Proving Uniformly Continuous Extension of Function ##f## in Metric Space ##E'##

[Lee33]

Homework Statement

Let $S\subset E$ where $E$ is a metric space with the property that each point of $S^c$ is a cluster point of $S.$ Let $E'$ be a complete metric space and $f: S\to E'$ a uniformly continuous function. Prove that $f$ can be extended to a continuous function from $E$ into $E'$ and that this extended function is also uniformly continuous.

2. The attempt at a solution

This is a complicated problem for me and I want to know if my reasoning is correct?

Let $s\in S^c$ then since $s$ is a cluster point of $S$ there exists a sequence $s_n \in S$ such that $s_n \to s$ thus $s_n$ is a Cauchy sequence. Since $f$ is a uniformly continuous function and $E'$ is complete then $f(s_n)$ is also a Cauchy sequence that converges in $E'$ to some $x,$ thus we have $\lim f(s_n) = x.$

Now we want to show that $f$ can be extended to $E$ and that this extension is uniformly continuous, in order to do that we must have that for any sequence in $E$ that converges to $s$ then the image sequence must converge to the same limit $x.$ Therefore we define our extension to be $\lim g(y_n) = x$ for $y_n \in E.$

Now to prove that this is indeed an extension we must show two things: that is uniformly continuous and that the image sequence converges to the same limit $x$ thus $\lim g(y_n) = x.$

Is my reasoning correct? Thank you for your time!

 

[mighty2000]

Your reasoning is mostly correct, but there are a few small mistakes. Let me walk you through the proof step by step.

First, we have to show that the extension of $f$, which we will call $g$, is well-defined. This means that if we have two sequences $y_n, z_n \in E$ that converge to the same limit $s \in S^c$, then $\lim g(y_n) = \lim g(z_n)$. You have correctly stated that since $s$ is a cluster point of $S$, there exists sequences $y_n, z_n \in S$ that converge to $s$. However, you cannot just say that $\lim f(y_n) = x$ and $\lim f(z_n) = x$ - this is what you are trying to prove! Instead, you should say that since $f$ is uniformly continuous, for any $\epsilon > 0$ there exists a $\delta > 0$ such that if $d(y_n, z_n) < \delta$, then $d(f(y_n), f(z_n)) < \epsilon$. But since $y_n$ and $z_n$ both converge to $s$, we can find an $N$ such that for all $n \geq N$, we have $d(y_n, z_n) < \delta$. Therefore, we have $d(f(y_n), f(z_n)) < \epsilon$ for all $n \geq N$, and since $f(y_n)$ and $f(z_n)$ both converge to $x$, we can conclude that $\lim f(y_n) = \lim f(z_n) = x$.

Next, you have to show that $g$ is uniformly continuous. This means that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $d(y_n, z_n) < \delta$, then $d(g(y_n), g(z_n)) < \epsilon$. You can use a similar argument as above to show this. Let $\epsilon > 0$ be given. Since $f$ is uniformly continuous, there exists a $\delta > 0$ such that if $d(y_n, z_n) < \delta$, then ##d