The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Formal approach to (countable) prisoners and hats problem.

I've found this nice puzzle about AC (I'm referring to the countable infinite case, with two colors). The puzzle has been discussed before on math.SE, but I can't find any description of what is ...
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2answers
486 views

Dedekind's theorem on an integrally closed algebra over a field without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951. Can we prove the following theorem without Axiom of Choice? Theorem Let $A$ be an integrally ...
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1answer
4k views

Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
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1answer
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Follow up on “Proof of $X \times X \hookrightarrow X$ implies $[X]^2 \hookrightarrow X$”

This is a follow up on this earlier question of mine. We have the following statements: (HSO) For every infinite set $X$ there exists an injection $f: X \times X \hookrightarrow X$ (HSU) For every ...
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1answer
54 views

Can you verify this proof of the Schroeder-Bernstein theorem?

I'm a freshman in college and my professor challenged us to find a proof of this theorem. Please don't give me the answer but please verify if this proof works or, if not, if it is the start of a ...
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1answer
57 views

Does the existence of a recursive sequence that involves an arbitrary choice at every step require the axiom of choice?

Say I want to define a sequence $(x_n)$ recursively, and at each step I make an arbitrary selection for $x_{n+1}$ out of some nonempty pool of acceptable candidates dependent on $x_n$. Does the ...
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1answer
84 views

Proving equivalence of Axiom of Choice

I am working on the following question concerning the axiom of choice and one of its many equivalences. Advice as to whether I am on the right track would be appreciated. As a preface, I have looked ...
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1answer
68 views

Proving Equivalence of Two Version of Axiom of Choice

I am working on an assignment that requires proving the equivalence of two versions of the axiom of choice. (1st form): For any relation $R$, there is a function $H \subseteq R$ with dom $H =$ dom ...
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1answer
89 views

Some properties of Polish space

Let $X$ be a separable complete metric space. I wonder if following properties hold in ZF. Limit Compact ⇒ Compact Does there exist a function$f$ such that $f(E)$ is closed and $f(E)\subset E$, for ...
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1answer
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Basis in topology and AC

Let S ⊂ ℘(X). Let T be the coarsest topology on X which contains S. Then we call T the topology generated by S. Let S ⊂ ℘(X). Then we can easily prove using (generalised distributive law) the ...
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1answer
122 views

Noetherianess of a locally noetherian affine scheme without axiom of choice

I use the definition of a noetherian ring given by Qiaochu in this: A commutative ring is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is ...
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Choice in a Particular HOD Type Model

Let $V \models ZFC$. Let $P$ be a forcing and $G \subseteq P$ be generic over $V$. Let $x \in V[G]$. Let $M$ be the class of set that are hereditarily definable (in $V[G]$) using as parameters $x$ ...
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287 views

Proving equivalence of a tree-based version of Countable Choice for families of finite sets.

In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: Each of the following statements imply those beneath it. The countable union of ...
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0answers
282 views

A few standard results (on metrizability and relative separation strength) without choice?

I've been going back over some results from Munkres's Topology, and I'm curious about some things.... I know that Choice principles have some connection to the separation axioms (in ZF, at ...
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0answers
92 views

Connectedness of parts used in the Banach–Tarski paradox

A quote from the Wikipedia article "Axiom of choice": One example is the Banach–Tarski paradox which says that it is possible to decompose the 3-dimensional solid unit ball into finitely many ...
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One more AC equivalence question

Is "Every vector space admits a norm" weaker than AC? I know that the statement follows from "Every vector space has a basis", which is equivalent to AC.
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0answers
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How do I prove that $S_A\cong S_B\implies |A|=|B|$?

Let $A,B$ be infinite sets such that $S_A\cong S_B$. (Symmetric groups are group isomorphic) How do I prove that $|A|=|B|$? The only proof I know uses Axiom of choice. (That is, using AC to give ...
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0answers
206 views

Is Kunen's claim about non-equivalent forms of Axiom of Choice, true?

