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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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 ...
7
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1answer
160 views

What is the significance of having Prime Ideal Theorem in models for failure of Axiom of Choice?

Prime Ideal Theorem says: PIT: Every ideal on a Boolean algebra can be extended to a prime ideal. It follows from Axiom of Choice but is weaker than it. In many cases I saw that people check ...
2
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1answer
58 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
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Countable choice and totally bounded metric spaces

Can we prove that the following statement is equivalent to the axiom of countable choice (CC)? If every sequence in a metric space $X$ has a Cauchy subsequence, then $X$ is totally bounded. ...
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99 views

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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296 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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314 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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105 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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96 views

(Non-Hopfian) groups that only have quotients that are themselves or the trivial group.

A group is non-Hopfian provided it is isomorphic to a proper quotient. The classic, finitely presented, example of such a group is the Baumslag-Solitar group $$BS(2,3)= \langle x,t \mid t^{-1}x^2 t ...
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48 views

Undetermined game of length $\omega_1+\omega$, without choice

On the following page, Taranovsky is talking about his "Determinacy Maximum" axiom: http://web.mit.edu/dmytro/www/DeterminacyMaximum.htm He also justifies the choice of the name, by pointing out that ...
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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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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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248 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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85 views

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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178 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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109 views

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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137 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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45 views

Over ZF does “every non-seperable Hilbert space has an orthonormal basis” imply “there exists a non-Lebesgue measurable set”?

I know from this question that it's an open problem whether or not the existence of a dense orthonomral basis for every real or complex Hilbert space $(\text{B}_\text{orth})$ implies the axiom of ...
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45 views

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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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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156 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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52 views

Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

Does the ring of formal power series $\mathbb{F}[[x]]$ as a vector space over $\mathbb{F}$ admit a basis without assuming the Axiom of choice, at least in some special cases of $\mathbb{F}$? I'm ...
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53 views

$Gal(\bar{\mathbb Q}/\mathbb Q)$ 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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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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391 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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28 views

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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95 views

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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79 views

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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80 views

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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79 views

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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102 views

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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95 views

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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135 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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252 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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50 views

Choice of a skeleton

Suppose we are in presence of a strong enough axiom of choice (e.g., choice for conglomerates). I know that any category has a skeleton, but I would like to know if I can choose a skeleton which ...
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134 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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35 views

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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96 views

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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42 views

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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87 views

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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53 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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79 views

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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54 views

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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174 views

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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57 views

A question regarding a theorem of Erdos and Hajnal

Consider the following theorem of Erdos and Hajnal: Definition. For any set $x$, a function $f$ is called ${\omega} $-Jonsson iff $f$: $^{\omega}x$ $\rightarrow$ x and whenever $y$$\subseteq$$x$ and ...
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every infinite bounded subset of X has an limit point in X and CC

Related to questions and definitions used in question: Nearest point property and WeistrassB copactness of each infinite bounded set Proof of one side of theorem: every bounded infinite bounded ...
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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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61 views

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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43 views

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 ...