108 Chapter 2Q = {(x, y): a::; x::; a+ l, b::; y::; b + l}, l > 0 being the side length of Q
and (a, b) the bottom left vertex of Q. In view of Theorem 2.15 it suffices
to prove that Q is compact.
Suppose that Q is not compact: i.e., suppose that there is an open cover-
ing UaEI Ga of Q that contains no finite subcovering. Subdividing Q into
four equal squares with side length^1 / 2 l it follows that at least one of these
four squares will fail to be covered by tlie union of a finite number of the sets
Ga· Let Qi be one such square and call ( a1, bi) its bottom left vertex (if
there are several squares with the same property, for definiteness we choose
first those to the left, then the one below, if necessary). Repeating the pro-
cess with Q 1 we obtain a closed square Q 2 with side length l/2^2 , bottom
left vertex ( a 2 , b 2 ), and such that no finite collection of the Ga covers Q 2 •
Proceeding in the same manner we obtain, by induction, for each positive
k a closed square Q k with side length l /2k, bottom left vertex ( ak, bk),
and such that no finite subcollection of the GO/ covers Qk. Since the real
sequences {ak} and {bk} are monotonic and bounded, they are convergent.
Let limak =a and lim bk= b. Then it follows that lim(ak, h) =(a, b).
But ( ak, bk) E Q for each k and Q is closed. Hence (a*, b*) lies in Q also.
Since (a*,b*) E Q there is some open set, say Ga 0 , covering (a*,b*), i.e.,
(a*, b*) E Gao for some ao E I. Since Gao is open there is a neighborhood
N 0 ((a*,b*)), 8 > 0, such that Ns((a*,b*)) C Gao (Fig. 2.9).If we choose k large enough, say k > J(, so that
-/2l 1 8
2k < 2and(a*, b*)Fig. 2.9