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The Axiom of Choice reads:

The product of a collection of non-empty sets is non-empty.

As you know well, this axiom is equivalent to many other statements. A few examples (probably the most known) are the following:

  • Zorn's lemma: Let $(P,\ge)$ be a poset in which every chain has an upper bound. Then $P$ has a maximal element.

This is used in a number of different places to prove very powerful theorems. A couple of examples I can think of right now are: the proof that every ring has a maximal ideal (commutative algebra) and the Hahn-Banach theorem (functional analysis).

  • Well ordering theorem: Every set can be well-ordered (i.e. it admits a total order such that every subset has a least element).
  • Tychonoff's theorem: Every product of compact topological spaces is compact.

This is used for example in the proof of Alaoglu's theorem (again functional analysis).

Wikipedia gives various more. Among them:

  • Tarski's theorem: If $A$ is an infinite set, then there is a bijection $A\to A\times A$.

(As @Wicht dutifully commented, Tarski's theorem is the name given to the implication $|A|=A\times A \Rightarrow$ Axiom of Choice, while the other implication was known before the proof of this fact by Tarski.)

  • Every vector space has a basis.

My question is: what are other useful statements which are equivalent to the axiom of choice? Where are they used, and to prove what?

P.s.: By "useful statements" I mean either statements which are important theorems by themselves (as Tychonoff's theorem) or that are directly used in (simple versions of) the proof of important theorems (as Zorn's lemma).

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    $\begingroup$ There's like... a huge book of statements equivalent to the axiom of choice. Do you want them all, or just a few? $\endgroup$ – Asaf Karagila Jan 6 '14 at 21:02
  • $\begingroup$ There is a website dedicated to collect all this. Have you tried Google? $\endgroup$ – Pedro Tamaroff Jan 6 '14 at 21:02
  • $\begingroup$ (As it stands, I am tempted to close as too broad. If you can narrow it down, then perhaps this can be reasonably answered.) $\endgroup$ – Asaf Karagila Jan 6 '14 at 21:03
  • $\begingroup$ @AsafKaragila I am interested mainly in statements equivalent to the axiom of choice which are either kind of "big theorems" by themselves (like Tychonoff's theorem) or are directly used in the proof of such "big theorems" (an obvious example of this is Zorn's lemma). I will edit my question to make this clear. $\endgroup$ – Daniel Robert-Nicoud Jan 6 '14 at 21:05
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    $\begingroup$ @Wicht Ok, thanks. I will add a note to the post. $\endgroup$ – Daniel Robert-Nicoud Jan 8 '14 at 22:26
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From the top of my head, some "big" equivalents to the axiom of choice.

  1. Every commutative ring with a unity has a maximal ideal.
  2. Every surjection has an injective inverse.
  3. Every free abelian group is projective.
  4. Every divisible abelian group is injective.
  5. Skolem-Lowenheim theorems in model theory.
  6. Every two cardinals are comparable. This is used often in "hiding" where we have two sets and we know that one necessarily injects into the other.
  7. Every spanning set includes a basis (which is stronger than just the existence of a basis).
  8. Every set is included in $L[A]$ for some $A$. Where $L[A]$ is a model of $\sf ZFC$. This can be used to prove certain things in set theory.

There are many many many more, and one can scrounge through Rubin & Rubin Equivalents of the Axiom of Choice II to find many of them including proofs.

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    $\begingroup$ I suppose 2. means that every surjective $f:A\to B$ allows a $g:B\to A$ such that $f\circ g$ is the identity mapping on $B$. That is kind of a right-inverse to $f$. There will surely not be a left-inverse in general. $\endgroup$ – Jeppe Stig Nielsen Jan 8 '14 at 22:42
  • $\begingroup$ Clearly. Yes... $\endgroup$ – Asaf Karagila Jan 8 '14 at 22:43
  • $\begingroup$ Isn't the Banach-Tarski Paradox also equivalent to AC? $\endgroup$ – user 170039 Jan 24 at 5:17
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    $\begingroup$ @user170039: Not even remotely. It's a local statement about the real numbers. It also follows from the Hahn–Banach theorem, which is much weaker than the ultrafilter lemma, which doesn't even imply countable choice. $\endgroup$ – Asaf Karagila Jan 24 at 7:03

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