Does anyone have an elegant/easy to remember proof of this? I can prove bounded easily, and I have notes on why its closed, but they're kind of messy and confusing!
By how we defined them, each contains only a finite number of terms of a sequence (a countable infinte set) so when we get our finite subcover we can get at most a finite number of terms of the sequence in the union of a finite number of balls, but these union contains which contains the sequence.
Here are traditional proofs for both properties.
Suppose that is a limit point of but .
this is true .
The collection covers .
So finite subcollection also covers .
But note that . That is a open set that contains and no other point of . Contradiction.
For bounded, there is finite collection covering .
Let . It is easy to show that is a bound for .
Here's what I think is the simplest proof that any compact set is bounded.
(Assuming that A is in a metric space, of course.)
Let p be any point in the compact set, A. Let be the open ball centered at p with radius n. Certainly every point in A has some distance from p and there exist an integer larger than that distance. That is, the set of all such open balls is an open cover of A. Since A is compact, there is a finite subcover of A, so there is a largest "N". Show that, if x and y are any points in A, d(x,y)< 2N.
To show that any compact set, A, is closed, show that its complement is open. (Again, in a metric space.)
Let p be a point in the complement of A. For any q in A, let B(q) be the open ball, of radius 1/2 the distance from p to q, centered on q. Let C(q) be the open ball, of radius 1/2 the distance from p to q, centered on p (note that C(q), though centered on p, is still "indexed" by q). The set of all open balls, B(q), for all q in A, is an open cover for A. Since A is compact, there exist a finite subcover, . Look at the corresponding collection of open sets . Since p is in each of them it is in there intersection. Further, since this is a finite collection, its intersection is an open set. Finally, since every member of A is in one of the sets , it is not in the corresponding and so not in there intersection.
That is, the intersection of the all the is an open set, containing p, which contains no member of A. That means that p is an interior point of the complement of A and, since p could be any member of the complement of A, the complement of A is open and A itself is closed.