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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!

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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.

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  • $\begingroup$ Nice counter example. Would you say, however, that an equilibrium point of a continuous time system is backward invariant? $\endgroup$
    – Dina Abdelhadi
    Jan 9, 2021 at 23:11
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    $\begingroup$ @Dina: it depends somewhat delicately on what hypotheses you put on the system. If the system is the flow of a Lipschitz-continuous vector field then by the Picard-Lindelof theorem the flow exists uniquely for some open interval around $0$; in particular, time evolution can be inverted uniquely for small time, so fixed points flow back to themselves. More generally this implies that distinct integral curves of the vector field do not intersect. Without this hypothesis there are counterexamples: math.stackexchange.com/a/3001584/232 $\endgroup$
    – Qiaochu Yuan
    Jan 10, 2021 at 2:39

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