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Let $T\colon X\to X$ and $Y:=\bigcap_{n=1}^{\infty}T^nX$.

Do we have $T\colon Y\to Y$? I think no

$$ T(Y)\subset\bigcap_{n=2}^{\infty}T^nX\not\subset \bigcap_{n=1}^{\infty}T^nX=Y. $$

Conversely, we have $Y\subset \bigcap_{n\geq 2}T^nX$.

A bit confused since it is said that there is some function from Y to Y and this function is even called T.

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    Use that $T X\subseteq X$ and apply $T$ on both sides. – 2017-01-28
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    Oh you are right... so indeed $T(Y)\subset Y$ hence $T_{|Y}\colon Y\to Y$ is allright. – 2017-01-28

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It is not so trivial after all. In general you need not have $TY=Y$. ($TY\subset Y$ is trivially true).

Example: We will define $X$ as a subset of ${\Bbb N}_0^2$ and a dynamical system on it.

Set $\Omega_0=\{(0,0),(0,1),...\}$ and for $n\geq 1 : \Omega_n = \{ (n,1),...,(n,n)\}$. We let $X=\bigcup_{n\geq 0} \Omega_n$. The dynamics is as follows:

$T(0,k)=(0,k+1)$ (a right shift) and for $2\leq k\leq n : T(n,k)=(n,k-1)$ (a left shift at level $n$). Finally $T(n,1)=(0,0)$, $n\geq 1$.

Then $Y=\bigcap T^n X=\Omega_0$ but $TY$ does not contain the point $(0,0)\in Y$.

Perhaps an intuitive explanantion of what is going on may help: $T$ defines a right shift on $\Omega_0$. Now, for every $n\geq 1$ the chain $\Omega_n$ is nilpotent (becomes empty after $n+1$ iterations), so will not belong to $Y$. On the other hand $(0,0)$ is in the image of $T^n X$ for every $n\geq 1$, so $Y=\Omega_0$. But $(0,0)$ has no preimage in $Y$.

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    My question only was if it is correct to say $T\colon Y\to Y$ and I guess thats okay since $T(Y)\subset Y$. Or I guess what was really meant is if we restrict $T$ to $Y$ then $T_{|Y}\colon Y\to Y$ (since originally, T is defined on X). – 2017-01-28
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    Ok, but in the beginning there was a question also about $TY=Y$ which I found intriguing. – 2017-01-28
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    You are right. And thank you very much for your great example. – 2017-01-28