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Does ZF (without choice) prove the following?

$\forall \ ordinal \ \alpha \ \exists f( f:\alpha \times \alpha \hookrightarrow \alpha)$

where: $ \alpha\times \alpha $ is the cartesian product of $\alpha$ by itself; and $\hookrightarrow$ signifies injection.

Zuhair
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    Yes, $|\alpha\times \alpha|=|\alpha|$ for ordinals in ZF. (This is usually proven with the pairing function.) – spaceisdarkgreen Nov 09 '20 at 18:28
  • but the pairing function is form ordered pairs of naturals to naturals, can this be generalized over all ordinals? – Zuhair Nov 09 '20 at 18:46
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    Yes. https://mathoverflow.net/questions/112079/what-is-g%C3%B6dels-pairing-function-on-ordinals (Also, the usual proof of the result that uses this function is available in e.g. Jech or Kunen.) – spaceisdarkgreen Nov 09 '20 at 18:54
  • @spaceisdarkgreen, Thank you! – Zuhair Nov 09 '20 at 19:04

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