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?
Best Answer
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.