MT2002 Analysis

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## The boundedness theorem

This result explains why closed bounded intervals have nicer properties than other ones.

Theorem

A continuous function on a closed bounded interval is bounded and attains its bounds.

Proof
Suppose f is defined and continuous at every point of the interval [a, b]. Then if f were not bounded above, we could find a point x1 with f (x1) > 1, a point x2 with f (x2) > 2, ...
Now look at the sequence (xn). By the Bolzano-Weierstrass theorem, it has a subsequence (xij) which converges to a point α ∈ [a, b]. By our construction the sequence (f (xij)) is unbounded, but by the continuity of f, this sequence should converge to f (α) and we have a contradiction.
The proof that f is bounded below is similar.

To show that f attains its bounds, take M to be the least upper bound of the set X = { f (x) | x ∈ [a, b] }. We need to find a point β ∈ [a, b] with f (β) = M . To do this we construct a sequence in the following way:
For each nN, let xn be a point for which | M - f (xn) | < 1/n. Such a point must exist otherwise M - 1/n would be an upper bound of X. Some subsequence of (x1 , x2 , ... ) converges to β (say) and (f (x1) , f (x2) , ... )→ M and by continuity f (β) = M as required.
The proof that f attains its lower bound is similar.

Here are some examples to show why you must have a closed bounded interval for this result to work.

1. Interval not closed
The function f: (0, 1]→ R defined by f (x) = 1 /x is continuous but not bounded.
The function f: [0, 1)→ R defined by f (x) = x is continuous and bounded but does not attain its least upper bound of 1.

2. Interval not bounded
The function f: [0, ∞)→ R defined by f (x) = x is continuous but not bounded.
The function f: [0, ∞)→ R defined by f (x) = x/(1+x) is continuous and bounded but does not attain its least upper bound of 1.

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JOC September 2001