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

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## 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

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

Aiming for a contradiction, suppose 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.

$\blacksquare$