# Quick Proof rigor check.

1. Feb 22, 2009

### Quantumpencil

1. The problem statement, all variables and given/known data
Let f be a real uniformly continuous function on the bounbed set E in R. Prove that f is bounded on E.

2. Relevant equations

3. The attempt at a solution

Since f is uniformly continuous, $$\left|f(x)-f(y)\right|< \epsilon$$ if $$\left|x-y\right|< \delta$$. Since E is bounded, there exists some maximal distance between x and y; let M be the upper limit and P be the lower limit of E. Then

$$\left|x-y\right|< \\left|M-P\right|<$$. If M, P are not in E, then define two sequences {x_n} and {y_n} such that lim n->infinity {x_n}=M and lim n->infinity {y_n} = P.

Then the definition of uniform continuity implies that if we let $$\delta = abs(M-P)$$, so that abs(x-y) < delta for all x, y then abs(f(x)-f(y))< epsilon for all f(x), f(y), with epsilon equal to lim n->infinity {f(x_n)}=M and lim n->infinity {f(y_n)}

I feel fairly solid about this, but I'm finding the difference between absolute continuity and continuity a bit confusing and would like a rigor/correctness check.

Thanks.

2. Feb 23, 2009

### phreak

Here's where your proof goes wrong:

You can't just pick a delta like this. You can only pick an epsilon and then prove existence of delta using the definition of uniform continuity. Here's how I would prove it:

Choose arbitrary epsilon > 0. Then there exists delta>0 such that as long as |x-y|<delta, |f(x)-f(y)|<epsilon. Choose N so large that N delta > |M-P|. Now, choose arbitrary x0 in E. Then we must have f(x0) - epsilon N < f(x) < f(x0) + epsilon N for all x in E (check this). This proves the theorem.

3. Feb 23, 2009

### Quantumpencil

Hm, ok. So that is the same idea I think; or at least that makes complete sense and I was just being silly.

Thanks.