Uniform Continuity Theorem

1. Feb 11, 2009

DMN

I'm having some trouble understanding the proof for uniform continuity. I'm using the book Introduction to Real Analysis by Bartle and Sherbert 3rd Edition, page 138, if anyone has access to it. The Theorem states:

I understand the proof up to the part where it says it is clear that $$(u_n_k)$$ also converges to z. I dont know why showing this inequality shows that the subsequence converges to z. Also why is the difference between the two sequences less then 1/n?

2. Feb 11, 2009

phreak

Let $$(x_{n_k})$$ converge to z. Then for each $$\epsilon>0$$, there exists a K such that $$|x_{n_k}-z| < \epsilon/2$$ for all k>K, by definition of convergence. Now, by assumption, we have that $$|x_n - u_n| < 1/n$$. Therefore, we may pick M>K such that $$|x_{n_k} - u_{n_k}| < \epsilon/2$$ for all k>M. This implies that for each $$\epsilon>0$$, there exists N (=M) such that $$|u_{n_k}-z| \le |u_{n_k} - x_{n_k}| + |x_{n_k}-z| < \epsilon/2 + \epsilon/2 = \epsilon$$. It's rather intuitive, if you think about it hard enough. If two sequences get arbitrarily close to each other as $$n\to \infty$$, and moreover, one of the sequences converges to an element z, then the other sequence will as well.

The reason that the distance between the two sequences is less than 1/n is because that's how we make them (i.e., it's an assumption). We assume that f is NOT uniformly continuous. Uniform continuity means that for every $$\epsilon > 0$$, and every two sequences $$x_n, u_n$$, we can find an N such that $$|x_n - u_n| < 1/n$$ implies $$|f(x_n)-f(u_n)|< \epsilon$$ for all n>N. The converse of this statement (that is, NON-uniform continuity), says that we CAN find an $$\epsilon>0$$ such that no matter how close the sequences get, the differences of their images under f are still going to be greater than $$\epsilon$$. The 1/n measures how close the two sequences get, and we want to see what happens when they get arbitrarily close.

Last edited: Feb 11, 2009