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Questions tagged [axiom-of-choice]

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.

-6 votes
1 answer
124 views

Actual infinitesimals for solving Vitali paradox

Has anyone tried to use actual infinitesimals to solve paradoxes about non-measurability? In Vitali paradox, for example, they divide a set with measure 1 into $\infty$ subsets of zero measure and ...
Марат Рамазанов's user avatar
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 ...
Paul Blain Levy's user avatar
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 ...
Lave Cave's user avatar
  • 243
10 votes
1 answer
613 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 ...
Mike Battaglia's user avatar
1 vote
0 answers
85 views

Everywhere-defined unbounded operators between Banach spaces

In this post, it is said that there are no constructive examples of everywhere-defined unbounded operators between Banach spaces; every example furnished must use the axiom of choice. This seems like ...
Damalone's user avatar
  • 141
-2 votes
1 answer
162 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 ...
Zuhair Al-Johar's user avatar
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 ...
MCarl's user avatar
  • 93
5 votes
1 answer
238 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 $X_1$, $X_2$...
JoshuaZ's user avatar
  • 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 ...
Noah Schweber's user avatar
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 ...
Zuhair Al-Johar's user avatar
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 ...
Ollie's user avatar
  • 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 ...
Dominic van der Zypen's user avatar
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}}}}}...
François G. Dorais's user avatar
12 votes
1 answer
838 views

Partition of unity without AC

Several existence theorems for partition of unity are known. For example (source), Proposition 3.1. If $(X,\tau)$ is a paracompact topological space, then for every open cover $\{U_i \subset X\}_{i \...
BonBon's user avatar
  • 223
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 ...
Calliope Ryan-Smith's user avatar
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 $...
Maria S.'s user avatar
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 ...
A. Bailleul's user avatar
  • 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 ...
Dmytro Taranovsky's user avatar
7 votes
0 answers
180 views

Large cardinals beyond choice and HOD(Ord^ω)

Are Reinhardt and Berkeley cardinals (and even a stationary class of club Berkeley cardinals) consistent with $V=\mathrm{HOD}(\mathrm{Ord}^ω)$ ? It seems natural to expect no, but I do not see a proof....
Dmytro Taranovsky's user avatar
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 ...
Dmytro Taranovsky's user avatar
5 votes
1 answer
270 views

Would strengthening Foundation and Choice in NBG, make it equi-consistent with MK?

This is a follow up to an earlier question about strengthening of foundation in relation to proving the consistency of ZFC. It was shown that it would achieve that, but it may fail short of MK. Here, ...
Zuhair Al-Johar's user avatar
5 votes
1 answer
230 views

Long chains of Dedekind finite sets

This is a variation on this question with amorphous cardinals replaced with dedekind finite sets. Dedekind finite sets are sets that have no countable subset, and it is well known that this is a ...
Ynir Paz's user avatar
  • 384
13 votes
1 answer
549 views

Long chains of amorphous cardinalities

An amorphous set is an infinite set that cannot be partitioned into 2 infinite subsets. An amorphous cardinality is the cardinality of an amorphous set. Working in $\sf ZF$, it is consistent that ...
Ynir Paz's user avatar
  • 384
12 votes
1 answer
506 views

Building the real from Dedekind finite sets

It is well known that the real numbers can be countable union of countable sets by starting with GCH and taking a finite support permutations while collapsing all of $\aleph_n$ for natural $n$. The ...
Holo's user avatar
  • 1,654
3 votes
1 answer
255 views

A possible ${\sf (ZF)}$-theorem in the spirit of the $3$-set-lemma

The number $3$ plays an interesting role in the following statement: $\newcommand{\S}{\sf(S_3)}\S$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in ...
Dominic van der Zypen's user avatar
4 votes
1 answer
199 views

