Question #217049

Prove that any uniformly continuous function from a dense subset of a metric space X to a complete metric space Y has a uniformly continuous function from X to Y


1
Expert's answer
2021-07-27T10:56:27-0400

We have: (X,d) and (Y,e) are metric spaces, f:SYf:S\to Y

For each ϵR+\epsilon\isin R^+ , let δϵ\delta_{\epsilon} be the largest real number in (0,1] such that for all a,bSa,b\isin S (dense subset of a metric space X) with d(a, b)<δϵ\delta_{\epsilon} we have e(f(a), f(b))<ϵ\epsilon . Then, for each xXx\isin X , let Bx,ϵ=bX[x;δϵ)B_{x,\epsilon}=b_X[x;\delta_{\epsilon}) . Suppose zXz\isin X . Because d(a,b)<δϵd(a,b)<\delta_{\epsilon} for all a,bSBz,ϵa,b\isin S\land B_{z,\epsilon} , we have also e(f(a),f(b))<ϵe(f(a),f(b))<\epsilon , whence diam(f(Bz,ϵ))ϵdiam(f(B_{z,\epsilon}))\le\epsilon . Because S is dense in X, each of the balls Bz,ϵB_{z,\epsilon} has non-empty intersection with S, so each of the sets f(Bz,ϵ)f(B_{z,\epsilon}) is non-empty. Because Y is complete, the nest {f(Bz,ϵ)ϵR+}\{\overline{f(B_{z,\epsilon})}|\epsilon\isin R^+\} of non-empty closed subsets of Y has singleton intersection. Set f~(z)\tilde{f}(z) to be the sole member of that intersection; it is equal to f(z)f(z) if zSz\isin S , so, when this action performed for each zXz\isin X , f~\tilde{f} is an extension of f to X.

We must show that f~\tilde{f} is uniformly continuous.

Let γR+\gamma\isin R^+ and suppose that u,vXu,v\isin X satisfy d(u,v)<δγ/3/2d(u,v)<\delta_{\gamma/3}/2 . Then the open set Bu,γ/3Bv,γ/3B_{u,\gamma/3}\land B_{v,\gamma/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)f(a)\isin f(B_{u,\gamma/3})\land f(B_{v,\gamma/3}) . By definition, we have both f~(u)f(Bu,γ/3)\tilde{f}(u)\isin f(B_{u,\gamma/3}) and f~(v)f(Bv,γ/3)\tilde{f}(v)\isin f(B_{v,\gamma/3}) . We have shown above that these sets have diameter not exceeding γ/3\gamma/3 , forcing both e(f(a),f~(u))γ/3e(f(a),\tilde{f}(u))\le\gamma/3 and e(f(a),f~(v))γ/3e(f(a),\tilde{f}(v))\le\gamma/3 and yielding e(f~(u),f~(v))2γ/3<γe(\tilde{f}(u),\tilde{f}(v))\le 2\gamma/3<\gamma . So f~\tilde{f} is uniformly continuous on X.


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