Brouwer's Fixed Point Theorem/One-Dimensional Version/Proof Using Connectedness

Proof
By Subset of Real Numbers is Interval iff Connected, $\left[{a \,.\,.\, b}\right]$ is connected.

there is no fixed point.

Then $f \left({a}\right) > a$ and $f \left({b}\right) < b$.

Let:
 * $U = \left\{ {x \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) > x}\right\}$
 * $V = \left\{ {x \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) < x}\right\}$

Then $U$ and $V$ are open in $\left[{a \,.\,.\, b}\right]$.

Because $a \in U$ and $b\in V$, $U$ and $V$ are non-empty.

By assumption:
 * $U \cup V = \left[{a \,.\,.\, b}\right]$

Thus $\left[{a \,.\,.\, b}\right]$ is not connected, which is a contradiction.

Thus, by Proof by Contradiction, there exists at least one fixed point.