Lately, I have done an exercise in the book "Introduction to Dynamical Systems" by Brin and Stuck.
Exercise 1.4.5: Assume that all entries of some power of $A$ are positive. Show that in the product topology on $\sum_A$ and $\sum_A^+$, periodic points are dense and there are dense orbits.
This question was asked in this post Periodic points and dense orbits
The proof for "periodic points are dense" was indicated in the post. Now, I just want to make sure I understand the idea of the proof for the second part.
Assume the alphabet $A_m$ has $m$ elements from 1 to $m$. By our assumption, there is an integer $k$ such that all the entries of $A^k$ is greater than 0. I construct a sequence $y$ as follows:
For any $i,j \in A_m$, if there is a path from $i$ to $j$ after $n$ steps. Let $w$ is a word contains these $n$ symbols. Then consider the sequence of all such $w$'s. For example, for $i=1$ and $j=1$, if there is a path from $1$ to $1$ after 1 step, let $y_1=1, y_2=2$. If there is a path from $1$ to $1$ after 2 steps, let $y_3=1$, $y_5=1$ and so on...
Then we can prove this sequence is dense in $\sum_A^+$.
But I have a question here. Is it possible to construct this sequence?
Any help is appreciated.
I'm not sure I understand your construction, but I'll propose a different solution. I will denote the power in your question by $p\in \mathbb{N}$.
Since any shift of a sequence in $\Sigma_A$ or in $\Sigma_{A}^+$ stays in the subshift, you can move it to the $0$ index.
Since finite valid words at $0$, with respect to $A$, induce a basis to the product topology, it is enough to have a sequence containing all valid words.
Enumerate all possible finite words by $\{ w_n \}_{n=1}^\infty$. Now you need to build a sequence with all these words.
For example, put $w_1$ at $0$. Every word $w_n$ starts with a letter $a_n$ and ends at $b_n$. Since $A^p$ has strictly positive entries, there is some valid word $u(b_1,a_2)$ of length $p-2$ such that $b_1 u(b_1,a_2)a_2$ is a valid word. You can then conclude that the concatenation, $w_1u(b_1,a_2)a_2$, is valid. Then $w_1u(b_1,a_2)w_2$ is a valid word as well. You can then recursively construct the locally valid sequence
$$ w_1u(b_1,a_2)w_2u(b_2,a_3)w_3u(b_3,a_4)w_4..., $$
by iterated concatenations. Using the previous arguments, the constructed sequence will have all finite valid words. Therefore, it will have a dense orbit in $\Sigma_A^+$. Similarly, you can construct from this method a sequence $x\in \Sigma_A$ with a dense orbit.