# Uniform Continuity

• Nov 13th 2010, 09:00 AM
zebra2147
Uniform Continuity
Here is a proof that I am having trouble getting started with. I'd appreciate any help.

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

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, 12:45 PM
Drexel28
Quote:

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

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

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

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

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

Try to connect these three.
• Nov 13th 2010, 03:22 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 $x_0$ is a limit point in $D$ then for $x_0\in\overline{D}$ there must be a sequence of points that are contained in $D$ that converge to $x_0$.

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

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 $x_0$ where $x_0\in \overline{D}$. So, $\overline{D}$ 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, 03:32 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 $x_0$ is a limit point in $D$ then for $x_0\in\overline{D}$ there must be a sequence of points that are contained in $D$ that converge to $x_0$.

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

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 $x_0$ where $x_0\in \overline{D}$. So, $\overline{D}$ 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 $\overline{D}$ is complete (any closed subspace of a complete space is complete), but the point is this.

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