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I know how to prove that $^\omega$$\omega$ is uncountable, but how can we prove that it is, in fact, equinumerous to $\Bbb R$? Or at least that $^\omega$$\omega$ is equinumerous to $^\omega$$2$?

Asaf Karagila
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Joe
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  • $2 < |\Bbb N| < |2^\Bbb N| \ |2^\Bbb N| \le |\Bbb N^\Bbb N| \le |(2^\Bbb N)^\Bbb N| = |2^{\Bbb N \times \Bbb N}| = |2^{\Bbb N}| \ |\Bbb N^\Bbb N| = |2^\Bbb N| = |\Bbb R|$ – DHMO Apr 13 '17 at 18:32
  • Ahh right, ok.. I knew it was something simple like this. Thanks. – Joe Apr 13 '17 at 18:39
  • Try proving each theorem used. – DHMO Apr 13 '17 at 18:40

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