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.

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Field automorphisms of projective spaces without the axiom of choice

Suppose P is a projective space over the field $k$. If P has finite dimension $n$, we can fix a base. Relative to this base, the full automorphism group of P can be described by the action on the ...
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Nonlinear automorphisms of projective spaces and the axiom of choice

Let $k$ be a field and $\mathbf{P}$ a projective space over $k$. If we accept the axiom of choice (AC), then $\mathbf{P}$ has a basis and a dimension $m$, and if $m$ is finite, the automorphism group ...
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Does a well-pointed topos with enough projectives satisfy the internal axiom of chioice?

If yes, then I am also wondering if being well-pointed can be weakened to boolean (i.e. this is in the context of using Set as our metalogic so that well pointed Topoi are boolean). If not, then any ...
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A question about the axiom of dependent choice

Let $\mathrm{NBG}^-$ be $\mathrm{NBG}$ minus the Axiom of Choice for Classes (including sets)). Further let $\mathrm{DC}$ be the Axiom of Dependent Choice for sets and $\mathrm{DC}^\omega$ be Bernays ...
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4 votes
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A question regarding the Hahn-Banach theorem and Banach limits

Set theorists typically prove the existence of Banach limits (EBL) using the Ultrafilter Theorem or, its equivalent, the Boolean Prime Ideal Theorem (BPI). Analysts, on the other hand, typically prove ...
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Does the partition principle imply (DC)?

For sets $x, y$ we write $x\leq y$, if there is an injection $\iota: x \to y$, and we write $x \leq^* y$ if either $x = \emptyset$ or there is a surjection $s: y \to x$. In ${\sf (ZF)}$ we have that $...
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Can Reinhardt cardinals be compatible with Choice in absence of Extensionality?

Is the proof of existence of Reinhardt (and higher) cardinals violating Choice dependent on Extensionality in an essential manner? What I mean is if we work in $\sf ZFA$ would it be possible to have a ...
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finite image has finite preimage without the axiom of choice [duplicate]

Is it necessary to use the axiom of choice, to prove, that finite image of function $f: X\to Y$ implies the existance of restriction of a function which has the same image and finite domain? I can ...
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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 ...
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Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$?

For any set $X$ and cardinal $\mu \neq \emptyset$, we denote by $[X]^\mu$ the collection of subsets of cardinality $\mu$. If $\kappa, \mu \neq \emptyset$ are cardinals and $f: [X]^\mu\to \kappa$ is a ...
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Why is choice needed in Ellis' Lemma?

Ellis Lemma on idempotent elements asserts that: Lemma (Ellis). Every compact semigroup has an idempotent. The proof below is excerpted from Todorcevic's Introduction to Ramsey Spaces, Lemma 2.1. ...
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13 answers
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Unnecessary uses of the axiom of choice

What examples are there of habitual but unnecessary uses of the axiom of choice, in any area of mathematics except topology? I'm interested in standard proofs that use the axiom of choice, but where ...
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Is there an "anti-choice axiom"?

Edit: I have now rewritten the question in terms of well-ordering. Before, had I used the axiom of choice directly, which made the question unclear. The anti-foundation axiom is a nice thing: It not ...
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Can a countable union of two-element sets be uncountable?

I am thinking about the Axiom of Choice and I am trying to understand the Axiom with some but a little progress. Many questions are arising in my head. So, I know that there exists a model of ZF set ...
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Completeness of the space $L^p$ and the Axiom of Countable Choice

I am thinking about the proof that the usual $L^p$ spaces are complete. So, let $(X,\mathcal{F},\mu)$ be a measure space and let $p\in[1,+\infty)$. Important: by a measure I mean a nonnegative $\sigma$...
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What would be examples of modified Scotts cardinals with less structure?

The intention behind this posting is to arrive at a definition of cardinality that can work in $\sf ZF$, and at the same time doesn't exhaust the whole of $V$. The known definition uses Scott's trick, ...
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Does existence of those cyclic cardinals imply a kind of Choice?

