# Forward orbits and backwards orbits same closures

Assume that $T: X \to X$ is a homeomorphism of a compact metric space to itself, and let $\mu$ be a $T$-invariant Borel probability measure on $X$. I want to show that for almost every $x \in X$ the forward and backwards orbits of $x$ have the same closures, that is $$\overline{\{T^n x : n \geq 1 \}} = \overline{\{T^{-n} x : n \geq 1\}}$$ for almost every $x$ and that the closures contain $x$.

So, I first want to start with the last statement, $x \in \overline{\{T^n x : n \geq 1 \}}$.

So, $X$ is totally bounded. Take $\epsilon_n = \dfrac1{2^n}$ so and the take one of the cover sets $B_1(x, \epsilon_1)$, we know now by Poincaré recurrence that there exists a subset $F_1$ of $B_1$ such that for $x_1 \in F_1$, $T^{n^1_k} x_1$ always lands in $F_1$. By compactness, passing to a subsequence if necessary, $T^{n^1_k} x_1$ converges to $y_1 \in \overline F_1$. $\overline F_1$ is compact and closed, hence totally bounded, so again I can find a cover with $\epsilon_2$ as the ball radius and I can pick the first ball and $x_2\in F_2$. So then I have $T^{n^2_k} x_2\to y_2$, if I continue like this (needs verification) I think I can find $x \in \bigcap \overline F_n$ (Cantor nesting lemma) and then $y_n$ should converge to $x$. Since I only did all of this countably many times the union should give me a set of full measure. (For the sequence I take the diagonal sequence)

This seems to me overly complicated and a bit fishy so I would like to know if I'm on the right track, am I missing something clear?

This is a homework assignment, so I would appreciate it if I only get hints.

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First, cover $X$ with finitely balls of radius $1$: $B_{1,1},\ldots ,B_{1,m_1}$. Then, by the Poincaré Recurrence Theorem, for each $B_{1,k}$, almost every $x\in B_{1,k}$ is such that $T^j(x)\in B_{1,k}$ for infinitely many positive integers $j$. Hence, for almost every $x\in X$, there is some $k$ such that $x\in B_{1,k}$ and $T^j(x)\in B_{1,k}$ for infinitely many positive integers $j$. Proceeding as you do, we see that there is an $x_1\in \overline{B_{1,k}}$ such that $x_1\in \overline{\left\{ T^n(x)|\, n\geq 1\right\}}$. Then, just as you said, we can cover $\overline{B_{1,k}}$ with finitely many balls of radius $1/2$: $B_{2,1},\ldots ,B_{2,m_2}$. Proceeding as before, we see that, for almost every $x$ for which we can find such an $x_1$, we can obtain an element $x_2$ contained in $\overline{\left\{ T^n(x)|\, n\geq 1\right\}}$ within a distance of $1/2$ from $x$. Proceeding inductively, we may construct $x_n$ contained in $\overline{\left\{ T^n(x)|\, n\geq 1\right\}}$ within a distance $1/n$ of $x$ for almost every $x$ for which we can find such an $x_{n-1}$. Thus, for almost every $x\in X$, we can construct such an $x_n$ for all $n\geq 1$. As this sequence converges to $x$, it follows that $x\in \overline{\left\{ T^n(x)|\, n\geq 1\right\}}$ for almost every $x$.