Question

What is the boundedness theorem?

Answer 1

A continuous function defined on a closed interval has an upper (and lower) bound.

Explanation:

Probably the simplest boundedness theorem states that a continuous function defined on a closed interval has an upper (and lower) bound.

Proof by contradiction

Suppose `f(x)` is defined and continuous on a closed interval `[a, b]`, but has no upper bound.

Then:

`AA n in NN, EE x_n in [a, b] : f(x_n) > n`

Since the sequence of `x_n`'s lies in a bounded interval, it is dense at some point in the closure of the interval. Since the interval is closed, that must be at some point `c` actually in the interval `[a, b]`.

Since the sequence of `x_n`'s is dense at `c`, there is some monotonically increasing sequence `n_k in NN` such that `x_(n_k) -> c` as `k->oo`.

Now `f(x)` is continuous at `c`, so:

`lim_(x->c) f(x) = f(c)`

which is bounded.

But:

`lim_(k->oo) x_(n_k) = c" "` and `" "lim_(k->oo) f(x_(n_k)) = oo`

is unbounded.

...contradiction.

So there is no such `f(x)` lacking upper (or lower) bound.