For any set $A$ such that $A$ has a bijection with $A\times A$ , prove that there's a bijection between $P(A)$ and $P(A) \times P(A)$, where $P(A)$ is the power set of $A$.

  • $\begingroup$ There will be a bijection between $P(A)$ and $P(A \times A) $ Does it help ? $\endgroup$ – user90041 Nov 10 '13 at 19:26


Recall that if $A$ and $B$ are disjoint then $\mathcal P(A)\times\mathcal P(B)$ has a bijection with $\mathcal P(A\cup B)$. Show that the condition of $A$ implies that $A$ and $A\times\{0,1\}$ have a bijection.

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  • $\begingroup$ i was not aware of the fact u stated cud u please explain why it happens ,i.e f A and B are disjoint then P(A)×P(B) has a bijection with P(A∪B). $\endgroup$ – Aditya Nambiar Nov 11 '13 at 15:56
  • $\begingroup$ There's a really obvious bijection between the two sets, you should be able to come up with on your own. $\endgroup$ – Asaf Karagila Nov 11 '13 at 15:59
  • $\begingroup$ for finite sets its clear, but for infinite sets it doesn't seem to be clear to me $\endgroup$ – Aditya Nambiar Nov 11 '13 at 16:10
  • $\begingroup$ Because you insist to think about it in terms of "number of elements" and not the elements themselves. $\endgroup$ – Asaf Karagila Nov 11 '13 at 16:14
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    $\begingroup$ U rock!! Thanks alot! $\endgroup$ – Aditya Nambiar Nov 11 '13 at 16:47

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