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Let $(X,d)$ be a metric space. If $K⊆X$, and $K$ is a closed set. Does that mean any Cauchy sequence in $K$ converges in $K$?

If no, could someone give an example?

If yes, then what is the difference between complete sets and closed sets?

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  • $\begingroup$ No. Take $(\mathbb Q,d)$. Then $\mathbb Q$ is closed but not every Cauchy sequence converges. E.g take a sequence $(x_n)$ in $\mathbb Q$ converging in $\mathbb R$ to $\pi\notin\mathbb Q$. $\endgroup$ – drhab Aug 13 '15 at 15:11
  • $\begingroup$ Great, What about if the inclusion is strictly? $\endgroup$ – Fabian Aug 13 '15 at 15:13
  • $\begingroup$ My example shows that the case will not be changed if the inclusion is strict. $(-\sqrt11,\sqrt11)$ is closed in $\mathbb Q$ but you can let $x_n\in(-\sqrt11,\sqrt11)$ $\endgroup$ – drhab Aug 13 '15 at 15:14
  • $\begingroup$ Aha, Thank you so much. $\endgroup$ – Fabian Aug 13 '15 at 15:18
  • $\begingroup$ If $(X,d)$ is complete metric space, and $K\subseteq X$ is closed in $X$, then it is complete. Proof. $\endgroup$ – Ennar Aug 13 '15 at 15:20
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The rationals are closed in the rationals, but obviously not complete. For strict inclusion, take $\mathbb{Q} \cap [0,1]$ in $\mathbb{Q}$. Also not complete, but closed.

Incidentally, complete implies closed. Just look at limit points and sequences.

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  • $\begingroup$ Great, What about if the inclusion is strictly? $\endgroup$ – Fabian Aug 13 '15 at 15:13

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