We have: (X,d) and (Y,e) are metric spaces, f:S→Y
For each ϵ∈R+ , let δϵ be the largest real number in (0,1] such that for all a,b∈S (dense subset of a metric space X) with d(a, b)<δϵ we have e(f(a), f(b))<ϵ . Then, for each x∈X , let Bx,ϵ=bX[x;δϵ) . Suppose z∈X . Because d(a,b)<δϵ for all a,b∈S∧Bz,ϵ , we have also e(f(a),f(b))<ϵ , whence diam(f(Bz,ϵ))≤ϵ . Because S is dense in X, each of the balls Bz,ϵ has non-empty intersection with S, so each of the sets f(Bz,ϵ) is non-empty. Because Y is complete, the nest {f(Bz,ϵ)∣ϵ∈R+} of non-empty closed subsets of Y has singleton intersection. Set f~(z) to be the sole member of that intersection; it is equal to f(z) if z∈S , so, when this action performed for each z∈X , f~ is an extension of f to X.
We must show that f~ is uniformly continuous.
Let γ∈R+ and suppose that u,v∈X satisfy d(u,v)<δγ/3/2 . Then the open set Bu,γ/3∧Bv,γ/3 contains both u and v and is thus not empty; because S is dense in X, it contains some point a of S. Therefore f(a)∈f(Bu,γ/3)∧f(Bv,γ/3) . By definition, we have both f~(u)∈f(Bu,γ/3) and f~(v)∈f(Bv,γ/3) . We have shown above that these sets have diameter not exceeding γ/3 , forcing both e(f(a),f~(u))≤γ/3 and e(f(a),f~(v))≤γ/3 and yielding e(f~(u),f~(v))≤2γ/3<γ . So f~ is uniformly continuous on X.
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