ there is an infinite (but Dedekind finite) set $A$ of mutually Cohen re …
- 4,316
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, …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316
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 …
- 4,316