Continuous function that vanish at a point is dense Let $(E,d)$ be a metric space and $K$ a compact and $a\in K$
I have an algebra $A$ of $C(K,\Bbb{R})$ such that for all $f\in A, f(a)=0$ and it separates points.
I would like to prove that $\overline{A}=\{f\in C(K,\Bbb{R}): f(a)=0\}.$ If I denote $\{f\in C(K,\Bbb{R}): f(a)=0\}:=J_a$ 
How can I use Stone–Weierstrass theorem ? 
 A: I haven't seen this anywhere, but here's a very natural description of the closure of an arbitrary algebra of functions. I'll state it in terms of locally compact spaces, for the sake of generality.

Theorem: Let $X$ be a locally compact space and $A$ an algebra of real functions on $X$. Let $Z(A)=\left\{x\in X: fx=0\text{ for all }f\in A\right\}$ and $I(A)=\left\{(x,y)\in X\times X:f(x)=f(y)\text{ for all }f\in A\right\}$.
Then the closure of $A$ in $C_0(X)$ consists of those $f\in C_0(X)$ for which $f(x)=0$ for all $x\in Z(A)$ and $f(x)=f(y)$ for all $(x,y)\in I(A)$.

Proof. Let
$$B=\left\{f\in C_0(X):f|_{Z(A)}=0\text{ and }f(x)=f(y)\text{ for all }x,y\in I(A)\right\}$$
By going to the one-point compactification of $X$ if necessary (which will add at most one point to $Z(A)$), we can assume that $X$ is compact. We need to show that $\overline{A}=B$.
It is not hard to check that $B$ is closed and $A\subseteq B$, so $\overline{A}\subseteq B$.
Now note that $I(A)$ is a closed equivalence relation on $X$. Consider the quotient space $K=X/I(A)$. Every $f\in B$ factors through $K$, which gives us an isometric homomorphism $B\to C(K)$. Moreover, $Z(A)$ is either empty or an equivalence class of $I(A)$. In either case, either the image of $B$ in $C(K)$ will be all of $C(K)$ (if $Z(A)$ is empty) or it will be the algebra of functions which vanishes at a single point (namely, the class $Z(A)$), and the image of $A$ in $C(K)$ will be separating and will vanish at the same (single) point. By Jochen's comment, the image of $A$ is dense in the image of $B$ and therefore $A$ is dense in $B$ (because the embedding $B\to C(K)$ is isometric).QED

EDIT: This implies Bishop's theorem, as below:
Suppose $A$ is a closed algebra of functions of $C(X)$, with $X$ compact. Let's say that a subset $S\subseteq X$ is \emph{$A$-constant} if for every $a\in A$, $a|_S$ is constant.
Suppose $f\in C(X)$ is such that for every maximal $A$-constant set $S$, there exists $a\in A$ with $f|_S=a|_S$. We need to show that $f\in A$.
Note that maximal $A$-constant sets are precisely the equivalence classes of $I(A)$. It is then easy enough to see that $f(x)=0$ for all $x\in Z(A)$ and $f(x)=f(y)$ for all $(x,y)\in I(A)$, therefore, by the theorem above, $f$ is in the closure of $A$, which is $A$ itself.
