Complete+bounded homeomorphic to incomplete+unbounded

Question

I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) \cong \mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:

Are there homeomorphic metric spaces $M$ and $N$ such that $M$ is both complete and bounded, but $N$ is neither complete nor bounded?

Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!

Answer 1

You could let $M$ be $\mathbb R$ with the metric $d(x,y)=\min(1,|x-y|)$ and $N$ be $(0,\infty)$ with the usual metric.

Answer 2

Yes. Let $M$ be $\mathbb N$ with the discrete metric, and let $N= \{1/n:n\in \mathbb N\}\cup \{2,3,4,\dots\}$ with the usual $\mathbb R$ metric.