Let $C:=\{x\in\mathbb{C}:|x|=1\}$ be the complex circle and $f:C\rightarrow\mathbb{C}$ the mapping $x\mapsto 1/x$. From the Stone-Weierstrass theorem I get $\oint fdx =0$, but by Cauchy's theorem $\oint fdx =2\pi i$. It must be that I'm using the Stone-Weierstrass wrongly, but I don't see where. Can you help me?
The complex Stone-Weierstrass theorem says:
If $X$ is a compact Hausdorff space and $A$ is a closed subalgebra of $\mathcal{C}(X,\mathbb{C})$, which separates points, contains a non-zero constant function, and contains the conjugate of each of its functions, then $A$ equals $\mathcal{C}(X,\mathbb{C})$.
Here, $\mathcal{C}(X,\mathbb{C})$ is the linear space of continuous functions $X\rightarrow\mathbb{C}$, equipped with the supremum-norm. $C$ is compact and Hausdorff. The set of polynomials forms a closed subalgebra of $\mathcal{C}(C,\mathbb{C})$, which separates points, contains a non-zero constant function, and contains the conjugate of each of its functions. So, for all $\epsilon>0$, there exists a polynomial $p_\epsilon$, such that $|p(x)-f(x)|<\epsilon$ for all $x\in C$. Next, using the triangular inequality: $$ \left|\oint fdx\right|=\left|\oint f-p_\epsilon dx\right|+\left|\oint p_\epsilon dx\right| $$ By Cauchy's integral formula, $\oint p_\epsilon dx=0$. So, choosing $\gamma:[0,1]\rightarrow C$, with $t\mapsto e^{2\pi i t}$, $$ =\left|\oint f-p_\epsilon dx\right| =\left|\int_0^1 (f-p_\epsilon)\circ\gamma(t)\cdot\gamma'(t) dt\right| \leq\int_0^1\left| (f-p_\epsilon)\circ\gamma(t)\right|\left|\gamma'(t)\right| dt \leq \epsilon 2\pi $$ Since $\epsilon$ is arbitrary, it follows that $\left|\oint fdx\right|=0$ and therefore $\oint fdx=0$. But we know that $\oint fdx=2\pi i$. Where did I go wrong?