I think given any subset K in metric space X, K can be covered by X. Since X is open, K only need 1 open set to cover it. I am confused.
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3 Answers
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A subset $K$ of a metric space $X$ is compact if for every open cover of $K$, we can find a finite subcover.
The open cover $\{X\}$ is just one open cover, and it does indeed have a finite subcover. But the definition says we need any arbitrary open cover of $K$ to have a finite subcover. So just checking $\{X\}$ will not suffice.
Conversely, to show $K$ is not compact, we need only exhibit one example of an open cover that does not have a finite subcover.
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$\begingroup$ Can you give me one example for K is not compact in X ? Thank you very much! $\endgroup$– AndyShenSep 5, 2019 at 15:08
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$\begingroup$ Try $X = \mathbb{R}$ and $K = \mathbb{Z}$. Think about an open cover where each open set contains just one integer. $\endgroup$ Sep 5, 2019 at 15:14
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$\begingroup$ In $\mathbb{R}^n$ compact is equivalent to closed and bounded (not true in general metric spaces though!). So any subset of $\mathbb{R}^n$ that is not closed or is unbounded will not be compact. For instance, $\mathbb{R}$ itself is not compact in $\mathbb{R}$. Nor is $(0,1)$. See Why is an open interval not a compact set?. $\endgroup$– kccuSep 5, 2019 at 15:14
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I think you are misreading the definition of "compact". It says that every open cover has a finite subcover. Your question notes the obvious fact that there happens to be one particular open cover that is finite.
answered Sep 5, 2019 at 15:00
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A topological set $K$ is called compact if each of its open covers has a finite subcover. An open cover is a collection of subsets $U_i \subset K$ such that the union $\cup_i U_i = K$.
In your example, $X$ is not a subset of $K$, which is mandatory from the definition.
Furthermore, finding a single finite open cover is not enough: According to the definition, a set $X$ is compact if from ANY open cover has a finite subcover.
