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!
Best Answer
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.