If $x$ is a recurrent state of a discrete Markov chain and the probability to go from $x$ to $y$ is positive, then $y$ is recurrent Let


*

*$E$ be an at most countable Polish space and $\mathcal E$ be the Borel $\sigma$-algebra on $E$

*$X=(X_n)_{n\in\mathbb N_0}$ be a Markov chain with values $(E,\mathcal E)$, distributions $(\operatorname P_x)_{x\in E}$ and transition matrix $$p=\left(p(x,y)\right)_{x,y\in E}=\left(\operatorname P_x\left[X_1=y\right]\right)_{x,y\in E}$$

*$\tau_x^0:=0$ and $$\tau_x^k:=\inf\left\{n>\tau_x^{k-1}:X_n=x\right\}$$ for $x\in E$ and $k\in\mathbb N$ and $$\varrho(x,y):=\operatorname P_x\left[\tau_y^1<\infty\right]\color{blue}{=\operatorname P_x\left[\exists n\in\mathbb N:X_n=y\right]}$$


Suppose $x\in E$ is a recurrent state, i.e. $$\varrho(x,x)=1\;.$$ Let $y\in E$ with $\varrho(x,y)>0$ .

Clearly, since $\varrho(x,y)>0$, there exists a sequence $x_1,\ldots,x_n\in E$ with $x_n=y$ and $$\operatorname P\left[X_i=x_i\text{ for all }i\in\mathbb N\right]>0\;.$$ Moreover,
\begin{equation}
\begin{split}
1-\varrho(x,x)&=\operatorname P_x\left[\tau_x^1=\infty\right]\\
&\ge\operatorname P_x\left[X_1=x_1,\ldots,X_n=x_n\text{ and }\tau_x^1=\infty\right]\;,
\end{split}\tag 1
\end{equation}
but I don't understand why the last term is equal to $$\operatorname P_x\left[X_1=x_1,\ldots,X_n=x_n\right]\operatorname P_y\left[\tau_x^1=\infty\right]\;.\tag 2$$

Some authors state, that $(2)$ holds by "the" (elementary? weak? strong?) Markov property, but honestly, I don't see that.
As another important note: They assume $x_i\ne x$. That might be crucial for $(2)$, but that's the next problem: I don't understand why we can choose $n$ such that $x_i\ne x$. I've seen that authors define $n$ to be the infimum of $$\left\{n\in\mathbb N:\operatorname P_x\left[X_n=y\right]>0\right\}\;,$$ but that doesn't seem to guarantee, that $X$ doesn't return a number of time to $x$ before it goes to $y$.
EDIT:$\;\;\;$The last term in $(1)$ is equal to $$\operatorname P_x\left[\tau_x^1=\infty\mid X_1=x_1,\ldots,X_n=x_n\right]\operatorname P_x\left[X_1=x_1,\ldots,X_n=x_n\right]\;.$$ Intuitively, by the memorylessness of $X$, $\operatorname P_x\left[\tau_x^1=\infty\mid X_1=x_1,\ldots,X_n=x_n\right]$ should be equal to $\operatorname P_x\left[\tau_x^1=\infty\mid X_n=x_n\right]$ and since $x_n=y$, this probability should be equal to $\operatorname P_y\left[\tau_x^1=\infty\right]$.
How can we verify this formally?
 A: Let $$A:=\left\{X_1=x_1,\ldots,X_n=x_n\right\}\;.$$ As Calvin Khor pointed out, we can assume, that $x_i\ne x$. Thus, $$\left\{\tau_x^1<\infty\right\}\cap A=\underbrace{\left\{\exists k>n:X_k=x\right\}}_{=:B}\cap A\;.$$ Let $$\tilde X:=\left(X_{k+n}\right)_{k\in\mathbb N_0}$$ and $f$ be the indicator function of $\bigcup_{n\in\mathbb N}\left\{\pi_n=y\right\}$, where $\pi_n:E^{\mathbb N_0}\to E$ denotes the $n$-th coordinate map. Then, $$1_B=f\circ\tilde X\;.$$ So, we're able to conclude, that
\begin{equation}
\begin{split}
\operatorname P_x\left[\tau_x^1<\infty\mid A\right]&=\operatorname P_x\left[B\mid\ A\right]\\
&=\operatorname E_x\left[f\circ\tilde X\mid\ A\right]\\
&\stackrel{(3)}=\operatorname E_x\left[\operatorname E_x\left[f\circ\tilde X\mid\ \sigma(X_0,\ldots,X_n)\right]\mid\ A\right]\\
&\stackrel{(4)}=\operatorname E_x\left[\operatorname E_{X_n}\left[f\circ X\right]\mid\ A\right]\\
&=\operatorname E_x\left[\operatorname P_{X_n}\left[\tau_x^1<\infty\right]\mid\ A\right]\\
&\stackrel{(5)}=\operatorname E_x\left[\operatorname P_y\left[\tau_x^1<\infty\right]\mid\ A\right]\\
&=\operatorname P_y\left[\tau_x^1<\infty\right]\color{blue}{\operatorname E_x\left[1\mid\ A\right]}\;,
\end{split}
\end{equation}
where I've used the tower property in $(3)$, the weak Markov property in $(4)$ and $\left\{X_n=y\right\}\subseteq A$ in $(5)$. Since $$\operatorname P_x\left[\tau_x^1=\infty\mid A\right]=1-\operatorname P_x\left[\tau_x^1<\infty\mid A\right]\;$$ we should be done.
A: I believe this is what they had in mind. For brevity $A_n:=[X_1=x_1,…,X_n = x_n]$.
\begin{align}P_x(τ_x^1 = ∞ ∩ A) &= P_x(τ_x^1 = ∞  |A) P(A) \\&\overset{\star}{=} P_x(τ_x^1 = ∞ |X_n = y)P(A) \\&= P_y(τ_x^1 = ∞)P(A)
\end{align}
Where $\star$ is where the markov property is needed. I'm not sure how to fully justify this.
Regarding the choice of $n$: if $x_i=x$ for some $i$, why don't we just terminate the sequence then? 
