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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?

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    $\begingroup$ Your reasoning is flawed; there is no reason to consider these open balls. Instead, you need to take your cover $\{G_\alpha\}$ and show that you only need finitely many of them to cover the set. In this case, if $F$ is finite and $F \subset \cup G_\alpha$, then each member of $F$ is in at least one particular $G_\alpha$. Do you see how you only need finitely many of $\{G_\alpha\}$ then? $\endgroup$ – User8128 May 25 at 6:19
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    $\begingroup$ No; you can’t use open balls. You need to find a finite set of elements of the given cover that already covers $F$. You know that $F\subseteq \cup G_{\alpha}$. That means, for example, that $a_1\in\cup G_{\alpha}$. That means that there exists an $\alpha_1$ such that $a_1\in G_{\alpha_1}$... $\endgroup$ – Arturo Magidin May 25 at 6:20
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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.

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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.

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    $\begingroup$ This should get a funniest answer price :D $\endgroup$ – Maximilian Janisch May 25 at 6:58
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    $\begingroup$ Hmm... I am not sure to understand how you cover $a_{22}$... $\endgroup$ – J.-E. Pin May 25 at 7:35

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