Weak Power Hypothesis and Dependent Choice

Consider in $\newcommand{\ZF}{{\sf (ZF)}}\ZF$ the following statement: Weak Power Hypothesis (WPH): if $X,Y$ are sets and there is a bijection between $\newcommand{\P}{{\cal P}}\P(X)$ and $\P(Y)$, ...
Dominic van der Zypen's user avatar
2 votes
0 answers
115 views

Adding partitions of one but not the other kind

Say that two partitions $(P_i)_{i\in I}, (Q_j)_{j\in J}$ are isomorphic iff there is a bijection $f: I\rightarrow J$ such that $\vert P_i\vert=\vert Q_{f(i)}\vert$ for all $i\in I$. (Note that in the ...
Noah Schweber's user avatar
7 votes
1 answer
400 views

How much choice is needed to prove the completeness of equational logic?

All the proofs of the completeness of (Birkhoff's) equational logic I have read seem to pick representatives for equivalence classes of terms and hence require the axiom of choice. Is AC (or a weak ...
ralphS16's user avatar
13 votes
1 answer
921 views

Cantor-Bernstein with "weakly injective" functions

Let us call a map $f: X \to Y$ between non-empty sets a "weak injection" if $f^{-1}(\{y\})\subseteq X$ is finite for every $y \in Y$. Recall that the (Schroeder-)Cantor-Bernstein-Theorem (...
Dominic van der Zypen's user avatar
17 votes
2 answers
1k views

Axiom of Choice for collections of Equinumerous sets

Let ACE (Axiom of Choice for Equinumerous sets) be the following choice principal: If $S$ is a set of non-empty sets such for any $X,Y\in S$ there is a bijection from $X$ to $Y$, then $S$ has a choice ...
Brian Pinsky's user avatar
8 votes
0 answers
188 views

Reference request: choiceless cardinality quantifiers

There is a substantial literature on the logic of cardinality quantifiers. (E.g., the quantifier $Q_\alpha$ where $M \vDash Q_\alpha x \, \varphi (x)$ iff $\vert \{a \in M : M \vDash \varphi[a] \} \...
Beau Madison Mount's user avatar
14 votes
1 answer
781 views

Does completeness of the theory of a bijection without finite orbits depend on choice?

Consider the following sentences in a first-order language with one unary function symbol $f$: $\forall x \exists y (fy=x)$ $\forall y\forall z(fy=fz\to y=z))$ $\forall x (\underbrace{f\dotsb f}_{n\...
George Hayduke's user avatar
1 vote
0 answers
274 views

Can the following definition of choice principle salvage the prior attempts?

In prior postings 1 , 2, I've presented a definition of choice principle as what is equivalent to a selection principle. However, it was proved that it is inadequate in the sense that it admits ...
Zuhair Al-Johar's user avatar
5 votes
1 answer
266 views

Does weak countable choice imply that the Cauchy reals are Dedekind complete?

Assuming the axiom of weak countable choice, is the set of modulated Cauchy reals Dedekind complete? The second theorem on this ncatlab page claims something equivalent, but it doesn't contain a proof ...
Christopher King's user avatar
2 votes
1 answer
579 views

Is there a strict limit on choice principles in $\sf ZFC$?

Is there a principle $\sf P$ that $\sf ZFC$ [or some suitable extension of it] proves to be a strict limit on choice principles? By a choice principle I mean a sentence (or scheme) that is equivalent ...
Zuhair Al-Johar's user avatar
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 ...
Alex Appel's user avatar
3 votes
2 answers
518 views

Is the Ordering Principle equivalent to a selection principle?

Working in the context of set theory $\sf ZF$, selection may be defined as a function from nonempty sets to their elements. Formally: $\operatorname {selective}(c) \iff \operatorname {function}(c) \...
Zuhair Al-Johar's user avatar
7 votes
1 answer
873 views

Logical strength of a statement about vector spaces

[Apologies if this is a really trivial question, I know virtually nothing about set theory, and the following came up while preparing undergraduate linear algebra lectures.] I'm asking about the ...
David Loeffler's user avatar
3 votes
1 answer
247 views

Is the Class Well Ordering principle "CWO" the maximal choice principle?

