A subset of the Cantor set that is homeomorphic to $\mathbb{N}^{\mathbb{N}}$

Asked Aug 04 '16 at 13:28

Active Aug 04 '16 at 13:28

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Let $\mathcal{C}$ denote the Cantor set and let $A\subset\mathcal{C}$ be the set of sequences with infinitely many 0's and infinitely many 1's.

Any hint on how to show that $A$ and $\mathbb{N}^{\mathbb{N}}$ are homeomorphic ?

Thanks in advance.

asked Aug 04 '16 at 13:28

valls

1

The $n $-th block of $k +1$ consecutive appearances of the same symbol corresponds to the number $k $ in the $n $- th position? – Andrés E. Caicedo Aug 04 '16 at 13:35
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Although the questions were different, this answer completely answers this question, as does this one. – Brian M. Scott Aug 04 '16 at 15:46

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