# Homework Help: Uniform continuity, bounded subsets

1. Jun 11, 2007

### redyelloworange

1. The problem statement, all variables and given/known data

Show that if f: S -> Rn is uniformly continuous and S is bounded, then f(S) is bounded.

2. Relevant equations

Uniformly continuous on S: for every e>0 there exists d>0 s.t. for every x,y in S, |x-y| < d implies |f(x) - f(y)| < e

bounded: a set S in Rn is bounded if it is contained in some ball about the origin. That is, there is a constant C s.t. |x|<C for every x in S.

3. The attempt at a solution

I understand the idea of the proof pretty well but I cannot write the correct mathematical interpretation of it down.

Basically, S is bounded, so it can be divided into segments (TA called them partitions which is confusing since S is in Rn not R).
Each segment can be made smaller than d. Then, by uniform continuity we know that f(segment) is smaller than e (bounded) so we can draw a ball around it.
Since there is a finite number of segments, there are a finite number of balls f(segment). Hence, we can draw a bigger ball around all of them, and thus f(S) is bounded.

How do I put this into math symbols?

Thanks for your help =)

2. Jun 12, 2007

### NateTG

The thing you want here is a cover, and not a partition.

Can you show that:
$$N_{\epsilon}(f(x)) \supset f(N_{\delta}(x))$$
and that
$$S \subset \bigcup_{\rm{finite}} N_{\delta}(x_i)$$

As a note: When you're applying a function to each of a set of objects, it's typical to refer to the result as the image of applying the function to that set. Something like:
$$N_{\epsilon}(f(x)) \supset Im_f(N_{\delta}(x))$$
might be better notation, because errors can occur when sets are treated like single-valued objects.

Last edited: Jun 12, 2007
3. Jun 14, 2007

### redyelloworange

What's N_e?

4. Jun 14, 2007

### NateTG

The epsilon neighborhood of something. It's the same idea as, say:
$$B(\epsilon,x)$$
The ball of radius $\epsilon$ around $x$.

I probably didn't make the best choices in notation there -- my apologies.