Extension of Uniformly Continuous Function into a Complete Metric Space

Let $\displaystyle E$ $ be a subset of $\displaystyle X $, and $\displaystyle f : E -> Y $ be uniformly continuous, where $\displaystyle Y$ is complete. Show that there exists a continuous function which maps the closure of $\displaystyle E $ to $\displaystyle Y $ such that $\displaystyle F(x) = f(x). $

So basically, we are trying to prove that a function can be extended to its domains closure, and still retain the same values on the original set.

It's late, so maybe I need to spend more time pondering this question, but I'm frankly a little stumped - of the continuity problems I have tackled so far from chapter 4 of Rudin, this is the first involving the hypothesis of a mapping into a *complete* metric space, so it suggests (at first to me anyway) a different method of proof.

I guess I would start off with first establishing that the closure of E is (obviously) a closed set, with a limit point in X (in fact, in E closure by definition). I can than approach this limit point with a convergent Cauchy sequence, and then the image of this sequence will again be Cauchy in Y and thus convergent in Y. These facts follow from the hypothesis that the function f is uniformly continuous, and Y is complete. Then I suppose I could somehow use the sequential characterization of continuity to conclude the theorem...but rigorizing this argument, and then actually proving that this "extension" big F is equal to little f on E, is something I'm not sure about.

Anyway, any other ideas, or modifications to mine, for this problem?

Re: Extension of Uniformly Continuous Function into a Complete Metric Space

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**TaylorM0192** Let $\displaystyle E$ $ be a subset of $\displaystyle X $, and $\displaystyle f : E -> Y $ be uniformly continuous, where $\displaystyle Y$ is complete. Show that there exists a continuous function which maps the closure of $\displaystyle E $ to $\displaystyle Y $ such that $\displaystyle F(x) = f(x). $

So basically, we are trying to prove that a function can be extended to its domains closure, and still retain the same values on the original set.

It's late, so maybe I need to spend more time pondering this question, but I'm frankly a little stumped - of the continuity problems I have tackled so far from chapter 4 of Rudin, this is the first involving the hypothesis of a mapping into a *complete* metric space, so it suggests (at first to me anyway) a different method of proof.

I guess I would start off with first establishing that the closure of E is (obviously) a closed set, with a limit point in X (in fact, in E closure by definition). I can than approach this limit point with a convergent Cauchy sequence, and then the image of this sequence will again be Cauchy in Y and thus convergent in Y. These facts follow from the hypothesis that the function f is uniformly continuous, and Y is complete. Then I suppose I could somehow use the sequential characterization of continuity to conclude the theorem...but rigorizing this argument, and then actually proving that this "extension" big F is equal to little f on E, is something I'm not sure about.

Anyway, any other ideas, or modifications to mine, for this problem?

Right. The basic idea is precisely as you stated. You can extend $\displaystyle f:E\to Y$ to $\displaystyle \widetilde{f}:\overline{E}\to Y$ by choosing, for each $\displaystyle x\in \overline{E}$, a sequence $\displaystyle \{e_n\}$ in $\displaystyle E$ converging to $\displaystyle x$ (you know one exists by definition of closure for metric spaces) and then defining $\displaystyle \widetilde{f}(x)=\lim f(e_n)$ where the right hand side makes sense since uniformly continuous functions are Cauchy continuous, and since $\displaystyle Y$ is completely we have that the desired limit exists.

From there you have to show that this mapping is well-defined, in the sense that it's independent of which sequence you choose that converges to a particular point. This isn't bad though. If $\displaystyle e_n,e'_n\to x$ then you know you can pick $\displaystyle n$ so large that they are both within ____ of $\displaystyle x$ and so by uniform continuity their images are with ____ of each other--ending in a $\displaystyle d(\lim f(e_n),\lim f(e'_n))<\varepsilon$ for all $\displaystyle \varepsilon$ and so they're equal. Note that this automatically gives you that $\displaystyle \widetilde{f}_{\mid E}=f$ (i.e. that $\displaystyle \widetilde{f}$ extends $\displaystyle f$) since for each $\displaystyle e\in E$ you can choose $\displaystyle \{e_n\}$ to be the constant sequence $\displaystyle e$.

Showing its' continuous isn't bad then.

Does that help? If you're curious you can look on this blog post of mine, where I prove something (ever so) slightly more general.

Re: Extension of Uniformly Continuous Function into a Complete Metric Space

Let me sleep on this tonight and write up a solution in the morning, with your extra thoughts in mind. Then I'll read your blog post afterward - it's a bit longer than I expected! On another note, I suppose it's time to boost my confidence a little more with this material - it has just been a literal crash course in "real" mathematics all quarter long for me; I'm a physics major, and so the only mathematics I've been exposed to prior to now has been largely computational.