Brouwer's Fixed Point Theorem/One-Dimensional Version

Theorem
Let $f: \closedint a b \to \closedint a b$ be a real function which is continuous on the closed interval $\closedint a b$.

Then:
 * $\exists \xi \in \closedint a b: \map f \xi = \xi$

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 $\closedint a b$, it follows that the image of $f$ is a subset of $\closedint a b$.

Thus $\map f a \ge a$ and $\map f b \le b$.

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

Then by the Combined Sum Rule for Continuous Functions, $\map g x$ is continuous on $\closedint a b$.

But $\map g a \ge 0$ and $\map g b \le 0$.

By the Intermediate Value Theorem, $\exists \xi: \map g \xi = 0$.

Thus $\map f \xi = \xi$.