# How does one think in terms of the definition of compactness?

(noob question; I'm rather confused by the compactness definition and haven't found any help on the internet yet)...

So the definition given in my textbook for compactness is: a set X is compact if, for every open covering $\{U_i|i \in I\}$ of X, there exists a finite subset $J$ of $I$ such that $\{U_j|j \in J\}$ is also a covering of X.

I can understand that e.g. $[0,1] \subset \mathbb{R}$ is compact, because one could have some open interval $(0-\epsilon,1+\epsilon)$ ($\epsilon \in \mathbb{R}$ small) which contains $[0,1]$ for any $\epsilon \gt 0$ and is therefore a covering of the interval; but any iteration would then terminate, as we land on a closed subcover.

But then one could make the same argument about any bounded but merely non-open (but not necessarily closed, e.g. a half-open finite) interval. So I must be thinking about this in the wrong way - could anyone help me out, please? Hints/references most welcome.

(Bonus noob question: why exactly must it be a finite subset? What changes when e.g. we take a covering consisting of $N$ subsets of X and take $N \rightarrow \infty$?)

Addendum to the bonus noob question: I'm not quite sure how to elucidate what I'd like to get out of it, so ignore it. It's presumably just there to exclude some pathological cases or in order to coincide with the intuitive notion of compactness, but I'm just not sure how.

• How would you apply your algorithm to the open cover $[0,\frac{1}{n})$ of $[0,1)$? – Najib Idrissi Sep 29 '14 at 15:07
• To address the bonus noob question, if you have a covering consisting of $N$ subsets of $X$, it's already finite and so the definition of compactness supplies you with no more information than you already have. – Lee Mosher Sep 29 '14 at 15:10
• @NajibIdrissi If we define $[0,1)$ as an open interval, my logic evidently falls down. Now I think about it, it does not contain all of its boundaries and therefore it could be an open set - I went the other way and assumed, because it contained some boundary it was therefore closed (or at least not open)! Noob cannon overload – A. C. Sep 29 '14 at 15:27
• @LeeMosher This is true. I'll edit the noob question; I wasn't quite clear. You get some bonus. – A. C. Sep 29 '14 at 15:33
• Just want to say thank you all for the answers and comments; I need to take some time to get my head around them and then I will accept one. – A. C. Sep 29 '14 at 15:49

It's hard to get an intuitive handle on compactness at first. One way to view a compact set is as one that behaves, topologically, as if it were finite. I don't know if that actually helps, but I'll try to explain it anyway.

Certainly every finite set is compact, no matter the topology. (You should prove this.) And it is possible that finite sets are the only compact sets, if the topology is right. (You should prove this too.) So the claim that compactness is something like finiteness is not completely insane.

But let's consider an infinite compact set, say $[0,1]$ in the usual topology. Since we are doing topology, we would like to understand points by what open sets they are in. So for each point $p$ of $[0,1]$, pick an open set $G_p$ containing $p$, and let $G$ be the collection of all $G_p$. The family $G$ is an uncountable collection of sets! But because $[0,1]$ is compact, we can find a simpler way to understand $[0,1]$. There is a finite subset $F\subset [0,1]$ for which each point of $[0,1]$ is in $G_p$ for some $p$ in the finite subset $F$. If $x\in G_p$ means that $x$ is ‘close to’ $p$, then every point of $[0,1]$ is ‘close to’ some element of the finite set $F$. And no matter how complicated the original family $G_p$ is, we can always simplify it to a finite family of the $G_p$ and every element of $[0,1]$ will be in one of the members of the simpler family.

I would like to add that your claim that you can understand why $[0,1]$ is compact is suspect, because your explanation of why $[0,1]$ is compact is missing the important part. Obviously $[0,1]$ can be covered by $(0-\epsilon, 1+\epsilon)$. That is not the reason that $[0,1]$ is compact. Every set can be covered by a finite family of open sets (you should prove this), so this property is not interesting. The “subfamily” part of compactness is crucial and you can't leave it out. But your explanation did leave it out. (This is a very common misunderstanding among people who are new to compactness.)

