# Understanding compactness and how it relates to finiteness

It isn't intuitively obvious to me how compactness relates to finiteness, even though I often hear that they are very closely related.

My definition of compactness is: a set A is compact if, given an open cover of A (collection of open sets such that A is a subset of the union of this collection) then for all open covers there exists a finite subcover (a finite amount of open sets which cover A)

It seems to me that there only needs to be a finite amount of open sets which cover A, not that these open sets need to be bounded themselves. For example the reals themselves aren't bounded, yet the reals are an open set. To cover the reals, can't you just take the open cover to be the reals themselves and have a finite sub cover, or if not the reals themselves then any open unbounded set?

• I think there is a mistake in your definition of an open cover, it should be "$A$ is a subset of the union of this collection" – idok Dec 21 '17 at 16:41
• @idok Fixed, thanks. – jacobe Dec 21 '17 at 16:47

Compactness and finiteness are related because, for instance, every functions from a finite set into $\mathbb R$ has a maximum and a minimum and every continuous function from a compact space into $\mathbb R$ also has a maximum and a minimum.

Concerning the final paragraph, it is indeed true that $\{\mathbb{R}\}$ is a cover of $\mathbb R$ and that it has a finite subcover (itself). But being compact means that every open cover has a finite subcover, not that some open cover has a finite subcover. And $\mathbb R$ is not compact (with respect to the usual topology) because, for instance, the open cover $\{(-n,n)\,|\,n\in\mathbb N\}$ has no finite subcover.

• Ah I see, so I wasn't following the definition exactly. It makes sense now, but it's hard to wrap my mind around EVERY open cover of a set has a finite sub cover. So for clarification, say, for the set (0,1), even though the open set, for example, {(-5,5)} is a cover of (0,1) and has a finite subcover (itself), one must show there is a finite subcover for ALL open covers, which i haven't done? – jacobe Dec 21 '17 at 16:55
• And indeed $\{(0+1/n,1-1/n) : n \in \Bbb N\}$ is an (open) cover with no finite subcover – eepperly16 Dec 21 '17 at 16:56
• @jacobelm Yes. The set $(0,1)$ would be compact if every open cover had a finite subcover, but that's not the case. – José Carlos Santos Dec 21 '17 at 17:03
• @eepperly16 I see. But for the reals, it is a closed set, and I know factually that not every open cover has a finite sub cover, but it's hard for me to see why that that is because it is unbounded. Is it because, for example, you can make a finite open ball around every point, which is an open cover, but there doesn't exist a finite subcover to that open cover? – jacobe Dec 21 '17 at 17:06
• In my answer, I provided an example of an open cover of the reals without a finite subcover. – José Carlos Santos Dec 21 '17 at 17:14

Your def'n of compactness should say: $A$ is compact iff , given $any$ family $F$ of open sets such that $\cup F\supset A,$ there exists a finite $G\subset F$ such that $\cup G\supset A.$

Euivalently, $A$ is $not$ compact iff there exists an open cover of $A$ that has no finite sub-cover.

It is sometimes easier to show that some set is not compact, as we only need find a single open cover for which there is no finite sub-cover. For example $\{(-\infty,x):x\in \Bbb R\}$ is an open cover of $\Bbb R$ with no finite sub-cover. It takes more work to prove that $[0,1]$ $is$ compact.

The idea of compactness can not be appreciated unless we see its applications in proving certain non-trivial results. The fact that for a compact set we can replace any open cover by a finite subcover is important precisely because it is far easier to handle a finite number of things as compared to handling an infinite number of things. For instance we can always add a finite number of things (no need to worry for infinite series), or have a guarantee of existence of maximum /minimum of a finite set.

Consider for example that a function $f$ is continuous on a closed interval $[a, b]$. The fact that such a closed interval is compact is a non-trivial result which goes by the name Heine Borel Theorem. Now the function $f$ is bounded locally near each point of the interval $[a, b]$ or to put the matter more formally for each $x\in[a, b]$ we have a neighborhood $I_x$ of $x$ such that $f$ is bounded in $[a, b]$. But from this it is not obvious to conclude that $f$ is bounded on $[a, b]$ precisely because an infinite number of neighborhoods need to be analyzed. But by use of the fact that $[a, b]$ is compact we can just deal with the finite subcover consisting of a finite number of neighborhoods of type $I_x$ and then the function $f$ is bounded in $[a, b]$ because it is bounded in each of these finite number of chosen neighborhoods $I_x$ whose union contains $[a, b]$.

One can use compactness to prove the intermediate value property of continuous functions as well and also the fact that continuous functions attain their maximum/minimum values on a closed interval. There is another result about uniform continuity of continuous functions on a closed interval which does not have a proof without the idea of compactness. All of these applications involve arguments which necessarily require handling a finite number of things. You should try to have a look at such proofs (or better attempt to provide a proof yourself) in order to see how important the idea of compactness is.

You should also pay attention to the proof of Heine Borel Theorem which links it to other forms of completeness of real numbers.