THEOREM: Prove that Any Bounded subset of R has a supermium and Infimum?

Let S be a bounded and closed point set, and let f be a function defined on S which is continuous at each point of S. Then the values of f are bounded, or, as we customarily say, f is bounded on S.

The proof is rather simple. Suppose the values of f were not bounded. Then no finite interval of the real number scale contains all the values, and for each positive integer n there must be at least one point P_n in S such that |f(P_n)| > n. The sequence {P_n} is bounded, and must therefore contain a convergent sub sequenc. Denote the subsequence by {P_{n_i}} and its limit by Q. Since S is closed, Q belongs to S. Then, since f is continuous at Q, f(P_{n_i}) converges to f(Q). This is in contradiction to |f(P_{n_i})|, > n_i, which has the consequence that if |f(P_{n_i})| → ∞ as i→∞. To avoid the contradiction we are forced to conclude that f is bounded.