All Questions
Tagged with axiom-of-choice set-theory
436
questions
10
votes
1
answer
363
views
Hereditarily countable sets in Antifounded ZF
A set $x$ is hereditarily countable when every membership-descendant of $x$ (including $x$ itself) is countable.
In this paper, Jech proved in ZF that the class of all hereditarily countable sets is a ...
- 1,382
14
votes
2
answers
1k
views
Proof/Reference to a claim about AC and definable real numbers
I’ve read somewhere on this site (I believe from a JDH comment) that an argument in favor of AC (I believe from Asaf Karagila) is that without AC, there exists a real number which is not definable ...
- 243
10
votes
1
answer
614
views
Infinitary logics and the axiom of choice
Suppose we want to enhance ZF by allowing for infinitary formulas instead of just first-order ones in our axiom schema of separation and/or replacement. It seems that we don't need much power in our ...
- 4,857
-2
votes
1
answer
163
views
Is discriminative choice provable in ZFC?
Let $\phi$ be a formula defining an equivalence relation.
Definitions:
The $\phi$-cardinality of a set $X$ be the cardinality of $X/\phi$. That is, the cardinality of the set of all equivalence ...
- 10.2k
9
votes
1
answer
295
views
Complexity of definable global choice functions
It is well-known that $L$ has a $\Sigma_{1}$-definable global choice function; it is also known that there are other transitive class models of ZFC with this property.
I wonder about the complexity ...
- 93
5
votes
1
answer
239
views
Is a weak version of the three sets Lemma provable in ZF?
The Three Sets Lemma is the following Lemma:
Lemma: Let $f(x)$ be a function from $X$ to $X$ where $f(x)$ has no fixed points. Then there exists a partition of $X$ into three disjoint sets $X_1$, $X_2$...
- 6,819
10
votes
1
answer
488
views
Must strange sequences wear Russellian socks?
This is an attempt to make more precise a vague guess at the end of this answer of mine. We work in $\mathsf{ZF}$ throughout.
Say that a sequence $\mathcal{A}=(A_i)_{i\in\omega}$ of disjoint sets is ...
- 20.7k
6
votes
2
answers
458
views
Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?
Does $\sf ZFA + WOIPS$ prove $\sf AC$?
Where $\sf WOIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.
In $\sf ZF$, I ...
- 10.2k
3
votes
1
answer
149
views
Turning linear ordering into well-ordering
Work in ZF + DC. Assume that there is an uncountable $A \subseteq \mathbb{R}$ and a linear order $\prec$ on $A$ such that for every $x \in A$, $\{y \in A: y \prec x\}$ is countable. Must there exist ...
- 33
5
votes
1
answer
149
views
The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)
A set $X$ is Dedekind-infinite if there is an injective map $f: X\to X$ that is not surjective.
A set $X$ is dually Dedekind-infinite if there is a surjective map $f: X\to X$ that is not injective.
In ...
26
votes
1
answer
897
views
A cardinal inequality for finiteness
Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality:
$$2^{2^{2^{2^{\mathfrak{n}}}}}...
- 43.8k
6
votes
1
answer
160
views
Does $\mathsf{SVC}^\ast$ exist?
$\mathsf{SVC}(S)$ is the assertion that for all sets $X$ there is an ordinal $\eta$ and a surjection $f\colon\eta\times S\to X$. I would like to denote by $\mathsf{SVC}^\ast(S)$ the same assertion but ...
- 1,479
1
vote
1
answer
298
views
Proof of the axiom of choice for finite sets in ZF [closed]
Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$ satisfying $...
- 21
12
votes
1
answer
419
views
Did Gödel possess a proof of the independence of $\mathsf{AC}$?
We all know Gödel proved the consistency of the Axiom of Choice with $\mathsf{ZF}$ using his constructible universe, and Cohen proved the consistency of $\neg \mathsf{AC}$ using his new method of ...
- 1,303
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