Let me consider the matrix $$ M=\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$ We consider the associated markov shift $S:\Omega_M\rightarrow \Omega_M$ where $\Omega_M=\{\omega\in \{0,1,2,3\}^\Bbb{Z}: M(\omega_i,\omega_{i+1})=1~\forall i\in \Bbb{Z}\}$. I want to check if this markov shift is topologically mixing or not.
My claim is that it is not topologically mixing.
We say that $T:X\rightarrow X$ is topologically mixing if for all $U,V\subset X$ open nonempty, there exists $N$ s.t. for all $n\geq N$, $T^{-n}(U)\cap V\neq \emptyset$.
I wanted to take the following two sequences $u=...0100...$ where $u_1=0, u_2=1, u_3=0, u_4=0$ are fixed and similarly $v=...3233...$ where $v_1=3, v_2=2, v_3=3, v_4=3$ are fixed. Now define the set $$U:=\{\omega\in \Omega_M:\omega_1=0, \omega_2=1, \omega_3=0, \omega_4=0 \}$$ and $$V:=\{\omega\in \Omega_M:\omega_1=3, \omega_2=2, \omega_3=3, \omega_4=3 \}$$, then they are not empty and open. But now for every $N\in \Bbb{N}$ we can take $n=N$ and remark that $$S^{-n}(U)=\{\omega\in \Omega_M: \omega_{1-n}=0, \omega_{2-n}=1, \omega_{3-n}=0, \omega_{4-n}=0 \}$$ but then we see immediately that $S^{-n}(U)\cap V=\emptyset$ since there is no path $0$ or $1$ to $2$ or $3$ or vice versa using $M$. Therefore I would say $S$ not topologically mixing.
Does this work?
Yes, it works. Your Markov shift is not topologically mixing. It is not even irreducible since there is no connection between the first two states and the other states.