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 [tex]\epsilon > 0[/tex] and points x, z in [a, b] arbitrarily close to each other for which [tex]|f(x) - f(z)| \ge \epsilon[/tex]. It would then be possible for every n to choose points [tex]x_n, z_n [/tex] in [a, b] for which [tex]|f(x_n) - f(z_n)| \ge \epsilon[/tex] and [tex]|x_n - z_n| < 1/n[/tex]. Since the [tex]x_n[/tex] 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 [tex]z_n[/tex] would then also converge to c: since f is continuous at c, we would find that [tex]c = lim f(x_n) = lim f(z_n) [/tex] for n tending to infinity in the subsequence, which is impossible if [tex]|f(x_n) - f(z_n)| \ge \epsilon[/tex] for all n." I understand everything except one thing: Why is c equal to the limit of [tex]f(x_n)[/tex]? If you cut out that part and just say that since [tex]x_n[/tex] and [tex]z_n[/tex] both converge to c, then [tex]lim f(x_n) = lim f(z_n)[/tex] and there we have our contradiction.