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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Are countably infinite, compact, Hausdorff spaces necessarily second countable?

Let $X$ be a compact Hausdorff space. If $X$ is finite, then there are at most finitely many open sets and so it is second-countable. On the flipside if $X$ is uncountably infinite, it is possible ...
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The topological product of path-connected spaces is path-connected $\Rightarrow\sf AC$?

There is a very natural proof that the product of path-connected spaces is path-connected: Let $X=\prod_{i\in I}X_i$ be a product of path-connected spaces $X_i$. Given $(x_i)_{i\in I},(y_i)_{i\in ...
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Proving that axiom of choice is equivalent to this proposition

I'm trying to do the problem 8.2 of the book Notes on Set Theory of Moschovakis. I want to prove that Axiom of Choice is equivalent to this statement: $$\forall A\neq \emptyset,\, \forall ...
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A descending chain and choice axiom.

Let $R$ be a relation on a set $A$ such that for every $x\in A$ there exist a element $y\in A$ such that $x\mathrel{R}y$ the there exist a function $f\colon\omega\rightarrow A$ such that ...
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1answer
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Is this proof using the axiom of choice correct?

In Munkres' Topology, the axiom of choice is defined as follows: Given a collection $A$ of disjoint nonempty sets, there exists a set $C$ containing exactly element from each element of $A$; that ...
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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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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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Do I need axiom of choice in this question?

If $X$ is a finite set and $Y$ an infinite one, I'm trying to prove there is a injective function $f:X\to Y$. My solution: Let $X=\{x_1,\ldots,x_n\}$, for each $x_i$ choose $y_1,\ldots,y_n$ such that ...
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131 views

Why is Zorn's Lemma called a lemma?

By convention a Lemma is a technical intermediate step which has no standing as an independent result. Lemmas are only used to chop big proofs into handy pieces. (Quoted from here) I am wondering why ...
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if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$ [duplicate]

I'm trying to prove if $f:X\to Y$ is a surjective, then there is an injective function $g:Y\to X$. I know this intuitively, since $f$ is surjective, for every $y$ in $Y$ we need just to choice an ...
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Proving statements , that require Zorn's lemma , for countable case directly by well-ordering principle of natural numbers

We know that for countable sets , the existence of a choice function is a consequence of the well-ordering principle ; and it is also known that the results like "every vector space has a maximal ...
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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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1answer
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Additive function and continuity at a point

Does continuity at a point and Additive function imply continuity at all other points in a normed linear space. Is there some result like there exist a in field such that f(x) = ax for all x in normed ...
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Basic definition of continuity [duplicate]

Ltf(c+h) = f(c)(h goes to 0) if and only if Ltf(x) = f(c)(x goes to c). I am able to prove this fact using sequential criterion of continuity. But sequential criterion is dependent on Axiom of ...
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Open set in $\mathbb{R}$ as countable union of open intervals and which version of Choice [duplicate]

In proving, every non-empty open set in $\mathbb{R}$ is union of a countable collection of disjoint open intervals in $\mathbb{R}$. It seems to me this result is using some version of Choice(probably ...
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2answers
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$\omega_2$ is not a countable union of countable sets [duplicate]

Without using axiom of choice, can we show that $\omega_2$ is not a countable union of countable sets? I know this cannot be done for $\omega_1$.
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Is $\text{Hom}(\prod_p \Bbb Z/p\Bbb Z, \Bbb Q) = 0$ possible without choice?

That divisible abelian groups are precisely the injective groups is equivalent to choice; indeed, there are some models of ZF with no injective groups at all. Now, given that $\Bbb Q$ is injective, ...
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1answer
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Proof that the finite product of nonempty sets is nonempty without axiom of choice from ZF

How do you prove that for $X_{i} \neq \emptyset$, $i \in \{1,...,n\}$ that $\prod_{i=1}^{n} X_{i} \neq \emptyset$ only using the ZF axioms but not the Axiom of Choice? I would like to see a rigorous ...
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Injections from all ordinals into a set $X$

We are working in $\mathsf{ZF}$. Let $X$ be a set. Let $A$ be the class of all injections $f: \alpha \to X$ for arbitrary ordinals $\alpha$. I am quite sure that, in fact, $A$ is a set, since if ...
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Does $A\times A\cong B\times B$ imply $A\cong B$?

This is similar to What does it take to divide by $2$? about $(A\sqcup A\cong B\sqcup B)\Rightarrow A\cong B$ which is valid in $\textsf{ZFC}$ by using cardinalities and also in $\textsf{ZF}$ by some ...
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Counterexample to the Hausdorff Maximal Principle

The Hausdorff Maximal Principle states: Every partially ordered set $\left(X,\leqslant\right)$ has a linearly ordered subset $\left(E,\leqslant\right)$ such that no subset of $X$ that properly ...
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1answer
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Tychonoff principle iff Wellordering principle

Tychonoff Principle: Let $X_i$ for $i\in I$ be any sequence of non-empty sets indexed by the set $I$. Then the direct product $\prod_{i\in I}X_i$ is not empty, where $\prod_{i\in I}X_i$ is defined to ...
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Do an infinite set and its double have the same cardinality? [duplicate]

My question was inspired by this answer. Suppose $A$ is an infinite set. Does its double, $A\times\{0,1\}$, always have the same cardinality? In my head I quickly spotted a simple proof that the ...
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1answer
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General Distributive Law and Axiom of Choice

Where can I find the proof of the fact that general distributive law of union over intersection and intersection over union is equivalent to Axiom of Choice? The mathematical formulation of the ...
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Does $\aleph_0\cdot\kappa=\kappa$ for every $\kappa\ge\aleph_0$ hold in ZF?

