I have a Markov Chain $(X_n)$ that map the elements in a set $E$. We have the following relation of communicate,
$\forall i,j\in E$, i $communicate$ with j if $\exists$ $n\ge1$ such that $P( X_n = j | X_0 = i ) > 0$.
We assume that every element is recurrent.
It is well known that this is an equivalence relation, but I am not able to prove mathematically that the symmetric relation hold (intuitively it is ok).
Please, can anybody help me to prove it formally?
EDIT: I'm wrong, I have defined the relation 'lead to'. I want to prove that if we have just recurrent states, then we have an equivalence relation.
You have got the definition wrong. The property you have defined is '$i$ leads to $j$'. we say $i$ and $j$ communicate with each other if $i$ leads to $j$ and $j$ leads to $i$.
The relation you have defined is not symetric. For example, if $j$ is an absorbing state then it is possible that $i$ leads to $j$ but $j$ cannot lead to $i$.
Let $\{X_n\}$ be the MC and denote the transition matrix by $P$. Let $p_{ij}^{n}$ denote the $(i,j)$ element of $P^{n}$. Let $k$ be a positive integer and consider $P(X_k=j,X_n = i \, \text {for some} \, n>k|X_0=i)$. By Markov property this probability does not exceed $\sum_n p_{ij}^{k}P(X_n=i \, \text {for some} \, n|X_0=j)$. Suppose $j$ does not lead to $i$. Then $P(X_n=i \, \text {for some} \, n|X_0=j)=0$ for all $n$. Hence we get $P(X_k=j,X_n = i \, \text {for some} \, n>k|X_0=i)=0$. But recurrence of $i$ implies $P(X_n = i \, \text {for some} \, n>k|X_0=i)=1$. [ The chain returns to recurrent states infinitely many times with probaility $1$]. We can now conclude that $P(X_k=j|X_0=i)=0$ for each $k$. Thus $i$ does not lead to $j$.