Let (X, d) be a metric space.

Show that any finite subset is compact.

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

Show that any finite subset is compact. - January 11th 2010, 09:42 AMDrexel28
- January 11th 2010, 10:10 AMhebby
Hi

Thanks for the reply, could you explain what is this O small s? - January 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. - January 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 be some open cover of . Next, we let be arbitrary, and since*covers*there exists some open set, which we will call , in such that it contains . So, for each we take some corresponding and construct a new class . This is clearly a subcover, and since there is*at most*elements of it must be a finite subcover.

The two finer points that you should definitely answer yourself is why can I choose this and why is there*at most*elements in