# Tagged Questions

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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### The smallest infinity and the axiom of choice

The short version of this question is: which (natural) axiom should be added to ZF so that the statement "$\aleph_{0}$ is the smallest infinity" becomes true? A set $A$ is called infinite if it can ...
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### $|A^2|=|A|$ for every infinite $A$ iff Axiom of Choice holds. [duplicate]

I've seen this assertion in a few comments around the site, and I found the answer to the $\rightarrow$ implication here. Does anyone know a (hopefully simple) proof of the $\leftarrow$ implication?
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### What is the first order formulation of Zorn's lemma in the language of set theory?

Very often in notes of courses in set theory you find the assertion that in ZF the Axiom of choice (AC) is equivalent to Zorn's Lemma (ZL) (which is equivalent to Well Ordering Principle which it ...
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### Proving implication on well ordered set implies AC

Consider the following statement: If $A$ is a well-ordered set such that every nonempty subset of $A$ has a maximal element, then $A$ is finite. I am trying to prove that this statement implies ...
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### Do the ZF-provable forcing principles differ from the ZFC-provable forcing principles?

In "The Modal Logic of Forcing", Joel David Hamkins and Benedikt Löwe show that the ZFC-provable forcing principles are exactly those of the modal logic S4.2 (interpreting $\Diamond \phi$ as asserting ...
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### Axiom of choice and the empty set

Could someone explain to me why it is important for a set to be non-empty when working with a choice function?
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### Why isn't there a total order of $\cal P(\Bbb R)$?

I have heard that, in some models of ZF, $\cal P(\Bbb R)$ has no total order. How could one prove this? I already know that $\Bbb R$ isn't necessarily well-ordered. I'm guessing that one could reduce ...
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### Context for Russell's Infinite Sock Pair Example

I wanted to verify the following considerations on the context of Russell's infinite sock pair conundrum. The conundrum pointed out that a rule for choosing from pairs of shoes is possible a-priori. ...
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### Axiom of choice in HoTT without sethood requirement

3.8.3 of the HoTT book gives the following as a variant of the axiom of choice: $\Pi$ (X : U) (Y : X $\rightarrow$ U), (isSet X) $\rightarrow$ ($\Pi$ (x : X), isSet (Y x)) $\rightarrow$ ($\Pi$ (x :...
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### When using an axiom scheme, are we implicitly using a choice principle?

I heard an interesting argument from a colleague recently that went something like this. Whenever we are using an axiom scheme, we are essentially choosing one of the instances of this scheme, and ...
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### Cardinality of the Euclidean topology and the axiom of choice

It is relatively straightforward to prove that the Euclidean topology has the same cardinality as the space itself. I have sketched a proof below. The proof seems to rely quite heavily on the axiom of ...
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### Is there a group with countably many subgroups, but is not countable in ZF?

Inspired by this question, although I don't think it was the OP's intention, hence this separate question: Is there a group $G$ with countably many subgroups, but is a not a countable group itself ...
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### Without the Axiom of Choice, does every infinite field contain a countably infinite subfield?

Earlier today I asked whether every infinite field contains a countably infinite subfield. That question quickly received several positive answers, but the question of whether those answers use the ...
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### If a certain kind of object exists and is unique, can we prove its existence and uniqueness without axiom of choice?

Suppose that using Zorn's lemma, we have proven that an object with some properties exists and then we've proven that such object is unique. Can we always conclude that we can prove the existence (...
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### A $\sigma$-algebra equivalent to the Lebesgue algebra under CC

Working in $\sf ZF$, is there a natural definition of an algebra $\Sigma$ with the following properties: $\Sigma$ is a $\sigma$-algebra on $\Bbb R$, i.e. it is closed under complement and countable ...
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### (Ordinal number)How to prove that epsilon-naught is countable without using Axiom of Choice? [duplicate]

How to prove that epsilon-naught is countable without using Axiom of Choice? or, Can we explicitly show that there is isomorpism between epsilon naught and subset of rational number?
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### Is there an explicit function / no-axiom-of-choice construction $f:[0,1] \to [0,1]$ so that uncountably many disjoint subsets map onto $[0,1]$?

So if I asked: "Is there an explicit function / no-axiom-of-choice construction $f:[0,1] \to [0,1]$ so that COUNTABLY many disjoint subsets of $[0,1]$ map onto $[0,1]$?" The answer would be yes, ...
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### Representation of $\mathbb{R}$, drop continuity assumption, Axiom of Choice.

A representation, for instance, of $\mathbb{R}$ is a group homomorphism $f: \mathbb{R} \to \text{GL}_n(\mathbb{R})$. If we assume that $f$ is continuous, then there is a very nice formula for all such ...
It seems that AC is hiding (maybe concealed?) even in some elementary results. An example: Theorem: Let $X \subseteq \mathbb R$ and let $x_0 \in \mathbb R$ be an accumulation point of $X$. Then ...
### Conditions that topologies must have if (only if) the condition “$G_\delta$ iff (open or closed)” holds?
Consider the class of topological spaces $\langle X,\mathcal T\rangle$ such that the following are equivalent for $A\subseteq X$: $A$ is a $G_\delta$ set with respect to $\mathcal T$ $A\in\mathcal T$...