Working in $\sf ZF - Fnd$, add the following axiom: AntiFoundation: $\forall x: x \neq \emptyset \to \exists! y: y \in y \land y \sim x$ where "$\sim$" stands for existence of a bijection. ...
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1 vote
1 answer
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Is existence of a cardinal that witness non-failure of GCH everywhere everyway, a theorem of ZF?

In an earlier positing to $\mathcal MO$, it appears that the answer to if the $\sf GCH$ can fail everywhere in every way is to the negative, this is the case in $\sf ZFC$, however it also appears that ...
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2 votes
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Is Definable Partition Principle not equivalent to AC over ZF?

Definable Partition Principle: If $\phi;\psi$ are formulas in which only the symbol $x$ occur free, then: $$A = \{x \mid \phi\} \land B=\{x \mid \psi\} \land B \, ||| \, A \to B \leq A$$ where $|||$ ...
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Injections without fixed-points and the Axiom of Choice

Consider the following statement in $\sf ZF$: (I) Whenever $X$ is a set with more than $1$ element, there is an injective map $\iota: X\to X$ such that $\iota(x) \neq x$ for all $x\in X$. The Axiom ...
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5 votes
1 answer
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Existence of surjection vs injection over $\sf ZF$

Consider the following statements in $\sf ZF$: (S) If $A, B$ are nonempty sets, then there is a surjection $s:A \to B$, or there is a surjection $t:B\to A$. (I) If $A, B$ are sets, then there is an ...
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11 votes
1 answer
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Is every set being cardinal definable consistent with ZF + negation of Choice?

Recall the definition of cardinal definable, where every set being cardinal definable is proved consistent relative to ZF + V=HOD. To re-iterate it: $Define: X \text { is cardinal definable} \iff \\\...
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9 votes
3 answers
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Terminology for a set that does not surject onto $\omega$ (in ZF)

Short question: Is there a standard term for a set $F$ such that there does not exist a surjection $F \twoheadrightarrow \omega$ (in the context of ZF)? More detailed version: Consider the following ...
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Is choice over definable sets equivalent to AC over axioms of ZF-Reg.?

If we add the following axiom schema to ZF-Reg., would the resulting theory prove $\sf AC$? Definable sets Choice: if $\phi$ is a formula in which only the symbol $``y"$ occurs free, then: $$\forall X ...
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2 votes
1 answer
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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 ...
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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 $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ ...
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8 votes
2 answers
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Continuous linear functionals and the Axiom of Choice

Can one prove without the Axiom of Choice that for every normed vector space $X$ there exist a nonzero continuous linear functional on $X$?
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6 votes
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Groups $G$ such that $\mathrm{Aut}(G) \simeq \mathbb{Z}/2\mathbb{Z}$

$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$First I hope my question belongs here, please let me know if it doesn't. It isn't too hard to show there is no groups $G$ such that $\Aut(G) ...
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Is this bi-hierarchy anti-Cantorian system consistent with AC?

In a posting to MO I've presented a bi-hierarchy anti-Cantorian system. The consistency of this system is not settled yet. However, if we assume that it is consistent, then would it be consistent with ...
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3 votes
1 answer
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Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of ...
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11 votes
1 answer
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Scott's trick without regularity

In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V_{α+1}\setminus V_α\mid α∈\mathrm{Ord}\}$, such a partition let ...
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Infinitary Ruzsa triangle inequality without Choice

If $X, Y$ are subsets of an abelian group, we denote $X - Y = \left\{ x - y \ \lvert \ x \in X, \ y \in Y \right\}$. Question: Let $A, B, C$ be non-empty subsets of an abelian group $G$. Is the ...
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15 votes
1 answer
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Examples of vector spaces with bases of different cardinalities

In this question Sizes of bases of vector spaces without the axiom of choice it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different ...
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6 votes
0 answers
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Is $2^\mathfrak{m}<2^\mathfrak{n}\Rightarrow\mathfrak{m}<\mathfrak{n}$ equivalent to the axiom of choice?

Note that $2^\mathfrak{m}<2^\mathfrak{n}\Rightarrow\mathfrak{m}<\mathfrak{n}$ follows from the axiom of choice. Is it equivalent to the axiom of choice? A similar question. Remark. Lindenbaum ...
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10 votes
1 answer
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Does GCH for alephs imply the axiom of choice?

