Let (X, d) be a metric space.

Show that any finite subset is compact.

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- Jan 11th 2010, 09:27 AMhebbyCompact Set (topology)
Let (X, d) be a metric space.

Show that any finite subset is compact. - Jan 11th 2010, 09:42 AMDrexel28
Are you serious? Let $\displaystyle \mathcal{V}$ be an open cover for some finite subset $\displaystyle \mathcal{S}$ of $\displaystyle X$. There must exists some $\displaystyle O_s\in\mathcal{V}$ such that $\displaystyle s\in O_s$ for each $\displaystyle s\in\mathcal{S}$. Clearly then $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$ is a subcover of $\displaystyle \mathcal{S}$ and since $\displaystyle \mathcal{S}$ is finite so is $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$.

- Jan 11th 2010, 10:10 AMhebby
Hi

Thanks for the reply, could you explain what is this O small s? - Jan 11th 2010, 11:32 AMh2osprey
These "O-sub-s" things are open sets; the union of these sets forms an open cover for your finite set.

Recall that one of the properties of a compact set is that every open cover of that set admits a finite sub-covering. - Jan 11th 2010, 12:35 PMDrexel28
While , what

**h2osprey**said was undeniably correct, I think the finer points got lost.

To be compact*means*that every open cover has a finite subcover. Thus, to show that a set is compact one way is to take an arbitrary open cover and derive a finite subcover. So, the concept is we let $\displaystyle \mathcal{V}$ be some open cover of $\displaystyle \mathcal{S}$. Next, we let $\displaystyle s\in\mathcal{S}$ be arbitrary, and since $\displaystyle \mathcal{V}$*covers*$\displaystyle \mathcal{S}$ there exists some open set, which we will call $\displaystyle O_s$, in $\displaystyle \mathcal{V}$ such that it contains $\displaystyle s$. So, for each $\displaystyle s\in\mathcal{S}$ we take some corresponding $\displaystyle O_s\in\mathcal{V}$ and construct a new class $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$. This is clearly a subcover, and since there is*at most*$\displaystyle \text{card }\mathcal{S}$ elements of $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$ it must be a finite subcover.

The two finer points that you should definitely answer yourself is why can I choose this $\displaystyle O_s$ and why is there*at most*$\displaystyle \text{card }\mathcal{S}$ elements in $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$