I am trying to prove that $\mathcal{N}^\omega$ is homeomorphic to $\mathcal{N}$ where $\mathcal{N}$ is Baire Space $\omega^\omega$, of all sequences of natural numbers, $\langle a_n;n \in \mathbb{N}\rangle$ with the topology generated from basis sets of the form:
$O(s) = \{ f \in \mathcal{N}; s\subset f\} = \{\langle c_k ; k \in n\rangle : (\forall k < n) c_k=a_k \}$
Here is my suggested proof: It is a known fact that, there exists a bijection $f$ from $\omega^\omega$ onto $\omega$ (as shown in the picture below):
Since $\mathcal{N} \times \mathcal{N}$ is the space of all countable sequences of countable sequences, we can use the same map as a map from $\mathcal{N} \times \mathcal{N}$ to $\mathcal{N}$. For instance, Given the infinite countable sequence of sequences: $1,2,3,4,5,6,.....\\ 2,4,6,8,10,12,...\\ 3,6,9,12,15,18,...\\ 4,8,12,16,20,..\\ .\\.\\ n,2n,3n,4n,5n,...\\ .\\.\\$
it's image under $f$ is the sequence: $1,2,2,3,4,3,4,6,6,4,5,8,9,..$.
If what I have said so far is true, then it remains to show that the function I have described is a homeomorphism. So, we have to show that the image of an open set is an open set and vice versa. We will show that the image of a basis set is an open set and vice versa: Given a basic set in $\mathcal{N}^\omega$, It contains all the infinite sequences of infinite sequences with same prefix. Therefore, by construction, the resulting images will contain same prefix. The opposite direction is very similar.
What do you think of this proof? Is the idea true, is it formal enough, and if not, does anyone know how to state it in a more formal way?
Thank you! Shir
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