In a prior posting, the Class Well-Ordering principle "$\sf CWO$" was presented, which simply states that there is a well-ordering over all classes of $\sf MK$. On the other hand, it is ...
Zuhair Al-Johar's user avatar
3 votes
0 answers
190 views

Reverse-mathematical strength of Banach-Tarski

What is the reverse mathematical strength of the Banach-Tarski paradox? The usual proof of Banach-Tarski should carry out in $\mathrm{ZF}+\mathrm{AC}_\kappa$, where $\kappa$ is the supremum of the ...
C7X's user avatar
  • 1,696
5 votes
1 answer
421 views

Is there a class choice principle over MK that is equivalent to class well ordering over MK?

$\sf MKCWO$ is the theory obtained by adding a new primitive binary relation $\prec$ to the signature of $\sf MK$ and axiomatize that $\prec$ is a well order on classes, that is: $\textbf{Transitive:}...
Zuhair Al-Johar's user avatar
4 votes
2 answers
207 views

Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\lambda$?

Let $\lambda<\kappa$ be cardinals and consider the forcing $\operatorname{Col}(\lambda,\kappa)$ adding a generic surjection $\lambda\to\kappa$. More formally, $\operatorname{Col}(\lambda,\kappa)$ ...
Hanul Jeon's user avatar
  • 2,910
9 votes
0 answers
134 views

Locally presentable and accessible categories without the axiom of choice?

Is there a good reference for the study of locally presentable and accessible categories without the axiom of choice? For instance, it seems one will need to understand: What is a good notion of $\...
Tim Campion's user avatar
  • 62.6k
-1 votes
2 answers
333 views

An equivalent of the axiom of choice? [closed]

There is such a thing as a math course for relatively non-mathematically inclined people that is intended to challenge students' intelligence more than to teach them some mathematics. (It is true that ...
Michael Hardy's user avatar
3 votes
1 answer
226 views

Is Morse-Kelley set theory with Class Choice bi-interpretable with itself after removing Extensionality for classes?

Let $\sf MKCC$ stand for Morse-Kelley set theory with Class Choice. And let this theory be precisely $\sf MK$ with a binary primitive symbol $\prec$ added to its language, and the following axioms ...
Zuhair Al-Johar's user avatar
14 votes
1 answer
1k views

Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?

Kunen showed that Reinhardt cardinals are inconsistent in ZFC. But his proof is a bit technical for a non-set-theorist to follow. Berkeley cardinals are stronger than Reinhardt cardinals. You can ...
Tim Campion's user avatar
  • 62.6k
3 votes
0 answers
144 views

Can well-ordering of the universe due to global choice survive extensive failure of Extensionality?

That axiom of global choice leads to the well-ordering of the universe given the other axioms of Zermelo set theory is a famous result. Now, if we weaken the power set axiom to the axiom stating that ...
Zuhair Al-Johar's user avatar
12 votes
2 answers
411 views

Trading Choice for Comprehension (or Replacement)

This question is basically a request for clarification about a remark made by Sam Sanders in a comment to another question: IIUC what he's saying, there are statements that can be proved either with a ...
Gro-Tsen's user avatar
  • 30.8k
9 votes
3 answers
416 views

Exponentiation of Dedekind cardinals

Question: Let $\mathfrak n$ be an infinite cardinal. In ZF (set theory without the axiom of choice) can either of the implications $$\mathfrak n=\mathfrak n+1\implies2^\mathfrak n=2^{\mathfrak n+1}\...
bof's user avatar
  • 12.7k
6 votes
1 answer
277 views

The Parity Principle and $\mathbf{C}_2$ (choice for $2$-sets)

The Parity Principle states that if $X\neq \emptyset$ is a set, then there is $\mathcal B\subseteq \mathcal P(X)$ such that whenever $a,b\in \mathcal P(X)$ with $a\mathbin\Delta b = \{x\}$ for some $...
Dominic van der Zypen's user avatar

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