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Suppose a function $f : \mathbb{R}\rightarrow\mathbb{R}$ satisfies $f(f(f(x)))=x$ for all $x$ belonging to $\mathbb{R}$. Show that

(a) $f$ is one-to-one.

(b) $f$ cannot be strictly decreasing, and

(c) if $f$ is strictly increasing, then $f(x)=x$ for all $x$ belonging to $\mathbb{R}$.

Now, I've done the first part(the simplest that is) and I need someone to shed some light on the next two.

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(b) Suppose that $f$ is strictly decreasing, then $0 < 1 \Rightarrow f(0) > f(1) \Rightarrow f(f(0)) < f(f(1)) \Rightarrow f(f(f(0))) = 0 > 1 = f(f(f(1)))$. Contradiction.

(c) Suppose that $f$ is strictly increasing. Let $x \in \mathbb{R}$.

  • Suppose $f(x) > x$. Then $f(f(x)) > f(x) \Rightarrow f(f(f(x))) > f(f(x)) \Rightarrow x > f(f(x)) > f(x)$. This is absurd.

  • You can prove in the exact same way that $f(x) < x$ is absurd. Therefore $f(x) = x$.

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