0

I have just started reading Brin and Stuck's book on Dynamical Systems, and I am having trouble interpreting one of the exercises. The problem reads:

Show that the complement of a forward invariant set is backward invariant, and vice versa. Show that if $f$ is bijective, then an invariant set $A$ satisfies $f^t(A) = A$ for all $t$. Show that this is false, in general, if $f$ is not bijective.

I had no trouble with the first part, but I think I must be misinterpreting the second part.

I assume that when they say "$f$ is bijective" they mean that $\forall t: f^t$ is bijective. If this is the case, it appears to me that I have a counterexample to their second claim:

Consider the dynamical system $f^t: \mathbb{R} \to \mathbb{R}$ given by $f^t(x) = x+t$. Then $f$ is bijective, and the set $A = (0, \infty)$ is forward invariant, because $f^t(A) = (t, \infty) \subseteq A$. However, we see that $f(A) \not = A$.

Could anyone please clatify what is going on? Thank you.

William Elliot
  • 17,598
Ovi
  • 23,737
  • Your set $A$ is forward invariant but the exercise mentions "invariant set". – John B Aug 12 '20 at 17:53
  • @JohnB Oh, I thought by "invariant" they meant either forward or backward invariant. Is an invariant set a set that is both forward and backward invariant? – Ovi Aug 12 '20 at 18:02
  • It is, but that's not the usual definition (please check what is their definition, although it would be unlikely that Brin would use something nonstandard). That's something else that you need to prove. – John B Aug 12 '20 at 19:05
  • @JohnB I looked at their definition, and that's actually what they have. They say that a set $A$ is $f$-invariant if $f^t(A)\subseteq A$ for all $t$. Could you please tell me the usual definition? – Ovi Aug 13 '20 at 22:17
  • You forgot to write $t\in\mathbb R$ and that's the point of the exercise. The usual definition is $f^t A=A$ for all $t\in\mathbb R$. – John B Aug 13 '20 at 22:30
  • @JohnB Thanks for the definition. In my book it doesn't say $t \in \mathbb{R}$, it just says "A subset $A \subset X$ is $f$-invariant if $f^t(A) \subset A$ for all $t$; forward $f$- invariant if $f^t(A) \subset A$ for all $t \ge 0$; and backward f-invariant if $f^{-t}(A) \subset(A)$ for all $t \ge 0$." – Ovi Aug 14 '20 at 01:12

0 Answers0