How can an open set be compact? From what I understand, a compact set must be closed and bounded.

Also, is the following statement true or false?

Every subset of a metric space that is not closed, is not compact

I believe it's a true statement because a compact set must be closed.

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    $\begingroup$ All of this depends on the topology. Also recall that a set can be both closed and open. It is true that, in a metric space, a compact set must be closed. That doesn't mean it can't be open. $\endgroup$ – Randall May 2 at 15:49
  • $\begingroup$ Look up the definition of "topological space". You will see that it's not very restrictive. Hausdorff ($T_2$) spaces are certain kinds of topological spaces. Metric spaces are certain kinds of Hausdorff spaces. Compact subsets of a Hausdorff space must be closed. $\endgroup$ – DanielWainfleet May 2 at 17:39

You might be thinking of the Heine-Borel theorem, which says that every compact subset of the real line is a closed and bounded subset of the real line.

But "closed" and "open" do not quite behave like you might think.

For example, consider the topological space $X = [0,1] \cup [2,3]$, equipped with the subspace topology relative to the whole real line.

In this space $X$, its subspace $[0,1]$ is compact (because the subspace topology of $[0,1]$ relative to $X$ is identical to the subspace topology of $[0,1]$ relative to the real line).

On the other hand, $[0,1]$ is an open subset of $X$ (because it is the intersection of $X$ with $(-1/2,3/2)$ which is an open subset of the real line). As it turns out, $[0,1]$ is also a closed subset of $X$. So, it is both closed and open in $X$ (but it is not open in the real line, of course).


In metric spaces, a subset(seen as a subspace) is compact if, and only if, it is complete and totally bounded. If a subset is complete, it must be closed. Therefore, it it isn't closed, it can't be compact.


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