Is the Banach Algebra generated by a separable subset separable? Let $A$ be a Banach algebra and $S \subset A$ a subset. Suppose $B$ is the Banach algebra generated by $S$ and that $S$ is a separable subset of $A$.

Does this imply that $B$ is separable?

My intuition says the answer is yes, but I can't quite prove it. Here's my progress. For any  $T \subset A$, I define
$$
W(T):=\{ t_1\cdots t_m : m \in \mathbb{N}, t_j \in T\} \subset A
$$
By definition of Banach algebra generated by a set we have that
$$
B_1:=\left\{ \sum_{k=1}^n \lambda_k w_k  : n \in \mathbb{N}, \lambda_k \in \mathbb{C} , w_k \in W(S) \right\}
$$
is a dense subset of $B$. Let $D$ be a countable dense subset of $S$. I claim that
$$
B_2:=\left\{ \sum_{k=1}^n \lambda_k w_k : n \in \mathbb{N}, \lambda_k \in \mathbb{Q}+i\mathbb{Q} , w_k \in W(D) \right\}
$$
is a countable dense subset of $B$. That $B_2$ is countable is clear. I can't quite prove that is dense because I am having issues showing that an any element of $W(S)$ is arbitrarily close to an element in $W(D)$.
I would appreciate any help with this. It might as well be that my intuition is wrong or that I am trying to prove it in a complicated way.
Thanks in advance.
 A: Yes, it's true, and your proof idea works. The Banach algebra generated by $S$ is the closure of the (noncommuting) polynomials in $S$ as you say. So it suffices to show that any polynomial in $S$ (with complex coefficients) can be approximated by a sequence of polynomials in $D$ (with coefficients in $\mathbb{Q}(i)$), since then the closure of the latter contains the former and hence the closure of the former.
By linearity (and the continuity of addition and scalar multiplication), together with the density of $\mathbb{Q}(i)$ in $\mathbb{C}$, it suffices to show that a monomial $\prod_{i=1}^n s_i, s_i \in S$ can be approximated by a sequence of monomials in $D$. This follows from the continuity of multiplication: take for each $s_i$ a sequence $d_{i, j} \in D$ converging to $s_i$ and consider the sequence of monomials $j \mapsto \prod_{i=1}^n d_{i, j}$.
A: It's known that

*

*$\mathrm{cl}(B_1)=B$.

*$B$ is the smallest Banach algebra containing the set $S$.

On the one hand, $B_2 \subseteq B_1$ and therefore $\mathrm{cl}(B_2) \subseteq B$, by 1. above.
On the other hand, $\mathrm{cl}(B_2)$ is already a Banach algebra that contains $S$ (because $D$ is dense in $S$) and therefore $B \subseteq \mathrm{cl}(B_2)$, by 2. above.
Therefore, $\mathrm{cl}(B_2)=B$, so indeed $B$ is separable.
