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Let $A,B$ be two sets such that $2A \cong 2B$ (here $2A := A \coprod A$). Then $A \cong B$. This can be proven without the axiom of choice, which means that one can explicitly construct a bijection $A \to B$ out of a bijection $2A \to 2B$. This is non-trivial and interesting, see the wonderful paper by Conway, Doyle, also for generalizations. The construction is infinitary, and therefore the following question comes into my mind.

Question. Is the assertion also true in ZF - {Axiom of Infinity}?

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up vote 2 down vote accepted

The axiom of infinity is equivalent to there being some infinite set. So either all sets are finite, in which case the result obviously holds by simple induction, or the axiom of infinity holds and one can apply the infinitary proof.

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Hm, I don't understand. How do you define "infinite" at all without refering to $\mathbb{N}$? So what is the formalization of your first sentence in ZF? And how does your remark explicitly produce a proof of the assertion which does not use $\mathbb{N}$? –  Martin Brandenburg Oct 17 '12 at 8:07
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@MartinBrandenburg: A set is infinite if it not finite. A set is finite if it has the same cardinality as a natural number. A natural number is a successor ordinal that has only zero and successor ordinals as predecessors. Induction with natural numbers works even when the set of natural numbers forms a proper class- just like induction on ordinals. –  Michael Greinecker Oct 17 '12 at 8:13
    
Zero is not a successor ordinal, though. –  Zhen Lin Oct 17 '12 at 10:10
    
@ZhenLin That's true. So one has to add zero to this definition. –  Michael Greinecker Oct 17 '12 at 12:48
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