Understanding the Proof for Uniform Continuity on Compact Intervals

[Coffee_]

I would appreciate it if someone could explain the steps in the reasoning of the following statement. This is not a homework assignment or anything.

Let $U \subseteq R^{n}$ be compact and $f:U\to R$ a continuous function on $U$. However f is not uniformly continuous.
Then there exists an $\epsilon>0$ and sequences $a_n$ and $b_n$ such that if $|| a_n - b_n || < \frac{1}{n}$ then $||f(a_n) - f(b_n) \ge \epsilon$.

This statement will lead to proof that continuity on a compact interval means uniform continuity. However this is not proven yet so don't use that fact in your reasoning. Purely based on the definitions of uniform continuity.

 

[micromass]

So try to prove that any function $f:U\rightarrow \mathbb{R}^m$ (for any domain $U$) is uniform continuous if and only if for each two sequences $(a_n)_n$ and $(b_n)_n$ in $U$ holds that if $\| a_n - b_n\|\rightarrow 0$ then $\|f(a_n) - f(b_n)\|\rightarrow 0$.

 

[Coffee_]

Pick a random $\epsilon>0$. And let $a_n$ and $b_n$ be two sequences in $U$ so that $||a_n - b_n||$ converges.

Uniform continuity implies there exists a $\delta$ so that as long $||a_n - b_n|| < \delta$ it follows that $||f(a_n)-f(b_n)||< \epsilon$

Since $||a_n - b_n||$ converges I can always find a certain $N$ so that for $n \ge N$ holds that $||a_n - b_n|| < \delta$

So to summarize, I receive an $\epsilon$ , which in turn returns a unique $\delta$, and I can always find an $N$ so that for $n \ge N$ : $||f(a_n)-f(b_n)||< \epsilon$ which proves it one way.

 

[micromass]

That's correct.

 

[Coffee_]

Suggestions for the other direction of proof? It looks like it's a harder one. By the way I totally get the statement now.

 

[micromass]

Use contradiction. Assume that $f$ is not uniform continuous and use that to construct sequences $(a_n)_n$ and $(b_n)_n$.

 

[Coffee_]

Hello, I don't think this is worth posting another thread so I hope it's alright if I just ask here.

A similar problem to the last discussed matter. I need to show that these two statements are equivalent (note it's not about uniform continuity anymore though!)

1) $f:U->V$ is continuous on some open domain $D$

2) For any sequence where $x_n$ converges to $x$ on $D$ , the sequence $f(x_n)$ converges to $f(x)$ on $V$.

I tried for a bit with assuming that $f$ is not continuous like you mentioned for the previous problem but don't get far.

 

[mathwonk]

i think you can do this alone. the easy direction is 1 implies 2, assuming you are using the epsilon delta definition of continuity.

 

[Coffee_]

Yeah 1 implies 2 was no problem its the other way.

 

[mathwonk]

To pass from arbitrary e>0 to a sequence, one usually uses the fact that 1/n --> 0.

 

[WWGD]

I think you need 1st countability of the space to go from 2 to 1, or you need something like the space being a sequential space. I mean, this is not true of all spaces.

 

[micromass]

I think you need 1st countability of the space to go from 2 to 1, or you need something like the space being a sequential space. I mean, this is not true of all spaces.

We are working in $\mathbb{R}^n$...

 

[WWGD]

OK, sorry, I did not see that. Feel free to delete my previous post if that helps PF.