14

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?

  1. Every complete metric space is compact.
  2. Every complete metric space is locally compact.
  3. Every compact metric space is complete.
  4. Every locally compact metric space is complete.
  5. Every locally compact inner product space is of finite dimension.
M.Sina
  • 1,730
Sara
  • 614

2 Answers2

21

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

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

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

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

  5. True; see this question and this question.

Community
  • 1
Brian M. Scott
  • 631,399
  • 9
    I think the example illustrating why (2) is false is "weird". It'd be better to give a natural example of a complete metric space that is not locally compact. For instance, the space $C([0,1],{\mathbf R})$ of continuous real-valued functions on $[0,1]$, with metric $d(f,g) = \max_{x \in [0,1]} |f(x) - g(x)|$, is complete. It is not locally compact since any (nondiscrete) infinite-dimensional normed vector space over ${\mathbf R}$ is not locally compact (equiv., a (nondiscrete) locally compact normed vector space over $\mathbf R$ is finite-dimensional) – KCd Mar 24 '13 at 04:04
  • 4
    @KCd: But it’s very easy to give the irrationals a complete metric $-$ the two most common ways can be found here $-$ and it’s a bog-standard fact that they’re topologically complete. In fact they’re characterized as the unique topologically complete, nowhere locally compact, separable, zero-dimensional, metrizable space. (And from my point of view as a set-theoretic topologist, the irrationals are far more simpler and more natural than any function space.) – Brian M. Scott Mar 24 '13 at 18:30
  • @BrianM.Scott: the link in your comment is broken. :-( – André Caldas Feb 12 '22 at 09:52
  • 2
    @AndréCaldas: One way is to show that the irrationals are homeomorphic to $\Bbb N^{\Bbb N}$; the answers to this question give two proofs of that. If $\langle x_n\rangle_n,\langle y_n\rangle_n\in\Bbb N^{\Bbb N}$ are distinct, define the distance between them to be $\frac1m$, where $m=\min{n\in\Bbb N:x_n\ne y_n}$; it’s easily verified that this is a complete metric that generates the product topology. Another is to use the fact that the irrationals are a $G_\delta$ in $\Bbb R$ and the proof of the ... – Brian M. Scott Feb 12 '22 at 20:13
  • ... theorem that a $G_\delta$ subset of a complete metric space is completely metrizable: the usual proof actually shows how to construct a complete metric on it. – Brian M. Scott Feb 12 '22 at 20:13
0

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.

AB_IM
  • 6,888