I have been studying compact sets recently, and have been struggling a little to develop my intuition. I feel a little silly asking this question, because I know there is probably something very simple I am missing from a definition I've read.

So my issue is why sets in $\mathbb R$ closed and bounded sets must be compact. For example, the set $\{ \frac 1n, n \in \mathbb N \} $ is open, but it can be covered by the open set $(-1, 2)$. So if it can be covered by open sets such as this, why is it not compact?

The definition for compactness I've been working with is a set is compact if all open covers of that set have a finite subcover. Is the given set not compact because there is some open cover that doesn't have a finite subcover, even though there clearly are SOME finite subcovers?

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    $\begingroup$ Yes, that is exactly right. All sets have a finite open cover (namely the entire space), but compact sets are those such that all open covers can be refined to finite open subcovers. $\endgroup$ – Mees de Vries Nov 6 '17 at 10:43
  • $\begingroup$ Compare show that $\{ 1/n : n\in \Bbb N \}$ is not compact using open covers. $\endgroup$ – Martin R Nov 6 '17 at 10:44
  • $\begingroup$ That's basically the DeMorgan law for quantifiers. The opposite of "all open covers have a finite subcover" is "there exists an open cover that doesn't have a finite subcover" (actually there's a quantifier more here in "have a finite subcover". $\endgroup$ – skyking Nov 6 '17 at 11:03

Why do you say that $\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$ is open? It is neither open nor closed.

Every subset $A$ of $\mathbb R$ can be covered by an open cover: $\{\mathbb{R}\}$. The point about compacteness is that every open cover has a finite subcover. Take again the set $\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$. It is not compact because the cover $\left\{\left(\frac1{2n},\frac3{2n}\right)\,\middle|\,n\in\mathbb N\right\}$ is an open cover which has no finite subcover.

Note that an unbounded subset of $\mathbb R$ cannot be compact, because $\left\{(-n,n)\,\middle|\,n\in\mathbb N\right\}$ is an open subcover without a finite subcover. And if $A$ is a non-closed subset of $\mathbb R$, then $A$ cannot be compact because, if $x\in\overline A\setminus A$, then $\left\{\left(-\infty,x-\frac1n\right)\cup\left(x+\frac1n,+\infty\right)\,\middle|\,n\in\mathbb N\right\}$ is an open cover without a finite subcover.

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  • $\begingroup$ Yeah I just didn't stop and think for long enough about that example. I thought about the open $\frac 1n \to 0 $ side and not the closed side that starts with $1$. Whoops. $\endgroup$ – leob Nov 6 '17 at 12:14
  • $\begingroup$ Ok thank you, that clears things up a lot. I think I've just been missing a lot of solid examples from $\mathbb R$ to help build my intuition. $\endgroup$ – leob Nov 6 '17 at 12:17

You've already got a specific response to do with your set, so I'll try to help with the intuition side.

The most intuitive way for me to think about compactness is that's a generalisation of finiteness. If a set isn't bounded, it's certainly not finite, nor is it "compact", our generalisation of finite. Now I'll try to explain why being "closed" is also akin to being finite.

Consider a closed interval $[a,b] \subset \mathbb{R}$. Then no matter how much you try to scale it, you'll still end up with an interval of the form $[x, y] \subset \mathbb{R}$. You can never stretch it so that it's "the same size" as $\mathbb{R}$.

On the other hand, consider an open interval, like $(-\pi/2, \pi/2)$. This also looks pretty small compared to $\mathbb{R}$, but what if you scale it with $\tan$? $\tan[(-\pi/2, \pi/2)]$ is well defined, but the image is the entirety of $\mathbb{R}$. In this way, you can "scale" open intervals and get the entire real line. So in a handwavy way, open intervals are "bigger", or "less finite" than closed intervals.

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  • $\begingroup$ This seems arbitrary. If you take the extended real line $\mathbb R \cup \{\pm \infty\}$ the opposite is true. $\endgroup$ – Mees de Vries Nov 6 '17 at 11:47
  • $\begingroup$ Thanks for the comment. In the math units I've been taking intuition really doesn't get much time so I appreciate hearing some interesting ways to visualise this. $\endgroup$ – leob Nov 6 '17 at 12:28

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