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This question is within the context of my course on Dynamical Systems (book Introduction to Dynamical Systems, by Brin and Stuck).

Let $X$ be a non empty set and consider the map $f:X \to X$. For $n \in \mathbb{N}_{0}=\mathbb{N} \bigcup \left\{ 0 \right\}$, $f^{n}$ represents the $n$-fold composition, i.e., $f^{n}=f\circ \cdots \circ f$ (n times).

Given a subset $A \subset X$, we say it is $f$-invariant if $f^{n}(A) \subset A$ for all $n \in \mathbb{N}_{0}$.

I need to show that, if $f$ is a bijection, then the $f$-invariant set satisfies $f^{n}(A)=A$.

So far, given that $f$ is a bijection, I have been able to show that $f^{-n}(A^{c}) \subset A^{c}$ for all $n \in \mathbb{N}_{0}$, where $f^{-n}=f^{-1}\circ \cdots \circ f^{-1}$ (n times), but I haven't been able to go further than this to show that $A \subset f^{n}(A)$.

Any hint?

CA-Math
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  • Is $X$ finite? If not, a counterexample is $f(x)=x/2$, $A=[-1,1]$. –  Jan 29 '18 at 09:57
  • I think the condition of $f$-invariance is $f^t(A)\subseteq A,\forall t\in\mathbb Z$. Otherwise @Rahul provides a counterexample. – awllower Jan 29 '18 at 10:18
  • Hi guys, thank you very much (@Rahul) for pointing this out. I base my question on what I've read so far in the book I mentioned, where they are focusing on discrete-time dynamical systems, and where in particular $t$ can be in either $\mathbb{Z}$ or $\mathbb{N}$. Hence I was trying to prove for the latter set. @Rahul: If we restrict $X$ to be finite, then the claim can be proven? – CA-Math Jan 29 '18 at 10:29

1 Answers1

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I checked the book you mentioned. The definition of $f$-invariance is $$f^t(A)\subseteq A,$$ for any $t$ in the appropriate range. And when $f$ is bijective, then the range ought to be $\mathbb Z$. Hence the condition is $f(A)\subseteq A$ and $f^{-1}(A)\subseteq A$. From these two conditions we deduce easily that $f(A)=A$.

If we consider $t$ in $\mathbb N$, then $A$ should be finite for the statement to hold. If $A$ is not finite, then a counter-example is given by the comment of @Rahul. If $f(A)\subseteq A$ and $A$ is finite, then the statement also holds: as $f$ is bijective, the two sets $A$ and $f(A)$ have the same (finite) cardinality. Since two finite sets one of which is contained in another must be equal, $f(A)=A$.

Hope this helps.

awllower
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