# Uniform Continuity

• Nov 13th 2010, 08:00 AM
zebra2147
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 $\displaystyle f:D\rightarrow \mathbb{R}$ 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 $\displaystyle g:D\bigcup \{x_0\}\rightarrow \mathbb{R}$, 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.
• Nov 13th 2010, 11:45 AM
Drexel28
Quote:

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 $\displaystyle f:D\rightarrow \mathbb{R}$ 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 $\displaystyle g:D\bigcup \{x_0\}\rightarrow \mathbb{R}$, 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.
• Nov 13th 2010, 02:22 PM
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....
• Nov 13th 2010, 02:32 PM
Drexel28
Quote:

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 $\displaystyle D\cup\{x_0\}$.