I'm reading the Cuntz-Krieger essay "A class of $C^*$-algebras and topological Markov chains". I can't understand the following argument:
Let $I$ be a finite set and $A=(A(i,j))_{i,j\in I}$ be a binary matrix, i.e. $A(i,j)\in \{0,1\}$. The one-sided sub-shift $\sigma_A$ acts on the compact space $X_A=\{(x_k)_{k\in \Bbb{N}}\in I^{\Bbb{N}} |\ A(x_k,x_{k+1})=1\ \forall k\in \Bbb{N}\}$ by $(\sigma_A x)_k=x_{k+1}$.
Now, consider the algebra $C(X_A)$ of all continuous complex-valued functions on $X_A$. This algebra is generated by all functions of the form $\chi_i \circ \sigma_A^j={{\sigma_{A}}^*}^j(\chi_i)$ where $i\in I, j=0,1,2...$ and $\chi_i$ is the characteristic function of the cylinder set $z(i)=\{x\in X_A| x_1=i\}$.
I don't understand why they are the generators of the algebra. Any help would be appreciated, Thank you.
EDIT: From Stone-Wierstrass Theorem I know that If $span_{\Bbb{C}}\{\chi_i \circ \sigma_A^j | j\in \Bbb{N}, i\in I\}$ is closed in norm then it coincide with $C(X_A)$. That is because it's an associative algebra over the complex numbers ($\chi_i \circ \sigma_A^j\chi_k \circ \sigma_A^l$ is the same as taking the minimum between $\chi_i \circ \sigma_A^j$ and $\chi_k \circ \sigma_A^l$, which can be written as a linear combination of them). It's closed under $^*$ and it contains the constant function $1$, just take $max \{\chi_i\circ \sigma_A| i\in I \}$ (well-defined as $I$ is finite) and it separates points.
Thus, the closure of this unital $*$-algebra is the whole space $C(X_A)$. So, it's sufficient to show that it's already closed, but I can't see why that's true.
In a C$^*$-algebra paper, "generated by" usually means "generated as a C$^*$-algebra". So, for a subset $E \subset A $, if $A=C^*(E) $ this means that $A $ is the closure of the set of noncommutative polynomials in $E $ and $E^*$.
In your case, I don't think your set is closed, but you don't need it to be.