Just a sanity check in basic functional analysis Consider the algebra $C(S^1)$ of continuous functions $S^1 \to \mathbb C$ together with the $\|\cdot\|_\infty$ ($\sup$-norm). I am thinking that:
(?) The (sub-)algebra generated by $\rm{id}$ and $\overline{\cdot}$ (the complex conjugation) is the set of all polynomials. 
(??) The closure of this subalgebra generated by $z$ and $\overline{z}$ is the entire algebra $C(S^1)$ (because of Stone-Weierstrass)?.
Is that accurate? If my thinking is wrong I'd greatly appreciate any corrections. 
 A: Regarding the first question, yes, it is the algebra of polynomials, if you consider polynomials in $z$ and $\overline{z}$. This is also the algebra of trigonometric polynomials, generated by $\cos nt,\, n \geqslant 0$ and $\sin nt,\, n\geqslant 1$.
By Weierstraß' theorem, this algebra is dense in $C(S^1)$.
A: By Stone-Weierstrass, the sub-$C^*$-algebra generated by $z\mapsto z$ and $z\mapsto\bar{z}$ (which is to say, that generated by $z\mapsto z$, as complex conjugation is simply the involution of this algebra), which naturally separates points, is indeed dense in $C(S^1,\mathbb{C})$.
However, the fact mentioned above ($\int_{S^1}\frac{dz}{z}=2\pi i$ and $\int_{S^1}p(z)dz=0$ for all $p\in\mathbb{C}[z]$, implying that uniform approximation is impossible) shows that it must be a proper super-set of the polynomials.
In fact, it's not difficult to verify that the $C^*$-algebra of rational functions (which naturally contains the $C^*$-algebra generated by $z\mapsto z$) is also contained in it.
