Proof suppose that m is the lub of f on the interval

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Proof. Suppose that M is the l.u.b. of f on the interval [ a, b ]. Then, for each n N , there exists x n [ a, b ] such that | f ( x n ) - M | < n - 1 (otherwise y would be separated from the range of f and would not be the LEAST upper bound). The sequence { x n } has a subsequence { x n k } which converges to a limit c [ a, b ] as k → ∞ . Since f is continuous, we have f ( c ) = lim k →∞ f ( x n k ) = M . Similarly, considering the g.l.b. of f (or applying the above theorem to the function g ( x ) = - f ( x )), we obtain Minimum Theorem. If f : [ a, b ] R is continuous function on the closed bounded interval [ a, b ] then it attains a minimum value. Intermediate Value Theorem. If f : [ a, b ] R is continuous function on the closed bounded interval [ a, b ] and f ( a ) 6 d 6 f ( b ), then there exists c [ a, b ] such that f ( c ) = d . Proof. We proceed by subdividing the interval into halves repeatedly, starting with [ a 1 , b 1 ] = [ a, b ]. The idea in to construct intervals [ a n , b n ] such that [ a n +1 , b n +1 ] [ a n , b n ] for all n and b n +1 - a n +1 = ( b n - a n ) / 2 and f ( a n ) 6 d 6 f ( b n ) for all n . This is done inductively. If it has been done for some n then we make the next subdivision as follows. If f (( a n + b n ) / 2) > d then put [ a n +1 , b n +1 ] = [ a n , ( a n + b n ) / 2]. Otherwise f (( a n + b n ) / 2) < d and we put [ a n +1 , b n +1 ] = [( a n + b n ) / 2 , b n ]. Now, by the nested intervals theorem, there exists c [ a, b ] such that a n c and b n c . Also, we have f ( a n ) 6 d 6 f ( b n ) for all n . Using the continuity of f at c , we conclude that f ( c ) = lim n →∞ f ( a n ) 6 d 6 lim n →∞ f ( b n ) 6 f ( c ), so f ( c ) = d .
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