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**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 {B

_{i}}. Let C be the set of distances between pairs of center points of the elements of the subcover {B

_{i}}. 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 {B

_{i}} that contains u.