# Prove that for any $x \in [0,1]$, $f(x)=x$

## Attempt.

I was trying to argue by contradiction. Say there is a $$x_0 \in [0,1]$$ so that $$f(x_0) \neq x_0$$. WLOG, assume $$f(x_0) < x_0$$. Now, notice that $$f \circ f$$ is the inverse of $$f$$ by definition. so $$f$$ has to be monotonic. (increasing or decreasing).

Assume it is decreasing. That is, if $$a f(b)$$.

Now, we know there is some $$\alpha \in [0,1]$$ so that $$f(\alpha) = \alpha$$. No WLOG assume $$\alpha > x_0$$, then

$$f(\alpha) < f(x_0) \implies \alpha < f(x_0) < x_0$$

• Your assumption that $f$ is decreasing is a little risky, given that $f(1) > f(0)$. – user296602 Oct 9 '18 at 23:17