Compactness of finite sets In rudin's analysis books, he defines compactness as: A subset $K$ of a metric space $X$ is ${\bf compact}$ if every open cover of $K$ contains a finite subcover. More explicitly is that if $\{ G_{\alpha} \}$ is an open cover of $K$ then one can select finitely many indices so that $K$ is contained in $G_{\alpha_1} \cup ... \cup G_{\alpha_n } $
Rudin claims that a finite set is compact as an obvious fact. Im trying to verify this fact myself:
verification:
Let $F$ be a finite set and write it as $\{ a_1,...,a_n \}$ Take any open cover $\{ G_{\alpha} \}$ of $F$. We can consider the open balls centered at $a_i$ and let $B_i$ be such ball. Then, $B_1 \cup ... \cup B_n$ contains $F$. Now, this seems incomplete as we must show that this finite union of open balls is in $\cup_{\alpha } G_{\alpha}$. How do we check this?
 A: There is an open set that covers $a_1$
There is an open set that covers $a_2$
There is an open set that covers $a_3$
There is an open set that covers $a_4$
There is an open set that covers $a_5$
There is an open set that covers $a_6$
There is an open set that covers $a_7$
There is an open set that covers $a_8$
There is an open set that covers $a_9$
There is an open set that covers $a_{10}$
There is an open set that covers $a_{11}$
There is an open set that covers $a_{12}$
There is an open set that covers $a_{13}$
There is an open set that covers $a_{14}$
There is an open set that covers $a_{15}$
There is an open set that covers $a_{16}$
There is an open set that covers $a_{17}$
There is an open set that covers $a_{18}$
There is an open set that covers $a_{19}$
There is an open set that covers $a_{20}$
There is an open set that covers $a_{21}$
...
...
...
There is an open set that covers $a_{n}$
It's a long but finite list, so it's a finite subcover.
A: There's no guarantee that any of the $B_i$ are in the set $\{G_{\alpha}\}$. For instance, We could have $\{G_{\alpha}\} = \{F\}$.
The point of compactness is that you really find a subcover of any given cover. In this case, we can argue as follows: suppose we are given a cover $\{G_\alpha\}$ of $F = \{ a_1, \ldots, a_n \}$. For all $i$, pick some $G_{\alpha_i}$ which contains $a_i$ (we know such a set exists because $\{G_\alpha\}$ covers $F$). Then $\{G_{\alpha_1},\ldots,G_{\alpha_n}\}$ is a finite subcover.
Notice how compactness really is a consequence of the finiteness of $F$, i.e. a property of $F$, and we aren't using any properties of the cover, since we don't know what it looks like a priori.
