What is the relationship between completeness and local compactness?
Solution 1:
Note that the Baire space $\mathbb{N}^\mathbb{N}$ (which is homeomorphic to the space of irrationals) is a completely metrizable space which is not locally compact (in fact all compact subsets of $\mathbb{N}^\mathbb{N}$ have empty interior).
While locally compact metric spaces may not be complete (for example, as in Brad's answer, the open unit interval $(0,1)$ under the usual metric), these spaces will always be completely metrizable. There are at least two ways to see this result:
- Every locally compact metric space will be an open subset of its completion, and every G$_\delta$ subset of a complete metric space is completely metrizable.
- Every locally compact completely regular space is Čech-complete (i.e., is a G$_\delta$ subset of its Stone-Čech (or any other) compactification), and a metric space is completely metrizable iff it is Čech-complete.
Solution 2:
Neither implies the other. In addition to the examples you give, there are metric spaces which are complete and not locally compact (e.g. an infinite-dimensional Hilbert space) and metric spaces which are locally compact and not complete (e.g. the open interval $(0,1)$).