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!
Best Answer
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.