# Don't understand well the definition of compactness, or a compact set

I just learned compactness today in college analysis course (we are using baby Rudin for textbook). The Rudin's definition is

Open cover: By an open cover of a set $$E$$, in a metric space $$X$$, we mean a collection $$\{G_a\}$$ of open subsets of $$X$$ such that $$E \subset \cup_{\alpha}G_{\alpha}$$

Compactness: A subset $$K$$ of a metric space $$X$$ is said to be compact if every open cover of $$K$$ contains a finite subcover.

Professor said that open set like (0, 1) is not compact, but I just thought that why can't we just use (0, 1) itself as a finite subcover because (0, 1) is just one, which I mean here as "finite," set which actually covers (0,1).

I would greatly appreciate it if you could enlighten me so that I can truly understand compactness.

Thank you!

• Find me a finite subcover of the open cover by (1/3, 1) and (1/n,1/2) for all n. – user98602 Oct 12 '18 at 20:31
• The definition stipulates that any open cover has a finite subcover., so it's not up to you to choose an open cover.since there's nothing to choose – Bernard Oct 12 '18 at 20:31

## 2 Answers

It's important that there must be a finite subcover for every open cover. Otherwise yes, trivially every set has a finite open cover; simply take it to be a single open set consisting of the union of open neighborhoods centered at every point in the set.

One way to think of the finite subcover condition is that, for a compact subset of a metric space, an open cover has to necessarily "overshoot" the set. In doing so, it becomes impossible for the minimal subcover to be infinite.

Example 1. Take $$K = [0,1]$$. If we have an open cover of $$K$$ by some open balls, then necessarily there is going to be an open ball that contains the element $$1$$. But if an open ball contains $$1$$, then that open ball must actually extend past $$1$$, in particular that open ball must contain all points greater than $$1$$ and less than $$1 + \delta$$ for some $$\delta > 0$$.

Example 2. Take $$A = [0,1)$$. This set is not compact, because we could choose the open cover consisting of balls $$B_n = (-1, 1 - 1/n)$$ for $$n = 1, 2, \ldots$$. This open cover never "overshoots" $$A$$ on the right endpoint; rather, the balls $$B_n$$ get ever and ever closer to $$1$$. Thus it is impossible for any subcover to be finite; we really do need to take a sequence with $$n$$ going off to infinity.

• Thank you! I have just two questions more. 1) In example 2, is it right that $B_n=(−1,1−1/n)$ is an open cover but just doesn't have a finite subcover? 2) And we cannot use $B_n=(−1,1−1/n)$ as an open cover for $K=[0, 1]$ in example 1 because $B_n$ does not cover 1. Is this right? – Hunnam Oct 12 '18 at 21:39
• (1) Yes, there is no finite subcover. (2) Correct, no $B_n$ contains $1$, so they do not cover $[0,1]$. Whereas it does cover $(0,1)$ because for every number slightly less than $1$, for sufficiently large $n$ there is some $B_n$ containing it. We can already see here the connection between compact sets and the concept of limits. – Christopher A. Wong Oct 12 '18 at 21:46
• Thank you so much! Now I feel like I truly understand what compactness is haha. – Hunnam Oct 12 '18 at 22:13
• What is $x$ in your example 1? – Andrés E. Caicedo Oct 12 '18 at 23:15
• Sorry, that was a typo. – Christopher A. Wong Oct 18 '18 at 5:22

An open cover of $$(0,1)$$ need not have $$(0,1)$$ among its members; for instance $$\bigl\{(0,2/3),(1/3,1)\bigr\}$$ is an open cover. Of course this open cover is finite, so it obviously has an open subcover, namely itself.

What about an infinite open cover? Well the simplest choice is $$\bigl\{(1/2,1),(1/3,1),(1/4,1),\dotsc\bigr\}= \bigl\{(1/n,1):n>1, n\text{ integer}\bigr\}$$ Is this an open cover? Yes, because if $$x\in(0,1)$$, then there exists $$n>1$$ such that $$1/n. The sets are surely open.

Does it have a finite subcover? Suppose so; then it will be of the form $$\bigl\{(1/n_1,1),(1/n_2),\dots,(1/n_k,1)\bigr\}$$ for suitable integers $$n_1,n_2,\dots,n_k$$. But if $$m=\max\{n_1,n_2,\dots,n_k\}$$, then $$\bigcup_{i=1}^k\Bigl(\frac{1}{n_i},1\Bigr)= \Bigl(\frac{1}{m},1\Bigr)$$ and this is not the same as $$(0,1)$$, because $$\frac{1}{2m}\notin\Bigl(\frac{1}{m},1\Bigr)$$ but clearly $$\frac{1}{2m}\in(0,1)$$.

Thus there is at least an open cover of $$(0,1)$$ that doesn't admit finite subcovers. Hence $$(0,1)$$ is not compact.

You will also learn that a compact subset of a metric space is necessarily closed (not a sufficient condition, though). Then $$(0,1)$$ is not compact, because it is not closed in $$\mathbb{R}$$.

• "Does it have an open subcover?" should be "Does it have a finite subcover?" – symplectomorphic Oct 12 '18 at 20:58
• @symplectomorphic Lapsus calami, thanks! – egreg Oct 12 '18 at 20:59
• Thank you! Your explanation helped me a lot to understand. – Hunnam Oct 12 '18 at 22:13