MT2002 Analysis

 Previous page (The intermediate value theorem) Contents Next page (Images of intervals)

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 [a, b]. By our construction the sequence (f(xij)) is unbounded, but by the continuity of f, this sequence should converge to f() 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 [a, b] with f() = M . To do this we construct a sequence in the following way:
For each n N, 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 (say) and (f(x1) , f(x2) , ... ) M and by continuity f() = 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.

 Previous page (The intermediate value theorem) Contents Next page (Images of intervals)

JOC September 2002