Density of polynomials having "additive-separation of their variables"? Let $K$ be a compact subset of $\mathbb{R}^2$. Let $P$ be a polynomial in the variables $x$ and $y$. Given $\epsilon>0$, can we find two polynomials $P_1=P_1(x)$ and $P_2=P_2(y)$ such that $$ \sup_{x,y\in K}|P(x,y) - (P_1(x)+P_2(y))| \leq \epsilon ?$$ 
My feeling is No, but I don't see how to prove it...
EDIT : Concerning my feeling, I had in mind the case when $K$ is the unit disc. As mentioned by Nate Eldredge in the comment below, the question depends on $K$. I consequently ask the question assuming that there exists no relation like $y=f(x)$ parametrizing $K$, where $f$ is continuous where $x$ runs. 
 A: If $K$ contains the corners of the unit square, try taking $P$ with
$$P(0,0)=0, P(0,1)=0, P(1,0)=1, P(1,1)=-1.$$
An appropriate transformation should cover any $K$ with nonempty interior.
Edit: Davide Giraudo asks about the (uniform) closure of the polynomials of the form $P(x) + Q(y)$ (denote the set of all such polynomials by $\mathcal{P}$).  It is indeed given by the functions of the form $F(x,y) = f(x) + g(y)$ where $f,g$ are continuous. Let $\mathcal{A}$ denote the set of such functions.  The Stone-Weierstrass theorem gives $\mathcal{A} \subset \overline{\mathcal{P}}$, as Davide mentions.  For the other direction, note that $F \in \mathcal{A}$ if and only if it satisfies the equation
$$F(x,y) = F(x,y_0) + F(x_0,y) - F(x_0,y_0)$$
for some arbitrary fixed $(x_0, y_0) \in K$.  Since the right side is continuous in $F$ with respect to the uniform norm, it is easy to see that $\mathcal{A}$ is uniformly closed.  Thus $\overline{\mathcal{P}} = \mathcal{A}$.  This argument works for any compact $K$.
