16 questions linked to/from Axiom of Choice and finite sets
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### Naive understanding of choice axiom [duplicate]

It's written in many resources that if we consider only finite sets, that choice axiom can be skipped and be proven from other ZF axioms. I remember I've read the following explanation. If $A$ is a ...
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### Uses of Axiom of Choice [duplicate]

I am a first-year maths student but I occasionally drift away from our taught material. Some years ago I saw the ZFC axioms for the first time, but now that I am in college, and although the stuff I'...
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### Does the proof of rank-nullity theorem from Lang's “Linear Algebra” involves axiom of choice? [duplicate]

Disclaimer: I never had (yet c:) a rigorous exposure to set theory (independence proofs, and similar stuff...). I was wondering if, in the following proof of the rank-nullity theorem form Lang's "...
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### Can you explain the “Axiom of choice” in simple terms?

As I'm sure many of you do, I read the XKCD webcomic regularly. The most recent one involves a joke about the Axiom of Choice, which I didn't get. I went to Wikipedia to see what the Axiom of Choice ...
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### Why is the Axiom of Choice not needed when the collection of sets is finite?

According to Wikipedia: Informally put, the axiom of choice says that given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object ...
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### Do we need Axiom of Choice to make infinite choices from a set?

According to the answers to this question, we do not need choice to pick from a finite product of nonempty sets, even if each of the sets is infinite. The axiom of choice is required to ensure that a ...
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### Infinite sets with cardinality less than the natural numbers

Are there any infinite sets that have a lower cardinality than the natural numbers? Is there a proof of this?
Suppose I have a non-empty set $A$. How do I choose an element $x\in A$? More precisely, I believe I would like to find a formula $P(x,y)$ of ZF such that for every non-empty set $y$ there is ...