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My thinking is this: Let $(X, d)$ be a compact metric space, thus it is also sequentially compact, meaning that all sequences have a convergent subsequence. We also know that every compact metric space is complete.

Now we also know that a sequence is Cauchy iff it has a convergent subsequence, but since our space is complete, they should converge?

Example: Let $X = [0, 1] \subset \mathbb{R}$ and $d$ some metric. By Heine-Borel's theorem, $X$ is compact, since it is closed and bounded. Now, looking at the alternating sequence $(x_n)_{n = 0}^{\infty}$ given by:

$$ x_n = \begin{cases} 0, & \text{$n$ is even} \\ 1, & \text{$n$ is odd} \end{cases} $$

This sequence obviously doesn't converge. Where am I going wrong in this? Thank you for your help.

Alex Ravsky
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BlueberryKing
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    If you take every other term of your sequence, you have a convergent subsequence. – B. Goddard Jun 08 '20 at 15:43
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    It is not true that any sequence with a convergent subsequence is Cauchy. Read that linked question a little more carefully. – aschepler Jun 08 '20 at 15:45
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    The result in the linked question is that a sequence that we already know is Cauchy converges if and only if it has a convergent subsequence. If we only know that a sequence has a convergent subsequence, there is no reason to expect that it is Cauchy. In fact, since the space is compact this describes any sequence, convergent or not. – Matt Samuel Jun 08 '20 at 15:48
  • Ah, thanks guys! That makes sense. – BlueberryKing Jun 08 '20 at 16:45

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