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I'm having some trouble understanding the concept of compact set (I'm studying from Rudin's Principles of Mathematical Analysis).

Does every subset of a metric space have an open cover? Why?

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    $\begingroup$ Yes. Given a metric space $X$ and a subset $A\subset X$, then $X$ is trivially an open cover of $A$. The better question is, when does every open cover of a subset of $X$ have a finite subcover? $\endgroup$ – graydad Mar 26 '15 at 17:37
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    $\begingroup$ Trivially, $\{X\}$ with $X$ open is an open cover for any subset of $X$. Since we have a metric space, the whole space is such an open set. $\endgroup$ – AlexR Mar 26 '15 at 17:37
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The answer to your question is yes. In a metric space $X$, $X$ is open. Since (very reduntantly) every subset of $X$ is a subset of $X$, then $X$ functions as an open cover for each of its subsets. This observation is not useful though, and has nothing to do with compactness. A set $K\subset X$ of a metric space $X$ is compact iff every open cover of $K$ has a finite subcover. An open cover of $K$ could contain finitely many open sets, or it could contain infinitely many open sets. In the latter case, you have to know that you can choose finitely many open sets out of the infinite collection and still be able to cover $K$. If you cannot, then $K$ isn't compact.

It might be helpful to work through an example. I'm sure it's in your textbook, but can you find an open cover of $(0,1)$ in the metric space $(\Bbb{R},d)$ [where $d$ is the Euclidean distance metric] that does not have a finite subcover?

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A metric space $(X,d)$ is especially a topological space. One axiom for topological spaces $(X,\tau)$ is $X\in\tau$, i.e. $X$ is open. Thus for any $A\subset X$ a trivial open cover is $\{X\}$. Note that this argument works even for topological spaces, no need for a metric.

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