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I'm trying to remember a particular theorem of ZF but unfortunately my memory is quite incomplete.

  • The theorem is of the form (some set operation) is either (expected answer) or the empty set.
  • If the Axiom of Choice holds, (some set operation) is (expected answer).
  • I believe that if (some set operation) is (expected answer) for all infinite sets, then the Axiom of Choice holds -- that is, the statement is equivalent to AC.
  • The theorem has a formalized proof at Metamath.

I think it was something like $|S\times S|=|S|$ for infinite sets $S$, or $\bigcup S$ is... something... I'm just not sure. Does this ring a bell?

My apologies for the lack of well-defined information, and all the more for the possibility that some part of this is misleading. I figured that, short of MathOverflow, this is just about the best place I could come to find my answer.

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up vote 3 down vote accepted

I believe what you are looking for is that the Cartesian product of non-empty sets is non-empty. It is equivalent to AC. Without AC you cannot be sure.

Tarski's theorem, which you quote in the form of cardinals: for infinite $S$ there is a bijective map between $S×S$ and $S$ is also an equivalent.

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Which, if either, of these has my "empty set" form when AC does not hold? – Charles Oct 21 '10 at 22:00
I thought the first, in its mention of the product(a set operation) is non-empty(expected) vs empty. Without AC the product could be empty, but is not required to be. I posted the second due to the match with what you posted, thinking the name might help. There are links to many equivalents at Hope this helps. – Ross Millikan Oct 21 '10 at 22:11
@Charles: The axiom of choice is equivalent to the assertion that if $\{A_i\}_{i\in I}$ is any nonempty family of nonempty sets ($I\neq\emptyset$, and for each $i\in I$, $A_i\neq\emptyset$), then $\times_{i\in I}A_i\neq\emptyset$. If AC does not hold, then there exists a particular nonempty family of nonempty sets $\{B_j\}_{j\in J}$ such that the cartesian product $\times_{j\in J}B_i$ is empty. – Arturo Magidin Oct 22 '10 at 3:22
@Arturo: Thanks -- and +1 for Ross -- but these aren't what I was thinking of. Well, Tarski's theorem was what I wrote as my 'something like this', but it wasn't that. – Charles Oct 22 '10 at 4:02

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