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given dynamic system $(X, \mathcal{B}, F, \mu), \mu \circ F^{-1}=\mu, F $ is mixing, $ A \in \mathcal{B}, s.t. \mu(A) >0 $.

consider dynamic system $(X\times X, \mathcal{B}\otimes \mathcal{B}, F\times F, \mu \times \mu)$, then $ F \times F $ is mixing too, hence ergodic, $(\mu \times \mu )(A \times A )>0$.

fixed delay time $n_0 \in \mathbb{N}$, define $R(x)=\min \{n: F^n(x) \in A \} $, and stopping times recursively: for all $ (x, x') \in X \times X$,

$ \tau_1(x,x')= n_0 +R(F^{n_0}x) $

$ \tau_2(x,x')= \tau_1+n_0 +R(F^{n_0+\tau_1}x') $

$ \tau_3(x,x')= \tau_2+n_0 +R(F^{n_0+\tau_2}x) $

$ \tau_4(x,x')= \tau_3+n_0 +R(F^{n_0+\tau_3}x') $ ect.

then we know $ F^{\tau_1} \in A \times X, F^{\tau_2} \in X \times A$ ect.

can we prove:

for almost $ (x, x') $ wrt $ \mu\times \mu$, $ \exists \tau_n=\tau_n(x,x'), s.t. F^{\tau_n} (x,x') \in A \times A$? Thanks!

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Here is a revised answer, having correctly understood the question.

Let $A_1=A$ and $A_0=A^c$. For a sequence $\mathbf w=w_0,\ldots,w_{k-1}$, set $A_{\mathbf w}=A_{w_0}\cap F^{-1}A_{w_1}\cap\ldots\cap F^{-(k-1)}A_{w_{k-1}}$. That is, $A_{\mathbf w}$ is the set of points that is inside or outside of $A$ for $k$ steps according to $\mathbf w$.

Let $S=\{\mathbf w\colon |\mathbf w|=n_0+1;\, w_{n_0}=1\}$. Notice that $\bigcup_{\mathbf w\in S}A_{\mathbf w}=F^{-n_0}A$, so that there exists $\mathbf w\in S$ with $\mu(A_{\mathbf w})>0$. Now $\mu\times\mu$-almost every pair $(x,x')$ visits $A_{\mathbf w}\times A_{\mathbf w}$ infinitely often. On each visit, there is at least one common return to $A$. To see this, if $(F^mx,F^mx')\in A_{\mathbf w}\times A_{\mathbf w}$, then let $\tau_j<m$ be the last term in the sequence of $\tau$'s that is smaller than $m$. Then $\tau_j+{n_0}$ is inside the block $A_{\mathbf w}\times A_{\mathbf w}$ that starts at $m$, so that the first occurrence of $A$ is common.

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  • $\begingroup$ However, let's let $ N_1, N_2, ….$ be simultaneous time entering $ A$. Let's look at $\tau_n $, it is possible to have $ \exists m, N_m < \tau_n + n_0 <N_{m+1}$ , and $ N_m < \tau_{n+1}=\tau_n + n_0+ R(F^{\tau_n + n_0}) <N_{m+1}$, $ N_{m+1} < \tau_{n+1} +n_0 <N_{m+2}$ i.e $ n_0$ helps them to skip all $ (N_m)_{m \ge 1}$, then a common entry will not appear in either sequence? $\endgroup$
    – jason
    Commented Sep 9, 2018 at 3:45
  • $\begingroup$ if $n_0=0$, it does not cause such problem, just like you said, a common entry will appear in either sequence. $\endgroup$
    – jason
    Commented Sep 9, 2018 at 4:11
  • $\begingroup$ I see. You’re right - that’s more interesting. $\endgroup$ Commented Sep 9, 2018 at 6:22

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