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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 Pn in S such that |f(Pn)| > n. The sequence {Pn} is bounded, and must therefore contain a convergent sub sequenc. Denote the subsequence by {Pn_i} and its limit by Q. Since S is closed, Q belongs to S. Then, since f is continuous at Q, f(Pn_i) converges to f(Q). This is in contradiction to |f(Pn_i)|, > ni, which has the consequence that if |f(Pn_i)| → ∞ as i→∞. To avoid the contradiction we are forced to conclude that f is bounded.
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