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Prove that Cantor set $C$ is homeomorphic to the product of countably many $C$

It is easy to show that $C$ is homeomorphic to the product of finitely many copies of $C$ (Send $n$-digit to $C_n$). But it doesn't work for infinite product. Thanks!

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The Cantor set $C$ is homeomorphic to $\{0,1\}^\mathbb{N} = 2^\mathbb{N}$ (use ternary coordinates, which have two possible values $0$ and $2$, I use $2$ as shorthand for the discrete space $\{0,1\}$ ).

But then $$C^\mathbb{N} \approx (2^\mathbb{N})^\mathbb{N} \approx 2^{\mathbb{N} \times \mathbb{N}} \approx 2^\mathbb{N} \approx C$$

where the middle one is a standard shuffle (set theory bijection on coordinates which is always continuous on products by the universal property of mappings into products) and the next to last one follows as powers of the same cardinality are homeomorphic (and $\mathbb{N} \times \mathbb{N}$ is also countable).

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    I didn't catch why mappings from second one to third and vice versa are continuous. – 2017-02-15
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    If $f$ is in $(2^\mathbb{N})^\mathbb{N}$. Then $f(n) \in 2^\mathbb{N}$ for each $n$, so in fact $f$ is determined by all $f(n)(m)\in 2$, identify this with all maps from $\mathbb{N} \times \mathbb{N}$ into $2$ in the obvious way. – 2017-02-15