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From Rudin's Principles of Mathematical Analysis (p.37).

Theorem 2.33. Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.

Immediately after the following is written.

By virtue of this theorem we are able, in many situation, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces, or of closed spaces (every metric space $X$ is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric space.

I can't understand the sentence "every metric space $X$ is an open subset of itself, and is a closed subset of itself".

Actually, $(0,1)$ is metric space for metric $d(x,y)=|x-y|$, $\forall x,y\in (0,1)$. $(0,1)$ is just open subset of itself, but is not a closed subset of itself.

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  • $\begingroup$ $(0,1)$ is a closed subset of $(0,1)$; indeed every metric space $X$ is a closed subset of itself. The complement is $\emptyset$ which is open. $\endgroup$ – Lord Shark the Unknown Aug 13 '18 at 6:31
  • $\begingroup$ To check that $(0,1)$ is a closed subset of itself show that the complement $(0,1)\setminus (0,1)=\emptyset$ is open. The empty set is open, because it satisfies that every one of its points is inside a ball contained in the set. Since the empty set doesn't contain any points, there is nothing that could make the condition fail. $\endgroup$ – user583185 Aug 13 '18 at 6:31
  • $\begingroup$ Thank you, I can understand what you mean. $\endgroup$ – 백주상 Aug 13 '18 at 6:44
  • $\begingroup$ You must be careful to distinguish between a sub$set$ and a sub$space$. $\endgroup$ – DanielWainfleet Aug 13 '18 at 7:33
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To understand why this is true one must use the fundamental duality of openness and closeness. Note that if a set is open, it's complement is closed and vice versa. The complement of any metric subspace considered as a metric space in it's own right is the empty set. Clearly a metric space is closed with respect to itself since it contains all it's limits. This implies that the empty set is open in the metric space M. Also, the empty set is closed in M which implies M is an open subset of itself. Therefore every metric space is a clopen (open and closed subset of itself). Hope this helps.

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  • $\begingroup$ Thank you, I can understand what you mean. $\endgroup$ – 백주상 Aug 13 '18 at 6:44

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