It is know that for a metric space, it is locally compact and separable iff exist an equivalent metric where a set is compact iff it is closed and limited. So, locally compact and seperable metric spaces are topologically complete. The question is: there is any metric space that is locally compact and not topologically complete?
Another motivation for this question is: Hausdorff locally compact spaces are Baire spaces, as well topologicaly complete space. But there are metric spaces that are Baire, but not topologicaly complete. So, the natural question is that above.
Obs: A metric space $(X,d)$ is topologically complete if there is a metric $d′$ on $X$ such that $Id:(X,d) \rightarrow (X,d′)$ is a homeomorphism and $(X,d′)$ is complete.
Best Answer
Every locally compact metric space can be given a compatible complete metric. Suppose that $X$ is a locally compact metric space. Then $X$ is paracompact, so $X$ is a disjoint union of $\sigma$-compact locally compact spaces (There is a theorem proved in the book Topology by Dugundji that proves that every paracompact locally compact space is the disjoint union of $\sigma$-compact spaces. I prove this fact below.). Therefore we may without loss of generality assume that $X$ is $\sigma$-compact. Since we assume $X$ is $\sigma$-compact, there is a sequence of open sets $U_{n}$ such that $\overline{U_{n}}$ is compact and $\overline{U_{n}}\subseteq U_{n+1}$ for all $n$. From this sequence of open sets and Urysohn's lemma, there is a function $f:X\rightarrow[0,\infty)$ such that $\overline{U_{n}}\subseteq f^{-1}[0,n)\subseteq U_{n+1}$. Let $d$ be a metric on $X$, and define a new metric $d'$ on $X$ by letting $d'(x,y)=d(x,y)+|f(x)-f(y)|$. Clearly $(X,d')$ induces the original topology on $X$. I claim that $(X,d')$ is complete. Assume that $(x_{n})_{n}$ is a Cauchy sequence in $(X,d')$. Then the sequence $(x_{n})_{n}$ is bounded in $(X,d')$, so clearly the sequence $(f(x_{n}))_{n}$ is bounded as well. Therefore, there is some $N$ where $f(x_{n})<N$ for all $n$. In particular, since $f(x_{n})\subseteq f^{-1}[0,N)$, we have $x_{n}\in U_{N+1}\subseteq\overline{U_{N+1}}$ for all $n$. Since $\overline{U_{N+1}}$ is compact, and $x_{n}\in\overline{U_{N+1}}$ for all $n$, the sequence $(x_{n})_{n}$ has a convergent subsequence, so the sequence $(x_{n})_{n}$ itself must be convergent.
Added Later I will now prove that a paracompact locally compact space is a free union of $\sigma$-compact spaces since the proof is not too difficult.
Suppose that $X$ is locally compact and paracompact. Then let $\mathcal{U}$ be a cover of $X$ such that if $U\in\mathcal{U}$, then $\overline{U}$ is compact. Then let $\mathcal{V}$ be a locally finite open refinement of $\mathcal{U}$. I claim that each $U\in\mathcal{V}$ intersects only finitely many other elements in $\mathcal{V}$. Since $\mathcal{V}$ is locally finite, there is an open cover $\mathcal{O}$ of $X$ where each $O\in\mathcal{O}$ intersects only finitely many elements of $\mathcal{V}$. If $U\in\mathcal{V}$, then since $\overline{U}$ is compact, there are $O_{1},...,O_{n}\in\mathcal{O}$ where $U\subseteq\overline{U}\subseteq O_{1}\cup...\cup O_{n}$. However, since the sets $O_{1},...,O_{n}$ each only intersect finitely many elements of $\mathcal{V}$, the union $O_{1}\cup...\cup O_{n}$ can only intersect finitely many elements of $\mathcal{V}$. In particular, the set $U$ only intersects finitely many elements of $\mathcal{V}$. Now make the set $\mathcal{V}$ into a graph where we put an edge between $U,V\in\mathcal{V}$ if and only if $U\cap V\neq\emptyset$. Then each $U\in\mathcal{V}$ has only finitely edges. Let $(\mathcal{E}_{i})_{i\in I}$ be the set of all connected components of the graph $\mathcal{V}$. Then each $\mathcal{E}_{i}$ is a countable subset and clearly $\{\bigcup\mathcal{E}_{i}|i\in I\}$ partitions $X$ into clopen subsets. Therefore since each $\mathcal{E}_{i}$ is countable and each $U\in\mathcal{V}$ has compact closure, the each union $\bigcup\mathcal{E}_{i}$ is $\sigma$-compact.