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

Question

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!

Answer 1

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.

Answer 2

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 $1/n<x$. 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.

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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 $\mathbb{R}$.