# 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?

• 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? – User8128 May 25 at 6:19
• 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 $a_1\in G_{\alpha_1}$... – Arturo Magidin May 25 at 6:20

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.

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.

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