Let Consider the set $\Sigma = \{0,1,2,,,n-1\}^{\mathbb{Z}}$ consisting of all two-sided sequence of elements in $\{0,1,2,...,n-1\}$.
Define the full shift map $T:\Sigma \to \Sigma$ by $\sigma(...,x_{-1}\,x_o\,x_1\,x_2\,,...) = (...,x_1\,x_2\,,...)$, which is homeomorphim.
Let $A$ be an $n\times n$ binary matrix with entries $a_{i,j}$ for $i,j\in\{0,1,\ldots,n-1\}$, in which case
$$\Sigma_A\{\langle x_k:k\in\Bbb Z\rangle\in\Sigma_n:a_{x_k,x_{k+1}}=1\text{ for each }k\in\Bbb Z\}\,.$$
In other words, $\Sigma_A$ consists of the sequences in $\Sigma_n$ that are walks along with the graph with vertex set $\{0,1,\ldots,n-1\}$ and adjacency matrix $A$. The map $T:\Sigma_{A} \to \Sigma_{A}$ is a subshift of finite type.
Let $T:\Sigma \to \Sigma$ be a two side full shift on the compact metric space $\Sigma$, and let a subshift $\Sigma' \subset \Sigma$ with $\mu(\Sigma')>0$ for some some $T-$invariant measure $\mu$. The Poincaré recurrence theorem says for almost every $x\in \Sigma'$ there exists $r(x)\ge 1$ such that $T^{r(x)}(x) \in \Sigma'$. One can take $r(x)$ smallest, of course. The first-return map is $g(x)=T^{r(x)}(x)$. So, the return map is defined on a full measure subset of the measurable set $\Sigma'$. Is $\Sigma^{'}$ closed?
Assume that every element of $\Sigma'$ is dense i.e for every element $x\in \Sigma$, there is $x_{n} \in \Sigma'$ such that $x_{n} \to x$. Is that true that $T_{|\Sigma^{'}}$ transitive?
My answer:Here showed that the subshifts of finite type are compact.
Regarding transitivity, since the full shift is transitive, there is a point $p$, which has a dense orbit. On the other hand, every element of $\Sigma'$ is dense, so there is $(x_{n}) \in \Sigma'$ such that $x_{n} \to p$. Then, the result follows from the continuity.