Compactness, Local Compactness, Completeness Clearly, every compact metric space is locally compact. I always get confused when completeness is introduced into the picture. Which of the following are true? What are some easy counterexamples to those statements that are false?


*

*Every complete metric space is compact.

*Every complete metric space is locally compact.

*Every compact metric space is complete.

*Every locally compact metric space is complete.

*Every locally compact inner product space is of finite dimension.

 A: *

*False: the real line with the usual metric is a counterexample.

*False: the irrationals are completely metrizable and nowhere locally compact.

*True: a metric space is compact iff it is complete and totally bounded.

*False: $(0,1)$ with the usual metric is a counterexample.

*True; see this question and this question.
A: 1 and 2 are false.  Take $\ell^{\infty}$ be the set of uniformly bounded sequence on $\mathbb{R}$ with distance defined between any two sequences $(x_n)_{n=0}^{\infty}$ and $(\tilde{x}_n)_{n=0}^{\infty}$ therein by
$$
d\big((x_n)_{n=0}^{\infty},(\tilde{x}_n)_{n=0}^{\infty}\big) =\sup_n\, |x_n-\tilde{x}_n|
.  
$$
It is easy to see this space is complete; however, I'll focus on the fact that it is not locally-compact.  To see this let $\boldsymbol{0}$ denote the constant zero sequence $\boldsymbol{0}:=(0)_{n=0}^{\infty}$.  Since the open balls are a base for any metric topology then we only need to argue with open balls to show that $\ell^{\infty}$ is not locally-compact.  If it were locally compact, then there would be some $k>0$ such that the (open) set of sequences which are at a distance of $k$ would have compact closure; but such as set contains the countable totally-disconnected subset
$$
\{(k/2\delta_{i,n})_{n=0}^{\infty}\}_{i=0}^{\infty}
$$
which has no countable open cover.
