# 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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### 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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### 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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### 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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### 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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### 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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### 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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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 $\alpha(n+1) \subseteq \... 7 votes 2 answers 601 views ### Can second-order logic identify "amorphous satisfiability"? Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can ... 8 votes 1 answer 647 views ### If a vector space has a basis then its dual vector space has a basis Consider the following statement: If a vector space has a basis then its dual vector space also has a basis. It is not an axiom of ZF. It clearly follows from the Axiom of Choice. But it is also ... 5 votes 1 answer 202 views ### Countable choice for countable subfamily Axiom of Countable Choice (CC) states that for every countable family$\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$... 4 votes 1 answer 137 views ### Is Axiom of Choice for convex sets of distributions on naturals necessary? Take any family$(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]$such ... 8 votes 1 answer 188 views ### Does ZF + BPI alone prove the equivalence between "Baire theorem for compact Hausdorff spaces" and "Rasiowa-Sikorski Lemma for Forcing Posets"? Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o.$\mathbb{P}$(i.e.$\mathbb{P}$is a reflexive transitive relation) and for any countable family of dense subsets of$\mathbb{...
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 ...
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 ...