1. ## Uniform Continuity

Here is a proof that I am having trouble getting started with. I'd appreciate any help.

Let $\displaystyle D$ be a subset of $\displaystyle \mathbb{R}$. Let $\displaystyle f\rightarrow \mathbb{R}$ be uniformly continuous. Let $\displaystyle x_0$ be a limit point of $\displaystyle D$. Suppose $\displaystyle x_{0}\notin D$. Prove there is a continuous function $\displaystyle g\bigcup \{x_0\}\rightarrow \mathbb{R}$, such that $\displaystyle g(x)=f(x)$ for all $\displaystyle x\in D$.

I'll be happy to attempt the proof. If anyone could maybe just tell me exactly what direction I should be looking to go I would would appreciate it.

2. Originally Posted by zebra2147
Here is a proof that I am having trouble getting started with. I'd appreciate any help.

Let $\displaystyle D$ be a subset of $\displaystyle \mathbb{R}$. Let $\displaystyle f\rightarrow \mathbb{R}$ be uniformly continuous. Let $\displaystyle x_0$ be a limit point of $\displaystyle D$. Suppose $\displaystyle x_{0}\notin D$. Prove there is a continuous function $\displaystyle g\bigcup \{x_0\}\rightarrow \mathbb{R}$, such that $\displaystyle g(x)=f(x)$ for all $\displaystyle x\in D$.

I'll be happy to attempt the proof. If anyone could maybe just tell me exactly what direction I should be looking to go I would would appreciate it.
I'll give you three things to think about

1) If $\displaystyle x_0\in\overline{D}$ then there is a sequence of points in $\displaystyle D$ converging to it.

2) The image of Cauchy sequences under unif. cont. maps is unif. cont.

3) $\displaystyle \mathbb{R}$ is complete.

Try to connect these three.

3. To be honest we haven't talked about Cauchy sequences so I'm not sure how to connect all these.

However, I will do my best with what I have gathered form Wikipedia...
Since $\displaystyle x_0$ is a limit point in $\displaystyle D$ then for $\displaystyle x_0\in\overline{D}$ there must be a sequence of points that are contained in $\displaystyle D$ that converge to $\displaystyle x_0$.

Therefore, we have a sequence of points that are converging to a point $\displaystyle x_0$. Thus,(from what I gather about Cauchy Sequences), this sequence is a Cauchy sequence since the points are converging to $\displaystyle x_0$.

Then, (I got the following from Wikipedia) M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M or alternatively if every Cauchy sequence in M converges in M.
Here, our Cauchy sequences is converging to $\displaystyle x_0$ where $\displaystyle x_0\in \overline{D}$. So, $\displaystyle \overline{D}$ is complete?

Sorry this is might be kinda weak but I did what I could for never have learning about Cauchy Sequences or completeness....

4. Originally Posted by zebra2147
To be honest we haven't talked about Cauchy sequences so I'm not sure how to connect all these.

However, I will do my best with what I have gathered form Wikipedia...
Since $\displaystyle x_0$ is a limit point in $\displaystyle D$ then for $\displaystyle x_0\in\overline{D}$ there must be a sequence of points that are contained in $\displaystyle D$ that converge to $\displaystyle x_0$.

Therefore, we have a sequence of points that are converging to a point $\displaystyle x_0$. Thus,(from what I gather about Cauchy Sequences), this sequence is a Cauchy sequence since the points are converging to $\displaystyle x_0$.

Then, (I got the following from Wikipedia) M is said to be complete (or Cauchy) if every Cauchy sequence of points in M has a limit that is also in M or alternatively if every Cauchy sequence in M converges in M.
Here, our Cauchy sequences is converging to $\displaystyle x_0$ where $\displaystyle x_0\in \overline{D}$. So, $\displaystyle \overline{D}$ is complete?

Sorry this is might be kinda weak but I did what I could for never have learning about Cauchy Sequences or completeness....
Hmm, well $\displaystyle \overline{D}$ is complete (any closed subspace of a complete space is complete), but the point is this.

Picture this. Since $\displaystyle x_0\in\overline{D}$ there is a sequence of points $\displaystyle \{x_n\}$ in $\displaystyle D$ such that $\displaystyle x_n\to x_0$. Now, these points converge to $\displaystyle x_0$ but all we really care about is that they get really close to one another as $\displaystyle n$ gets big (they're Cauchy). But, it's a fact that if the $\displaystyle x_n$'s get really close to one another (Cauchy again) and $\displaystyle f$ is unif. cont then the values of the sequence $\displaystyle \{f(x_n)\}$ get really close to one another (Cauchy). But, it's a fact that says that all sequences of real numbers which get really close to one another (Cauchy) converge, and so in particular $\displaystyle f(x_n)$ converges to some $\displaystyle y$. So, what if we said that $\displaystyle f(x_0)=\lim_{n\to\infty}f(x_n)=y$. I mean, it's not obvious why this construction is independent of the choice of the sequence $\displaystyle \{x_n\}$, but it is. From there continuity (and in fact uniform continuity, although you don't need to prove that) are clear since if $\displaystyle \{d_n\}$ is any sequence in $\displaystyle D\cup\{x_0\}$ we have that (combining continuity on $\displaystyle D$ with the construction of $\displaystyle f(x_0)$) that $\displaystyle \displaystyle f\left(\lim_{n\to\infty}d_n\right)=\lim_{n\to\inft y}f(d_n)$ for any $\displaystyle \{d_n\}$ in $\displaystyle D\cup\{x_0\}$.