Why are fixed points of a discrete time system backward-invariant?

Question

Consider a discrete time system given by $x(t+1) = F(x(t))$.

A backward invariant set $S$ is one such that if the solution $x(t)\in S$ at time $T$, for all $t<T$, $x(t) \in S$.

I see why a fixed point is forward invariant, but why is it backward invariant? Is there proof that it cannot be reached from another point?

Thanks in advance!

Answer 1

This is false. For example, take $F(x) : \mathbb{R} \to \mathbb{R}$ to be the function $\text{max}(x - 1, 0)$; in other words, $F(x)$ subtracts $1$ until we get a number less than or equal to $0$ and then just outputs $0$. The unique fixed point of this function is $\{ 0 \}$, and there are trajectories such as

$$x(t) = \begin{cases} -t & \text{ if } t \le 0 \\ 0 & \text{ otherwise} \end{cases}$$

showing that $0$ is not backwards invariant (here I'm assuming $t \in \mathbb{Z}$).

It is, of course, true if $F$ is assumed to be invertible, since then $x(t) = F^{-1}(x(t+1))$, and $F(x) = x$ implies $F^{-1}(x) = x$, so we can run time evolution backwards uniquely in this case.