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Given a $C^{\ast}$-algebra $\mathcal{A}$ and a subset $S\subseteq\mathcal{A}$, denote by $C^{\ast}(S)\subseteq\mathcal{A}$ the minimal $C^{\ast}$-algebra in $\mathcal{A}$ containing $S$. This is what I mean by the $C^{\ast}$-algebra generated by $S$.

Is there a useful explicit description of $C^{\ast}(S)$?

This question is not well-posed, so I explain quickly what I am aiming at: In my situation the set $S$ is separable and I want to show that $C^{\ast}(S)$ was separable. It seems quite obvious but it becomes a mess to write down when you start taking unions, generated linear spaces and so on.

What I did was the following: let $V(S)$ be the closed subspace generated by $S\cup S^{\ast}$, where $S^{\ast}=\{s^{\ast};s\in S\}$. Then this should be separable (haven't done it formally). Similarly $V(V(S)V(S))$, the closed subspace generated by products should again be separable and contain also all the transposes of products. In my case $\mathcal{A}$ is unital and hence $V(V(S)^{n})\supseteq V(V(S)^{n-1})$ so I can take a countable increasing union, the closure thereof and should obtain the $C^{\ast}$-algebra generated by $S$. Now the countable union of the countabe dense subsets at each stage should again be dense and the question should be answered. However this looks horrible... So my question is: is there an easier argument?

MWL
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1 Answers1

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Denote $$ K=\mathbb{Q}+i\mathbb{Q}\\ P_n^m=\{p\in K[x_1,\ldots,x_n]:\deg p\leq m\}\\ C=\bigcup_{m,n\in\mathbb{N}}\bigcup_{p\in P_n^m}p\left((S\cup S^*)^m\right) $$ then $$ C^*(S)=\operatorname{cl}\left(\operatorname{span}(C)\right)\tag{1} $$ Now, we turn to seprability question. Since $^*:\mathcal{A}\to\mathcal{A}$ is an isometry, then $S^*$ is separable as image of seprable $S$ under distance preserving map. Then $S\cup S^*$ is separable as union of separable sets. For each $p\in P_n^m$ the set $p((S\cup S^*)^m)=\{p({\bf s}):{\bf s}\in (S\cup S^*)^m\}$ is separable as image of separable set under continuous map $p$. Now, since $K$ is countable, so does $P_n^m$, hence $\bigcup_{p\in P_n^m}p\left((S\cup S^*)^m\right)$ is separable as union of separable spaces. By the same argument $C$ is also separable. Since separability preserved under closures and linear span operation, then by $(1)$ $C^*(S)=\operatorname{cl}\left(\operatorname{span}\left(C\right)\right)$ is also separable.

The same argument shows that density character of $C^*(S)$ and $S$ are equal if the last one is infinite.

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Norbert
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