To show compactness of some set $S$, it is not enough to show that $S$ is covered by a finite family of open sets; this is trivial, because every set, compact or not, is covered by a finite family of open sets. Instead, the proof must go like this:

1. An adversary gives you $G$, which is a family of open sets for which $S\subset \bigcup G$
2. You find a finite subset $G'\subset G$ for which $S\subset \bigcup G'$.

If you can guarantee to succeed in step 2 regardless of which $G$ was given to you in step 1, you win, and $S$ is compact. Conversely, if you want to show that $S$ is not compact, you and the adversary switch roles: you produce a family $G$ in step 1, and try to foil the adversary, who plays step 2. You win if the adversary cannot find a finite $G'\subset G$ for which $S\subset\bigcup G'$.

I suggest you try this:

1. Show that $(0,1)$ is not compact: find the step 1 family $G$ of open sets so that your adversary can't win in step 2.
2. Show that $[0,1]$ is compact: let the adversary give you a family $G$ and show how to find finite $G'\subset G$ so that $G'$ still covers $[0,1]$.

If you can do those two basic exercises, you will have made a good start on understanding compactness.

• This is excellent, I think I have it now :). For starters, I was missing the key point (I had it the wrong way round) that to prove the set is compact, you have to prove you can always find some finite subcover. Cool. – A. C. Sep 29 '14 at 17:03
• So for any open (or half-open, or anything that doesn't fully contain its boundary, thanks @NajibIdrissi for making me realise that) interval in $\mathbb{R}$, one merely has to construct some family of subsets which can only cover it through some infinite limit. If then you have a closed interval, you've then got to shift the bounds of your subsets by a small epsilon - but then we've immediately introduced some large cut-off ($n=1/\epsilon$ for the usual example) which gets you the whole cover in a finite number of steps (though large). – A. C. Sep 29 '14 at 17:06
• Right. In $\Bbb R$ (and more generally in $\Bbb R^n$) there is a powerful theorem that says that the compact sets are exactly the ones that are bounded (and so have a boundary) and closed (and so contain that boundary). In more general spaces, this is not the case. – MJD Sep 29 '14 at 18:04

For instance, (in a metric topology) a compact is bounded: you want to show that for some $n$, $K\subset B(0,n)$ for an open ball $B(0,n)$.

Then consider the open covering $K\subset \cup B(0,n)$. Using compactness, you can find $k, n_1, \dots n_k$ such as you also have $$K\subset \bigcup_{l=1}^k B(0,n_l)$$ and then, $$K\subset B(0, \max n_l)$$

Play with some other properties and you will start to understand the usage of the definition.

For the closeness: consider $x\notin K$. You want to prove that there is $\epsilon>0$ such as $B(x,\epsilon)\subset K^c$, or in other words, that $$\inf\{d(x,y) : y\in K\}>0.$$

Consider the cover $$K\subset \bigcup_{y\in K} B(y, d(x,y))$$

Then you can also write $$K\subset \bigcup_{i=1}^n B(y_i, d(x,y_i))$$ and $$\inf\{d(x,y) : y\in K\} \ge \inf\{d(x,y) : y\in \bigcup_{i=1}^n B(y_i, d(x,y_i))\} \\ =\inf \{d(x,y_i): i= 1\dots n\} >0$$

A good way to understand compactness is to read and understand the proofs of various theorems that have compactness in the hypothesis and/or in the conclusion. I particularly recommend the following theorems:

• $[0,1]$ is compact (your understanding of the compactness of $[0,1]$ is very incomplete)
• The Heine-Borel theorem: a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.
• Given a continuous map of topological spaces $f : X \to Y$, if $X$ is compact then $\text{image}(f)$ is compact.
• If $Y$ is a compact Hausdorff space then every closed subset is compact.
• Given a continuous bijection of topological spaces $f : X \to Y$, if $X$ is compact and $Y$ is Hausdorff then $f$ is a homeomoprhism.