- #1
Rasalhague
- 1,387
- 2
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.
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.