Baire space is homeomorphic to countably many copies of itself

Question

On wikipedia I found that the Baire space $\mathcal{N}$ is homeomorphic to the product of a countable number of copies of itself, however, I haven't been able to find a proof.

The Baire space is defined as $\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ with the product topology, $\mathbb{N}$ being endowed with the discrete topology.

Can anyone help me, please?

Answer 1

Well, note that $\Bbb N \times \Bbb N$ has a bijection with $\Bbb N$, call it $h$, and let $g: \Bbb N \to \Bbb N \times \Bbb N$, $g(n) = (g_1(n), g_2(n))$ be the inverse, with $h(g_1(n), g_2(n)) = n$ for all $n$ etc.

Then map $f \in (\Bbb N^{\Bbb N})^{\Bbb N}$ to $\Phi(f) \in \Bbb N^{\Bbb N}$ by the rule $$\Phi(f)(n) = f(g_1(n))(g_2(n))$$ and show it's a bijection and a homeomorphsim in the product topologies.

This is ofetn summarized as

$$\mathcal{N}=\Bbb N^{\Bbb N} \simeq \Bbb N^{\Bbb N \times \Bbb N} \simeq (\Bbb N^{\Bbb N})^{\Bbb N}=\mathcal{N}^{\Bbb N}$$

where all homeomorphisms are natural (categorical), basically coordinate shuffles.

Answer 2

HINT: consider the left-shift map $$S:\mathcal{N}\rightarrow\mathcal{N}: f\mapsto \lambda x.f(x+1).$$

E.g. $S(0,1,2,3,4,5,...)=(1,2,3,4,5,...)$ and so on.

In more detail:

For each $i\in\mathbb{N}$, let $\mathcal{N}_i=\{f\in\mathcal{N}: f(0)=i\}$. Now think about what $S\upharpoonright \mathcal{N}_i$ does, topologically.