Proposition 3.43. If (X, d) is a metric space, then (X, Od ) is compact if and only if it is
sequentially compact.
Proof. Suppose that X is compact. Let (x i ) 2 X N be a sequence. Let AI = {x i : i 2 I + N}.
T
Note that if I  J, then AI AJ . We show that I 2N AI is non-empty. If it were empty,
then {UI := X \ AI : i 2 N} would an open cover and would thus have a finite subcover.
T
But this means that for some I
1, UI = X, which is non-sense. Let x 2 I 2N AI . For
each j 2 {1, 2, . . .} and each I 0, there exists an i 1 such that x i 2 B1/j (x). A diagonal
sequence argument constructs a sequence converging to x: Pick i j so that x i j 2 B1/j (X );
then lim j!1 x i j = x.
For the converse, suppose X is sequentially compact. Let {Ui : i 2 I} be an open cover.
Proposition. There is a " > 0 such that for any x 2 X there is an i 2 I such that
B" (x) ⇢ Ui .
Suppose not. Then for any n 2 1 + N we can find an x i such that B1/n (x n ) is not
contained in any of the Ui . By sequential compactness a subsequence (x nm ) converges, say
to x 2 X. Since {Ui : i 2 I} is a cover there is an i 2 I such that x 2 Ui . Since Ui is open,
there is an " > 0 such that B" (x) ⇢ Ui . For m sufficiently large 1/nm < "/2 and since
x = lim x nm we can also assume that x nm 2 B"/2 (x). But then by the triangle inequality
B1/nm (x nm ) ⇢ B"/2 (x nm ) ⇢ B" (x) ⇢ Ui .
This contradicts the choice of x nm however.
Proposition. Let " > 0 be as in the proposition. There is a finite set {x n : n = 1, . . . , N }
S
such that X = n B" (x n ). in particular, if i n 2 I is such that B" (x n ) 2 Uin , then
{Uin : n = 1, . . . , N } is a finite subcover.
Pick x 1 2 X. If B" (x 1 ) = X, we are done. If not, pick x 2 2 X \ B" (x 1 ). Again, if
B" (x 1 ) [ B" (x 2 ) = X, we are done. Otherwise, keep going and pick
[
x n+1 2 X \
B" (x k ).
k=1,...,n
If this process terminates, we are done. If not, we get a sequence (x n ). By sequential
compactness, a subsequence (x nm ) has a limit, say, x. Since (x nm ) converges it is Cauchy,
thus for m < k sufficiently large d(x nm , x nk )  "/2; but that contradicts the choice of
x nk .
⇤
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