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 (x_n). By the Bolzano-Weierstrass theorem, it has a subsequence (x_ij) which converges to a point [a, b]. By our construction the sequence (f(x_ij)) 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 x_n be a point for which | M - f(x_n) | < 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.

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.