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How can I prove that $|P(\Bbb R)\times \Bbb R|=|P(\Bbb R)|$?

I can use the following statements:

$$|A|<|P(A)| $$$$ |P(\Bbb N)|=|\Bbb R|$$$$ |\Bbb R\times\Bbb R|=|\Bbb R|$$$$ |\Bbb Z\times\Bbb Z|=|\Bbb Z|$$$$ |\Bbb Z|=|\Bbb N|=\aleph_0$$

I tried a lot, but I'm stuck.

thanks.

mle
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daniel
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  • @Pedro: When writing in Hebrew text (without reasonable mathematical rendering) it's more than common to write 0א. Remember that we write right-to-left. – Asaf Karagila Mar 16 '14 at 22:10
  • I certainly didn't expect that to be the explanation =D @Asaf – Pedro Mar 16 '14 at 22:15

1 Answers1

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HINT:

Prove that $|\mathcal P(\Bbb R)\times\mathcal P(\Bbb R)|=|\mathcal P(\Bbb R)|$. The proof is similar to that of $|\Bbb{R\times R}|=|\Bbb R|$. Then find some nice inequalities to bound $|\mathcal P(\Bbb R)\times\Bbb R|$ from above and below by sets of the same size as $\mathcal P(\Bbb R)$.

Asaf Karagila
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