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?

Solution 1:

The irrational numbers are not locally complete, but they are homeomorphic to the Baire space $\mathbb N^{\mathbb N}$, which can be given a metric turning it into a complete metric space.

(For example, endow $\mathbb N$ with the discrete metric and set $d(s, t) = \sum_{i=0}^\infty\frac{1}{2^i}d(s_i,t_i)$).

Solution 2:

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$$d(x,y)=|x-y|+\sum_{k=1}^\infty2^{-k}\inf\left(1,\left|\max_{i\leqslant k}\frac1{|x-q_i|}-\max_{i\leqslant k}\frac1{|y-q_i|}\right|\right).$$