Answer to Question #217049 in Real Analysis for Prathibha Rose

Question

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

Answer 1

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

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

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

Let $\gamma\isin R^+$ and suppose that $u,v\isin X$ satisfy $d(u,v)<\delta_{\gamma/3}/2$ . Then the open set $B_{u,\gamma/3}\land B_{v,\gamma/3}$ contains both u and v and is thus not empty; because S is dense in X, it contains some point a of S. Therefore $f(a)\isin f(B_{u,\gamma/3})\land f(B_{v,\gamma/3})$ . By definition, we have both $\tilde{f}(u)\isin f(B_{u,\gamma/3})$ and $\tilde{f}(v)\isin f(B_{v,\gamma/3})$ . We have shown above that these sets have diameter not exceeding $\gamma/3$ , forcing both $e(f(a),\tilde{f}(u))\le\gamma/3$ and $e(f(a),\tilde{f}(v))\le\gamma/3$ and yielding $e(\tilde{f}(u),\tilde{f}(v))\le 2\gamma/3<\gamma$ . So $\tilde{f}$ is uniformly continuous on X.