It is easy to show that for any (Dedekind) infinite cardinal $\kappa$ we have $\aleph_0+\kappa=\kappa$. Definition of an infinite cardinal is a cardinal such that $\aleph_0\le\kappa$. (I believe ...
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1answer
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Prove that Open Sets in $\mathbb{R}$ are The Disjoint Union of Open Intervals Without the Axioms of Choice

There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?
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Prove that $\forall\alpha\geq\omega$, $|L_\alpha|=|\alpha|$ without AC

Without using Axiom of Choice, prove that $$\forall\alpha\geq\omega,~~|L_\alpha|=|\alpha|,$$ in which $\alpha$ is an ordinals, $\omega$ is the set of natural numbers, $L_\alpha$ is the $\alpha$-th ...
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157 views

Are there logical arguments against modern $\sf ZFC$ set theory?

As of asking this question, my knowledge of set theory is quite pedestrian. I've read about it in numerous nontechnical papers and even worked through three chapters of Jech - Set Theory, but in terms ...
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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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Axiom of Choice: What exactly is a choice, and when and why is it needed?

I'm having trouble understanding the necessity of the Axiom of Choice. Given a set of non-empty subsets, what is the necessity of a function that picks out one element from each of those subsets? For ...
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Statements comparable with Axiom of Choice in ZF

Let $AC$ denote any fixed statement of the Axiom of Choice in $ZF$. Consider the set of statements $\phi$ in the language of $ZF$ such that either $ZF+\phi$ proves $AC$ or $ZF+AC$ proves $\phi$. The ...
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Adjunctions via Reflections and the Axiom of Choice

I have met two ways of defining adjunctions: via the triangle identities, and via reflections. Proposition 3.1.2 Let $F:\mathsf A \rightarrow \mathsf B$ be a functor and $B$ an object of $\mathsf ...
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Existence of finite sets of infinite set without using AC

Is it possible to prove that every infinite set $B$ has a subset of cardinality $n$, for every natural $n$, without using AC? I know how to prove this claim by induction. In the induction step I chose ...
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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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Examples of Extreme Anti-Choice Axioms

Axiom of Choice has many variants like the followings: There is a choice set for every family of non-empty sets. All sets are well-orderable. Of course in many cases one don't need AC to ...
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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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Why is the axiom of choice not taught from the start to mathematics undergraduates?

I've recently discovered that the following theorems require the axiom of choice to be proven: every surjective function has a right inverse. a real-valued function that is sequentially continuous ...
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Probabilistic implications of the existence of non-measurable sets

Measure theory and probability theory are deeply connected through the interpretation of subset measures on the sample space as probabilities of events. A major (and somewhat disturbing) result from ...
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ZF and The Cardinality of The Set of Finite Subsets

In a comment on one of my answers, I claimed that the abelian group generated by a set of $S$ generators, each of order two, could take on any infinite cardinality; this is equivalent to saying that, ...
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is there a known set in ZF, such that we can't find a well order on? [duplicate]

is there a known set in ZF, such that we cant find a well order on? since the axiom of choice $(AC)$ and it's negation is consistent with ZF, i wonder if we have a concrete example of a set $A$ that ...
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1answer
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Compact sets closed in Hausdorff spaces without choice?

An elementary proof that compact sets are closed in Hausdorff spaces involves making arbitrary choices based on the Hausdorff property. Is there a way to avoid invoking choice?
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A very quick way to prove a set is measurable. [duplicate]

All examples of non-measurable subset of $\mathbb{R}$ (in the Lebesgue sense) seem to need the axiom of choice in some way or the other. Hence, can we say: The set $A\subseteq \mathbb{R}$ is ...
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Can someone point out the flaw in my proof of AC?

I have a fake proof of the axiom of countable choice. Obviously it is not correct, but I cannot see my flaw. Forgive me, I am only learning set theory. Let $\{A_n : n \in \mathbb{N}\}$ be a countable ...
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Without AC, it is consistent that there is a function with domain $\mathbb{R}$ whose range has cardinality strictly larger than that of $\mathbb{R}$?

I stumbled across this question earlier, and the top comment on the bottom answer asserts two claims: Without the Axiom of Choice, It is consistent that there exists a function with domain ...
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A question dealing with cardinals, and axiom of choice.

I am given sets $A$,$B$ such that there exists $f:A\rightarrow B$ s.t. $f$ is onto $B$. I am trying to show that $B\le A$ Let $b\in B$, consider $\{a\in A \mid f(a) = b\}$, assuming axiom of choice, ...
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Axiom of Choice: Family of non-empty sets mutually disjoint or not?

I have noticed that the family of non-empty sets referred to in statements of the Axiom of Choice is sometimes required to be mutually disjoint, and sometimes not. Why is that?
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Mean value theorem and the axiom of choice

There's this theorem in Spivak's book of Calculus: Theorem 7 Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for ...
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AC and Tychonoff theorem

Although I have proof with me that Tynhonoff theorem implies AC. But I have some difficulties with it: 1. Do we define topology on empty set. If not then in proof of Tynhonoff theorem implies AC we ...
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How much choice is needed for the transfer principle?

To construct the hyperreals via ultrapower the Boolean prime ideal theorem apparently suffices. However, to prove the transfer principle for the extension $\mathbb{R}\subset{}^\ast\mathbb{R}$ ...
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1answer
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Collection of surjective functions implies axiom of choice

if I have this: (a) If $\left \{ f_i:A_i\rightarrow B_i|i\epsilon I \right \}$ is a collection of surjective functions then $\prod_{i\epsilon I} f_i: \prod_{i\epsilon I} A_i\rightarrow ...