I'm trying to prove that $\mathbb Q$ is homeomorphic to a subspace of $\mathbb N^{\mathbb N}$.
More precisely $\mathbb Q$ is homeomorphic to the sequence space $S$ of finite support over $\mathbb N$.
Actually I have been reading a proof of Sierpiński's topological characterisation of rational numbers, namely, every countable metric space $\left<X,d\right>$ without isolated points is homeomorphic to $\mathbb Q$, from an article by Krzysztof Chris Ciesielski. Here the author has given a homeomorphism between $X$ and $S$. I am trying to find a homeomorphism between $\mathbb Q$ and $S$. Then we have $X$ is homeomorphic to $\mathbb Q$.
Here is my idea : we can define a metric on $\mathbb N^{\mathbb N}$ so that when it is restricted to the subspace $S$, it gives the metric on $S$.
So first we topologize $\mathbb N^{\mathbb N}$ by defining a metric on it.
We identify each sequence $\left<\alpha_1,\alpha_2,\alpha_3,\ldots\right>$ with the unique real number : $\alpha_1+\dfrac1{\alpha_2+\dfrac1{\alpha_3+\dfrac1{\ddots}}}\tag*{(*)}$
And then we give a metric on $\mathbb N^{\mathbb N}$ as :
$$\text{distance of }\left<a_1,a_2,a_3,\ldots,\ldots\right>, \left<b_1,b_2,b_3,\ldots\ldots\right>=\text{usual distance between the real numbers arose by the identification }(*)$$
Now each rational number can be written in its terminating simple-continued-fraction form, and we can take this representation of rational numbers to be unique.
Then I identify each sequence $\left<\alpha_1,\alpha_2,\ldots,\alpha_k,0,0,\ldots\right>$ in $S$ with a unique rational number like :
$$\alpha_1+\dfrac1{\alpha_2+\dfrac1{\alpha_3+\dfrac1{\ddots+\alpha_{n-1}+\dfrac1{\alpha_n}}}}\tag*{(**)}$$
So if we give a metric to $S$ :
$$\text{distance of }\left<a_1,a_2,\ldots,a_m,0,0,\ldots\right>, \left<b_1,b_2,\ldots,b_n,0,0,\ldots\right>=\text{usual distance of the rational numbers arose by the identification }(**)$$
So this is an isometry and therefore is a homeomorphism.
I would like to know whether my idea is correct, and if not, what is the right way to think about this problem.
And I also like to know if there are any other explicit homeomorphisms between $S$ and $\mathbb Q$.
Best Answer
If you simply want to prove that $S$ is homeomorphic to $\Bbb Q$, there is a much simpler approach. $\Bbb N$ with the discrete topology is metrizable, and the product of countably many metrizable spaces is metrizable, so $\Bbb N^{\Bbb N}$ is metrizable, and hence so is its subspace $S$. $S$ is clearly countable, and it is easy to check that it has no isolated points, so it must be homeomorphic to $\Bbb Q$.
If you want an explicit homeomorphism, your idea works if you can prove that the metric on $\Bbb N^{\Bbb N}$ induced by the continued fraction embedding in $\Bbb R$ induces the product topology on $\Bbb N^{\Bbb N}$; that, however, takes quite a bit of work.