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I've got a problem with prove about cardinality of sets.
How can I prove that $\lbrace 0,1 \rbrace^\mathbb{N} \simeq \mathbb{N}^\mathbb{N}$?

Andrés E. Caicedo
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Wolfrimer
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    One way to think of an element of ${0, 1}^{\Bbb N}$ is as a sequence of $0$s and $1$s. You can similarly think of an element of $\Bbb N^{\Bbb N}$ as a sequence of positive integers. Can you think of a bijection between the set of sequences of the first type and the set of sequences of the second type? – MJD Jan 06 '15 at 17:36

3 Answers3

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Hint. Note that $$ \big\lvert \{0,1\}^{\mathbb N}\big\rvert\le \lvert {\mathbb N}^{\mathbb N}\rvert $$ and $$ \lvert {\mathbb N}^{\mathbb N}\rvert\le \big\lvert \big(\{0,1\}^{\mathbb N}\big)^{\mathbb N}\big\rvert=\big|\{0,1\}^{\mathbb N\times\mathbb N}\big|=\big|\{0,1\}^{\mathbb N}\big|. $$

Yiorgos S. Smyrlis
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Hint:

An injection from $\mathbb{N}^\mathbb{N}$ to $\{0,1\}^\mathbb{N}$ could be given by $(a_1,a_2,a_3,...)\mapsto \underbrace{1,1,...,1}_{a_1 \mbox{ times }},0,\underbrace{1,1,...,1}_{a_2 \mbox{ times }},0,\dots$

paw88789
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It's enough consider the following inequalities

$2^\omega\leq \omega^\omega \leq (2^\omega)^\omega = 2^{\omega\cdot\omega}=2^\omega$

Also, you can easily generalize this argument for infinite cardinal numbers.

Heracles
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