GCH for alephs means the statement that, for any aleph $\kappa$, there are no cardinals $\mathfrak{r}$ such that $\kappa<\mathfrak{r}<2^\kappa$. Does GCH for alephs imply the axiom of choice? ...
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Eigenvalues without the axiom of choice

Without the Axiom of Choice (AC), we can find models of ZF set theory in which some vector spaces have no base, and also models in which some vector spaces have bases of different cardinalities. The ...
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8 votes
2 answers
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SPOT as a conservative extension of Zermelo–Fraenkel

In Infinitesimal analysis without the Axiom of Choice, Hrbacek and Katz have shown that it is possible to formulate an axiomatic theory which provides a formalisation of calculus procedures which make ...
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Eigenbases without the Axiom of Choice

I understand that in ZF set theory without the Axiom of Choice (AC), it is consistent to have models in which there exist vector spaces over some (unspecified) field $k$ without a basis. So in ...
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Axiom of Choice and bases of $k$-vector spaces, $k$ fixed

I know that from ZF + the Axiom of Choice (AC) follows that every vector space has a basis. And, conversely, Blass proved that in ZF set theory, the assumption that every vector space has a basis ...
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1 vote
1 answer
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Strictly descending sequences of sets, the Partition Principle, and the Boolean Prime Ideal Theorem

In ${\sf ZFC}$ it can be easily proved that we cannot have infinitely descending sequences of cardinalities, that is, the following statement does not hold: (DescSeq) There is a set $A$ a map $\alpha:...
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35 votes
2 answers
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Can one show that the real field is not interpretable in the complex field without the axiom of choice?

We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number ...
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2 votes
1 answer
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Second-order strong minimality and amorphousness, take 2

Recently I asked a question about whether a second-order analogue of strong minimality could correspond to amorphous satisfiability (= having a model whose underlying set cannot be partitioned into ...
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3 votes
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"Matryoshka" sets and the Axiom of Choice

Consider the following two very similar statements in ${\sf ZF}$: (Mat_1) There is a set $A$ a map $\alpha: \omega \to {\cal P}(A)$ such that for all $n\in \omega$ we have $\alpha(n+1) \subseteq \...
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7 votes
2 answers
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Can second-order logic identify "amorphous satisfiability"?

Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can ...
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8 votes
1 answer
647 views

If a vector space has a basis then its dual vector space has a basis

Consider the following statement: If a vector space has a basis then its dual vector space also has a basis. It is not an axiom of ZF. It clearly follows from the Axiom of Choice. But it is also ...
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5 votes
1 answer
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Countable choice for countable subfamily

Axiom of Countable Choice (CC) states that for every countable family $\left\{A_i\right\}_{i=1}^\infty$ of nonempty sets there exists choice function $f \colon \mathbb{N} \to \bigcup_{i=1}^\infty A_i$ ...
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Is Axiom of Choice for convex sets of distributions on naturals necessary?

Take any family $(S_i)_{i∈I}$ such that each $S_i$ is a convex set of functions $f : ℕ→[0,1]$ where $\sum_{k∈ℕ} f(k) = 1$. By "convex" we mean that for any $f,g∈S_i$ and any $a,b∈[0,1]$ such ...
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8 votes
1 answer
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Does ZF + BPI alone prove the equivalence between "Baire theorem for compact Hausdorff spaces" and "Rasiowa-Sikorski Lemma for Forcing Posets"?

Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o. $\mathbb{P}$ (i.e. $\mathbb{P}$ is a reflexive transitive relation) and for any countable family of dense subsets of $\mathbb{...
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9 votes
1 answer
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Do saturated models require choice?

Let $T$ be a first-order theory, and suppose we want to build a saturated model $\mathbb U$ of $T$. That is, we want a model $\mathbb U$ of cardinality bigger than $|T|$, saturated in its own ...
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Can we choose a sequence of Hilbert spaces?

Let $n$ be a fixed natural number. Let $H$ be a complex Hilbert space and $H_1,\dotsc,H_n$ be closed subspaces of $H$. Set $H_0\mathrel{:=}H_1\cap H_2\cap\dotsb\cap H_n$ and let $P_i$ be the ...
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