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

Proof
By Subset of Real Numbers is Interval iff Connected, $\closedint a b$ is connected.

there is no fixed point.

Then $\map f a > a$ and $\map f b < b$.

Let:
 * $U = \set {x \in \closedint a b: \map f x > x}$
 * $V = \set {x \in \closedint a b: \map f x < x}$

Then $U$ and $V$ are open in $\closedint a b$.

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

By assumption:
 * $U \cup V = \closedint a b$

Thus $\closedint a b$ is not connected, which is a contradiction.

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