The full group C star algebra of a second-countable group is separable

Choose a countable dense subset $S\subseteq L^1(G)$ from [F1]. The zero class belongs to $L^1(G)$ by [F13], so $S$ is nonempty and [F6] gives a surjection $s:\mathbb{N}\to S$. Set $K=\{j(a)+ij(b):a,b\in \mathbb{Q}\}$ using the rational quotient and embeddings [F10, F11]; by [F11], this is the statement's set $\mathbb{Q}(i)$. By [F7] and [F8], $\mathbb{Q}\times \mathbb{Q}$ is at most countable and nonempty; [F6] gives a surjection $r_0:\mathbb{N}\to \mathbb{Q}\times \mathbb{Q}$. The map $(a,b)\mapsto j(a)+ij(b)$ is onto $K$, so composing it with $r_0$ gives a surjection onto $K$; [F6] then gives a surjection $r:\mathbb{N}\to K$. The formulas in [F11] and the field laws in [F9] show that $K$ is closed under addition and multiplication; [F12] gives closure under conjugation, and $1=j(1)+ij(0)\in K$. AC is assumed as required by [F1]–[F3]; the local enumerations use [F6] and require no additional choices.

Let $C=S\times \{0,1\}$ be the alphabet of factors, where $(s,0)$ evaluates to $q(s)$ and $(s,1)$ to $q(s{)}^{\ast }$. It is at most countable by [F8]. Let $W={\bigcup }_{n\ge 1}{C}^n$ be its nonempty finite words. Every $C^n$ is at most countable by [F5], and AC supplies the Countable Choice required by the union theorem there, so $W$ is at most countable. A word evaluates to the product of its factors, in their listed order. The record alphabet $R=K\times W$ is at most countable by [F8]; a pair records a coefficient and a word. Define $D$ to consist of all evaluations ${\sum }_{\mathrm{\ell }=1}^m{\alpha }_{\mathrm{\ell }}{w}_{\mathrm{\ell }}$ of finite lists from $R$, including the empty sum $0$. Thus it is exactly the finite $K$-linear combinations of nonempty words in $q(S)$ and their adjoints.

For each $m\in \mathbb{N}$, the finite power $R^m$ is at most countable by [F5], including its one-point empty-word case $m=0$. Under the Countable Choice supplied by AC, [F5] makes $E={\bigcup }_{m\in \mathbb{N}}{R}^m$ at most countable. It is nonempty, so [F6] gives a surjection from $\mathbb{N}$ onto $E$. Evaluation of a list is a well-defined map onto $D$; composing these maps gives a surjection $\mathbb{N}\to D$. Since $0\in D$, [F6] proves that $D$ is at most countable. No numerical coding of finite sequences is required.

The set $D$ is closed under addition because finite summand lists concatenate; it is closed under multiplication because distributivity gives a finite sum of concatenated words, with coefficients still in $K$ by step 1.1. For $\beta \in K$, multiplying a finite sum by $\beta$ replaces each coefficient ${\alpha }_{\mathrm{\ell }}$ by $\beta {\alpha }_{\mathrm{\ell }}\in K$, so $D$ is closed under $K$-scalar multiplication. Finally, $({\sum }_{\mathrm{\ell }}{\alpha }_{\mathrm{\ell }}{w}_{\mathrm{\ell }}{)}^{\ast }={\sum }_{\mathrm{\ell }}\overline{{\alpha }_{\mathrm{\ell }}}{w}_{\mathrm{\ell }}^{\ast }$; [F4] reverses each word and takes the adjoint of each factor, and [F12] and step 1.1 keep every coefficient in $K$. Hence $D$ is closed under adjunction. Since $1\in K$, each $q(s)$ is in $D$ by a one-letter word, so $q(S)\subseteq D$. By [F3], $\mathrm{\parallel }q(f)-q(s){\mathrm{\parallel }}_{C^{\ast }}\le \mathrm{\parallel }f-s{\mathrm{\parallel }}_1$; therefore $q(S)$ is dense in $q(L^1(G))$, and this image is dense in $C^{\ast }(G)$ by [F2]. Thus $D$ is dense. Since it is already closed under addition and $K$-scalar multiplication, its $K$-linear span equals $D$ and is norm-dense.