Let
- $E$ be a countable set equipped with the discrete topology and $\mathcal E:=\mathcal B(E)$
- $\pi$ be a probability measure on $(E,\mathcal E)$ with $$\pi(x):=\pi(\left\{x\right\})>0\;\;\;\text{for all }x\in E\tag1$$
- $q$ be a stochastic matrix$^1$ on $E$
Now, let $$p(x,y):=\begin{cases}q(x,y)\min\left(1,\frac{\pi(y)q(y,x)}{\pi(x)q(x,y)}\right)&\text{, if }x\ne y\text{ and }q(x,y)>0\\0&\text{, if }x\ne y\text{ and }q(x,y)=0\\1-\sum_{z\in E\setminus\left\{x\right\}}p(x,z)&\text{, if }x=y\end{cases}$$ for $x,y\in E$.
It's easy to see that $p$ is reversible$^2$ with respect to $\pi$ and hence $\pi$ is stationary$^3$ with respect to $p$.
How can we conclude that, if $q$ is irreducible$^4$ and $$\forall x,y\in E:q(x,y)>0\Leftrightarrow q(y,x)>0,\tag6$$ then $p$ is irreducible?
I'm not sure if this is a simple fact following from $(3)$ or $(4)$, or if we we need to make use of the actual shape of $p$.
$^1$ i.e. $q:E\times E\to[0,1]$ with $$\sum_{y\in E}q(x,y)=1\;\;\;\text{for all }x\in E.\tag2$$
$^2$ i.e. $$\pi(x)p(x,y)=\pi(y)p(y,x)\;\;\;\text{for all }x,y\in E.\tag3$$
$^3$ i.e. $$\pi p(y):=\sum_{x\in E}\pi(x)p(x,y)=\pi(y)\;\;\;\text{for all }y\in E.\tag4$$
$^4$ i.e. $$\forall x,y\in E:\exists n\in\mathbb N:q^n(x,y):=\sum_{z\in E}q^{n-1}(x,z)p(z,y)>0,\tag5$$ where $q^0(x,y):=1$ for all $x,y\in E$.