Example the function f x x 1 is continuous on the

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Example. The function f ( x ) = x - 1 is continuous on the closed interval [1 , ) but does not take the minimum value 0. Recall that intervals of the form ( a, b ), ( -∞ , b ), ( a, + ) and ( -∞ , + ) are said to be open. Bounded intervals of the form [ a, b ] are said to be closed. Warning: from the topological point of view, the unbounded interval ( -∞ , + ) is also closed. The following four theorems are only true for functions that are continuous on closed and bounded intervals. 1
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Theorem. If f : [ a, b ] R is continuous function on the closed bounded interval [ a, b ] then f is bounded. Proof. If f were unbounded then there would exist a sequence x n [ a, b ] such that | f ( x n ) | > n for all n . Since the sequence { x n } is bounded, by The Bolzano- Weierstrass Theorem it has a subsequence { x n k } which converges to a limit c as k → ∞ . Note that the limit c also belongs to the interval [ a, b ]. Indeed, if d 6∈ ( c - ε, c + ε ) then d < a or d > b . In either case there exists an open interval ( d - ε, d + ε ) about d which does not contain any points of the interval [ a, b ]. In particular, x n k 6∈ ( d - ε, d + ε ), which shows that the subsequence { x n k } cannot converge to such a point d . Since f is continuous, we must have f ( x n k ) f ( c ) as k → ∞ . However, f ( c ) is a finite number and | f ( x n k ) | > n k → ∞ as k → ∞ . The obtained contradiction proves the theorem. Maximum Theorem. If f : [ a, b ] R is continuous function on the closed bounded interval [ a, b ] then it attains a maximum value.
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