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I want to know if every complete metric space is not countable? I can't find a counterexample, so is it correct?

captain j
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4 Answers4

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This is true if you add the hypothesis that there are no isolated points.

hunter
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    Or, more generally, only finitely many isolated points. – Brian M. Scott Sep 07 '20 at 17:59
  • The empty space has no isolated points, is complete, and is not uncountable, so for this to be true, we must say "nonempty". @BrianM.Scott Every finite metric space has only finitely many isolated points (all points), is complete, and is not uncountable, so are you assuming the space is infinite? Or maybe we disagree about the definition of "not countable"; is a finite/empty set "not countable"? I think "not countable" means "uncountably infinite". – Jeppe Stig Nielsen Sep 08 '20 at 07:08
  • @JeppeStigNielsen: I think it virtually certain that the OP’s not countable should be understood as uncountable. Clearly hunter was assuming that the spaces under consideration are infinite, and since (a) the assumption is clear, and (b) that’s the only case of any real interest, I saw no reason to do more than point out the obvious extension. Frankly, it was a throwaway comment in passing. – Brian M. Scott Sep 08 '20 at 07:27
  • So: An infinite, complete metric space is uncountable if it has only finitely many isolated points. – Jeppe Stig Nielsen Sep 08 '20 at 07:46
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Take any complete metric space $X$ and any convergent sequence $(x_n)_{n\in\Bbb N}$ of element of $X$ and then the subspace$$\left\{x_n\mid n\in\Bbb N\right\}\cup\left\{\lim_{n\to\infty}x_n\right\}$$is complete and countable.

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Consider a countable space with the discrete metric.

J. W. Tanner
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If a set is finite, then it being countable and it being complete are both rather trivial. One can add a countable sequence going to a limit as long as one also adds the limit. In fact, one can do so countably many times.

Acccumulation
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  • For example the set: $$\left{ a+\frac{1}{b} \middle| a\in\mathbb{Z}, b\in\mathbb{Z}\setminus { 0 } \right}$$ is countable. With the usual metric, it is complete. It has (countably) infinitely many limit points. – Jeppe Stig Nielsen Sep 08 '20 at 07:38