Consider the following forms of the axiom of choice: $AC_1:\forall F\neq \emptyset~~~(\emptyset\notin F~\wedge~\forall x,y\in F~~~(x\neq y\rightarrow x\cap y= \emptyset))\rightarrow \exists C~\forall ...
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167 views

What are 'weak' forms of Urysohn's lemma, which do not require choice?

Reference: http://web.mat.bham.ac.uk/C.Good/research/pdfs/horror.pdf It is well known that the original proof of Urysohn's lemma uses a choice principle. (DC) What are weak forms of the Urysohn's ...
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0answers
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Why can countable recursive constructions not be done using countable choice?

Why can countable recursive constructions not be done using countable choice? For example, replace $Z$ by $\omega$ in the following theorem Why can't one prove it using countable choice? (The proof ...
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130 views

Is dependent choice necessary to prove every perfect compact Hausdorff space is uncountable?

The answer to Cardinality of a locally compact space without isolated point shows that AC is required to show that if $X$ is a compact Hausdorff space with no isolated points then $|X| \ge ...
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Construction of injective hulls without axiom of choice

Motivation: It is known that without the axiom of choice (AC), it is not provable that all categories of modules have enough injectives, let alone injective hulls. Still, there are examples of rings ...
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106 views

Fields whose algebraic closure cannot be constructed without the axiom of choice

One can show that the statement that every field has an algebraic closure requires the axiom of choice. However, for almost all "everyday" fields, it seems that one can actually produce an algebraic ...
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Strength of “Every finite dimensional subspace of a vector space has a complement”

Does the following choice principle have a name? Every finite dimensional subspace of a vector space has a complement. Equivalently, every line inside a vector space has a complementary ...
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144 views

Does the non-emptiness of the spectrum of an element of a Banach algebra depend on the Axiom of Choice?

One of the most basic results in functional analysis states that the spectrum of any element of a Banach algebra is non-empty. The proof, as most people might have seen, makes use of Liouville's ...
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Example of a maximal idea

Let $A$ be the set of bounded continuous functions from the set of real numbers to itself. Then $A$ is a ring under pointwise addition and multiplication. The set $I$ of all functions $f \in A$ ...
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$Gal(\bar{\mathbb Q}/\mathbb Q)$ without choice, and constructive Galois theory

By this question: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice? We have that over ZF, algebraic closures of $\mathbb Q$ aren't unique. Are their Galois groups as extensions over ...
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0answers
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What is the name of proofs with (without) Axiom of Choice

In many contexts we distinguish between proofs using AC and proofs which do not use AC. (To phrase this somewhat differently: If there is a proof without AC, this proof is usually preferred.) I would ...
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116 views

Choice-less Set Theory for Dummies

In almost every graduate set theory text there are some parts about equivalences of $AC$, its consequences, some axioms like $AD$ which imply $\neg AC$, some well-known axiomatic systems which $AC$ ...
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90 views

On the contradictory nature of large cardinals & choice-like axioms

Compare these two results: Theorem (Scott): $ZFC+V=L\vdash \nexists~\text{Measurable cardinal}$ Theorem (Kunen): $ZFC+AC\vdash \nexists~\text{Reinhardt cardinal}$ Now compare these two ...
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Separable Hahn-Banach and the axiom of choice

We had in our functional analysis course a proof for the Hahn-Banach theorem on a separable Banach space which doesn't need, according to our professsor, the axiom of choice. Yesterday I read the ...
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Differentiability and Monotonic Functions

I just read proof from Royden of theorem: 'Every Monotonic functions are differentiable almost everywhere.' But proof use Vitali Covering Lemma. But Vitali Covering Lemma is based on fact if we assume ...
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Axiom of choice in proof of Wigner's theorem?

In Appendix A of chapter 2 of "The Quantum Theory of Fields," vol. 1, Weinberg presents a proof of Wigner's theorem: given a symmetry transformation $T$ of rays, one can extend this to a symmetry ...
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317 views

Normal closure of field extension, axiom of choice

Update My previous proof was incorrect. This updated proof is inspired by the comment by 'MartianInvader'. Problem I can prove the statement 'Every algebraic extension $L:K$ has a normal closure ...
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Does the existence proof of $\mu$-completion require choice?

Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mathfrak{M}^*=\{E\subset X|\exists A,B\in\mathfrak{M} \text{ such that} A\subset E \subset B \text{ and} \mu(B\setminus A)=0\}$. How do i show that ...
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Theorem's relying on algebraic closures

When working with fields, it's a usual method to work on an algebraic closure of a field to obtain results about that field. In general (i. e. unless you're explicitly considering "well-behaved" ...
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132 views

Connected subset of a separable metric space is separable?

Continuity and the Axiom of Choice I have proved a small generalization of Brian's argument, that is, "If $f:X\rightarrow Y$ is sequentially continuous on $X$ and $X$ is separable, then $f$ is ...
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248 views

Noetherian condition on the ring of formal power series without Axiom of Choice

I use the definition of a Noetherian ring given by Qiaochu in this: A commutative ring is Noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is ...
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0answers
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Existence of minimal sub-systems

A topological dynamical system is a topological space $X$ together with a continuous function $f : \ X \to X$. In the following, I will assume that $X$ is compact and Hausdorff (in other words, I work ...
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First Order Model Existence without Weak Konigs Lemma or Choice

In studying Godel's Completeness Theorem and its various related formulations like the Model Existence Theorem and the Lowenheim-Skolem Theorem there is one rather subtle point that I have not yet ...
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divisible subgroup without axiom of choice

the theorem asserting that the divisible subgroup of an Abelian group is a direct summand depends on Zorn's lemma. in ZF without AC is there a construction which yields a model of an Abelian group ...
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Proof of “Every vector space has a basis $\implies$ AC” without mentioning von Neumann hierarchy

I am writing a short (30-50 pages) report on AC for an exam. I really would like to include the proof that "Every vector space has a basis $\implies$ AC". Actually, every proof I could find proves ...
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52 views

Hilbert space without the projection theorem

One succinct statement of the projection theorem in Hilbert space is $A+A^\bot=\scr H$, where $A\in\scr C$, the set of closed subspaces of $\scr H$. (We will also denote the set of all subspaces by ...
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Hilbert projection theorem without countable choice

All the proofs of the Hilbert projection theorem, existence part, that I have seen so far use countable choice (usually implicitly). Is this necessary? It seems like you might be able to leverage the ...
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Amount to choice necessary to prove instances of Tychonoff theorem

Let $I$ be a fixed nonempty set. I would like to know how much choice is necessary in order to prove that the product of any $I$-indexed family of compact topological spaces is compact (under the ...
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Set Theory , Konig's Lemma and Infinite Graph Theory

I am trying to understand the basics of Infinite Graph theory and various preconditions in Konig's Lemma. The texts I have studied tend to use the Axiom of Choice (usually Zorn's Lemma) as a tool of ...
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Is there a connection between contravariant functors and the axiom of choice?

Given that both can be seen as talking about reversing arrows between two objects: Is there a connection between contravariant functors and the axiom of choice? I'm initially motivated geometrical ...
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Choice-like axioms close to AC & rank-into-rank hypothesizes

Consider the following folklore theorems within ZF, Theorem 1: $V=L$ implies "$0^{\sharp}$ doesn't exist". Theorem 2: $V=L[U]$ implies "$0^{\dagger}$ doesn't exist". Theorem 3: $AC$ ...
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Existence of a basis in constructive vector spaces

As I was trying to review forgotten knowledge on Vector Spaces in wikipedia, I read that the existence of a basis follows from Zorn lemma, hence equivalently from the axiom of choice. Actually, the ...
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Is it possible to prove product of basis is a basis for the box topology without AC?

Let $\{(X_i,\tau_i)\}$ be a collection of topological spaces and $\mathscr{B}_i$ be a basis for $\tau_i$. Let $\mathbb{B}=\{\prod p_i : p\in \prod \mathscr{B}_i\}$. Then is $\mathbb{B}$ a basis for ...