Brouwer's Fixed Point Theorem/One-Dimensional Version

Theorem
Let $f: \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Then:
 * $\exists \xi \in \left[{a \,.\,.\, b}\right]: f \left({\xi}\right) = \xi$

That is, a continuous real function from a closed real interval to itself fixes some point of that interval.

Proof
As the codomain of $f$ is $\left[{a \,.\,.\, b}\right]$, it follows that the image of $f$ is a subset of $\left[{a \,.\,.\, b}\right]$.

Thus $f \left({a}\right) \ge a$ and $f \left({b}\right) \le b$.

Let us define the real function $g: \left[{a \,.\,.\, b}\right] \to \R$ by $g \left({x}\right) = f \left({x}\right) - x$.

Then by the Combined Sum Rule for Continuous Functions, $g \left({x}\right)$ is continuous on $\left[{a \,.\,.\, b}\right]$.

But $g \left({a}\right) \ge 0$ and $g \left({b}\right) \le 0$.

By the Intermediate Value Theorem, $\exists \xi: g \left({\xi}\right) = 0$.

Thus $f \left({\xi}\right) = \xi$.