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11 Proof: ====>: Firstnote that w(X, C(X)) ~ T. Let x EU E T and since X is completely regular, we can find f E C(X) suchthat f(x) = 0 and fluc = 1. Then the set V = {y E X : f(y) < 1} is a w(X, C(X))neighborhood of x and V~ U. Therefore U is w(X, C(X))-open and so T ~ w(X, C(X)), hence T = w(X, C(X)) = w(X, Cb(X)). Elements of Topology 23 -<=:=: Let C ~X closed and x ~ C. Since oc is weakly open, we can find U ~ cc with n U = n{y EX : Jfk(Y)- fk(x)J < 1}, k=l where fk E O(X). Let 9k(x) = min{1, lfk(z) - fk(x)J}, k E {1, ...

For example d and kd {k > 0} generate the same topology. A more interesting example is the following. If d is a metric compatible with the topology T, then so is . ( ) d(x,y) the metrzc d 1 x,y = l+d(x,y) for alt x,y EX. Note that d 1 ::; 1. REMARK Other equivalent metrics are given by: d2(x,y) = Jd(x,y), d3(x,y) = ln(l + d(x, y)), d4(x, y) = min{l, d(x, y)} (again note that d4 ::; 1}. p(r)], then ifd(x,y) =

0 such that d2(x,y) ::; Mdi(x,y) for alt x,y EX, then T(d2) ~ T(di).

24. 29 Every regular Lindelöf space X is normal. Proof: Let C1 and C2 be two nonempty disjoint closed sets in X. Because of the regularity of X, given any x E C 1 , we can find Ux E N(x) suchthat UxnC2 = 0. Similarly given any y E C2, we can find Vy E N(y) suchthat Vy n C1 = 0. The collection {Ux}xEC1 U {Vy}yEc2 U {Cf n C2} is an open cover of X. So we can find a countable subfamily {Un}n>l ~ {Ux}xEC1 , {Vn}n~l ~ {Vy}yEC2 such that C1 ~ Un~lUn and G; ~ Un>l Vn. Foreach n 2:: 1, let U~ = Un \ U~=l V k and v; = Vn \U~=l U k· Both sets are open, for all n, m 2:: 1, u~ n V~ = 0, Cl ~ Un>l u~ = u, c2 ~ Un>l v; =V and u n V= 0.