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.
  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.

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    $\begingroup$ 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) $\endgroup$
    – KCd
    Mar 24 '13 at 4:04
  • 1
    $\begingroup$ @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.) $\endgroup$ Mar 24 '13 at 18:30

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