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I'm trying to show this by using Stone-Weierstrass Theorem.

For clarity, let me state definitions and the theorem first :)

Definition: Uniform metric

Let $\overline{d}$ be the standard bounded metric induced by the standard metric $d$ on $\mathbb{C}$.

Define $\overline{\rho}(f,g)\triangleq \sup_{x\in X} \overline{d}(f(x),g(x))$.

Call $\overline{\rho}$ "the uniform metric on $\mathbb{C}^X$ induced by $d$".

(This is the definition given in Munkres-Topology)

Next,

Stone-Weierstrass Theorem (complex version)

Let $X$ be a locally compact space.

Let $C_0(X,\mathbb{C})$ be equipped with the topology induced by the uniform metric $\overline{\rho}$.

Let $\mathscr{A}$ be a subalgebra of $C_0(X,\mathbb{C})$

If $\mathscr{A}$ is self-adjoint, separates points and vanished nowhere, then $\mathscr{A}$ is dense in $C_0(X,\mathbb{C})$.

Now, let me illustrate what exactly my question is.

Set $X$ a compact subset of $\mathbb{C}$.

Since $X$ is compact, $C_0(X,\mathbb{C})=C(X,\mathbb{C})$

I want to know whether the set $P\triangleq \{f\in \mathbb{C}^X:f \text{ is a polynomial function}\}$ is dense in $C(X,\mathbb{C})$.

To apply the theorem, I tried to show that $P$ is self-adjoint, but I couldn't. Now, I don't really think that $P$ is self-adjoint, but if so, how do I prove this?

Even if $P$ is not self-adjoint, I believe it's true that $P$ is dense in $C(X,\mathbb{C})$. How do I show this?

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Depends of the interpretation of "polynomial function". Let be $z=x+iy$. If your polynomials are elements of $\Bbb C[x,y]$, then the self-adjointness condition is true. But $\Bbb C[z]$ isn't self-adjoint.

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Look at the case of $X = D^2$, the unit disc in $\mathbb{C}$. A uniform limit of polynomials on this domain is analytic, so the uniform closure is contained in the analytic functions on $D^2$.

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