Uniform Continuity Proof for Functions on Closed Intervals

[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.

 

[jostpuur]

Shouldn't it be

<br /> \lim_{n\to\infty} f(x_n) = f(c) = \lim_{n\to\infty} f(z_n)?<br />

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

 

[JG89]

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

 

[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.

 

[rochfor1]

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