The question I am trying to answer is this:
Given a complete metric space $(X,d)$, in which all of the points are accumulation points. Find if there is a one-to-one and onto map $f: X \rightarrow \mathbb{Q}$.
My initial thoughts are:
Given that $(X,d)$ is complete, we know that all Cauchy sequences in $X$ have a limit also in $X$.
We can easily prove that $\mathbb{Q}$ is not complete, since we can easily find a sequence in $\mathbb{Q}$ which has a limit in $\mathbb{R}\setminus\mathbb{Q}$.
So my thought is that there is no such map $f$. But how do I formally prove it?
If $X$ is a complete metric space in which every point is an accumulation point, it follows from Baire's category theorem that $X$ is uncountable.
Indeed, assume by contradiction that $X=\{x_1,x_2,\dotsc\}$. Then $X$ is the union of the closed sets $X_n=\{x_n\}$. It follows from Baire's category theorem that there is some $n$ such that $X_n$ has nonempty interior in $X$. It follows that $x_n$ is an isolated point of $X$.