Is locally completeness a topological property? I know that completeness itself is not a topological property because a complete and a not complete metric space can be homeomorphic, e.g. $\Bbb R$ and $(0,1)$.
However, both $\Bbb R$ and $(0,1)$ are locally complete (each point has a neighborhood that is complete under the induced metric). As all examples I know of are of this form, the naturally occuring next question is

Question: Is being locally complete a topological property?

Or the other way around: are there metric spaces which are homeomorphic, but one is locally complete and the other one is not?

Another way of proving that the irrationals can be made complete with respect to a metric $d$ which is equivalent to the usual one consists in providing such a metric. This can be donne as follows: let $(q_n)_{n\in\mathbb N}$ be an enumeration of the rationals. Then, if $x,y\in\mathbb{R}\setminus\mathbb Q$, define$$d(x,y)=|x-y|+\sum_{k=1}^\infty2^{-k}\inf\left(1,\left|\max_{i\leqslant k}\frac1{|x-q_i|}-\max_{i\leqslant k}\frac1{|y-q_i|}\right|\right).$$
