# Compact -> bounded

1. Apr 17, 2012

### Rasalhague

Compact --> bounded

In lecture 8 of Francis Su's Real Analysis online lecture series, he has a proof that a compact subset of a metric space is bounded: Given a metric space (X,d), if A is a compact subset of X, then every open cover of A has a finite subcover. Let B be a set of open balls of radius r, one centered on each point of A. This is an open cover of A, so it has a finite subcover {Bi}. Let C be the set of distances between pairs of center points of the elements of the subcover {Bi}. Let s = max C. Then B(x,s+2r) is an open ball that includes A.

My question: why the 2? Isn't B(x,s+r) also an open ball that includes A? If u is a point in A, then

d(x,u) <= d(x,w) + d(w,u) <= s + r

where w is the center of an open ball in {Bi} that contains u.

2. Apr 17, 2012

### xaos

Re: Compact --> bounded

set up a horizonal row of spheres tangent to eachother. then s=r*2*(k-1), k=number of spheres. the total diameter is s+2r=r*2*k=d. now draw B(x, d) and note: but where's x? is it in the center or somewhere else?

Last edited: Apr 17, 2012
3. Apr 17, 2012

### Rasalhague

Re: Compact --> bounded

Ah, I see now. I made x be the center of one of the open balls in my proof, although I forgot to mention that. He must have been been taking x to be an arbitrary point of A. Thanks, xaos.