# Proof of Uniform Continuity

1. Nov 24, 2008

### JG89

From my textbook, this is the proof given for a theorem stating that any function continuous in a closed interval is automatically uniformly continuous in that interval.

Proof: "If f were not uniformly continuous in [a, b] there would exist a fixed $$\epsilon > 0$$ and points x, z in [a, b] arbitrarily close to each other for which $$|f(x) - f(z)| \ge \epsilon$$. It would then be possible for every n to choose points $$x_n, z_n$$ in [a, b] for which $$|f(x_n) - f(z_n)| \ge \epsilon$$ and $$|x_n - z_n| < 1/n$$. Since the $$x_n$$ form a bounded sequence of numbers we could find a subsequence converging to a point c of the interval (using the compactness of closed intervals). The corresponding values $$z_n$$ would then also converge to c: since f is continuous at c, we would find that $$c = lim f(x_n) = lim f(z_n)$$ for n tending to infinity in the subsequence, which is impossible if $$|f(x_n) - f(z_n)| \ge \epsilon$$ for all n."

I understand everything except one thing: Why is c equal to the limit of $$f(x_n)$$? If you cut out that part and just say that since $$x_n$$ and $$z_n$$ both converge to c, then $$lim f(x_n) = lim f(z_n)$$ and there we have our contradiction.

Last edited: Nov 24, 2008
2. Nov 24, 2008

### jostpuur

Shouldn't it be

$$\lim_{n\to\infty} f(x_n) = f(c) = \lim_{n\to\infty} f(z_n)?$$

Looks like a mistake where c and f(c) are confused.

3. Nov 24, 2008

### JG89

That's EXACTLY what I was thinking. Glad to see I wasn't the only one thinking that :)

4. Nov 24, 2008

### mathwonk

given any open cover of a closed bounded interval, there is a positive number e such that every interval of diameter e is entirely contained in one of the open intervals. qed.

5. Nov 28, 2008

### rochfor1

Hopefully that interval is bounded too...otherwise things could get nasty.