Continuous decreasing function has a fixed point

Note that if $d$ is a fixed point of $f\circ f$ but not of $f$, then so is $f(d)$. So you can make pairs of fixed points.

Hint: Suppose $h$ only has finitely many fixed points. Call the set of fixed points $A$. Where does $f$ map $A$? (And what does uniqueness of the fixed point of $f$ then tell you?)

Added: Since the problem appears to be solved, here's how to complete the solution: define $$\begin{align}A_<&=\{x\in A\mid x<f(x)\},\\A_=&=\{x\in A\mid x=f(x)\},\\A_>&=\{x\in A\mid x>f(x)\}.\end{align}$$ Note that these three sets are disjoint and $A=A_<\cup A_=\cup A_>$. Note that $A_=$ consists of a single element $c$, the fixed point of $f$. Also note that $f$ maps $A_<$ bijectively onto $A_>$.

So $|A_<|=|A_>|=:k$ and $|A_=|=1$. We conclude that $|A|=2k+1$, which is an odd number.

Hint:

• Let $d = f(f(d))$.
• If $d$ is a fix-point of $f$, then there is only one such $d$.
• Otherwise $d \neq f(d)$; let $\hat{d} = f(d)$, then $$f(f(\hat{d})) = f(f(f(d))) = f(d) = \hat{d},$$ that is, $\hat{d}$ is another fix-point of $(f\circ f)$.

I hope this helps $\ddot\smile$