Let us consider the shift map $\sigma$ acting on the semi-infinite space of sequences of $k$ symbols, $\Sigma_k^+$, such that if $\omega = \omega_0\omega_1\omega_2\dots$, then $\sigma ^i \omega =\omega_i\omega_{i+1}\dots$ . We know that the Bernoulli mesure $\nu$ defined on cylinder sets $C^{\omega_0...\omega_n} := \{\eta \in \Sigma_k^+\ ;\ \eta_i = \omega_i \text{ for } 0\leq i\leq n\}$ is ergodic and invariant for $\sigma$, being $\nu(C^{\omega_0\dots\omega_n}) = p_0^{i_0}\cdot\dots \cdot p_n^{i_n},\ p_n>0$ where $\sum_{n}p_n = 1$ and symbol $\omega_k$ is repeated $i_k$ times in the sequence $\omega_0\dots \omega_n$.
Can we say that the set of elements such that a certain finite sequence appears in them has full measure? For example: if we consider only 2 symbols $\{0,1\}$ and let $A := \{\omega \in \Sigma_{2}^+\ ;\ \sigma^{i}\omega \in C^{12}\ \text{for some (finite) }i \in \mathbb{N}\}$. Is $\nu(A) = 1$? In this case we might want to consider $A^c$ which consists of the sequences \begin{equation} \begin{split} 2\dots 2\dots\\ 1\dots 1\dots\\ \underbrace{2\dots2}_{k}1\dots1\dots \end{split} \end{equation} with $k\in \mathbb{N}$. Since this is a countable union of points, can we conclude that $\nu(A^c) = 0$?
In general, I am actually trying to see that given a finite sequence of $k$ symbols $\eta_0\dots \eta_k$, then $\nu-$almost all $\omega \in \Sigma^{+}_k$ will visit the cylinder set $C^{\eta_0\dots \eta_k}$ infinitely many times under the shift map. Any help is more than welcome!