I'm trying to show that homeomorphic metric spaces need not have homeomorphic completions.
My idea was to find two homeomorphic metric spaces with one totally bounded and the other not. Then the completion of the totally bounded metric will be compact, and the other completion will not be compact.
I'm having trouble coming up with specific examples.
Best Answer
Summarising the comments, plus some more:
$X=\Bbb R$ has itself as its completion, while $Y=(0,1)\simeq X$ has $[0,1]$ as its completion. $\overline{Y}$ is compact while $X$ is not.
$X=(0,+\infty)$ has $\overline{X}=[0,+\infty)$ as its completion, non-compact; while $[0,1)= Y \simeq X$ has $[0,1]$ as its completion, compact.
$X=\Bbb Q$ has $\Bbb R$ as its completion, while $Y$: all end-points of the standard middle third Cantor set $0,1,\frac13,\frac23,\ldots$ are homeomorphic to $\Bbb Q$ (standard theorem) and $Y$'s completion is the Cantor set. (Very disconnected, as opposed to the reals).
As a generalisation of the last example: let $X'$ and $Y'$ be non-homeomorphic complete separable metric spaces (aka Polish spaces) both without isolated points and $X$ and $Y$ be countable dense subsets of $X'$ resp. $Y'$. Then as countable metric spaces without isolated points $X$ and $Y$ are homeomorphic and being dense in a Polish space the surrounding space $X'$ resp. $Y'$ is the resp. completion, and are non-homeomorphic. We can take $X'=\Bbb R^n$ for different $n$ e.g.