Here, I explain the problem for the case of $\mathbb{R}^2$. But, eventually we could extend this to any number of dimensions.
Consider any $\mathbb{P} \subseteq \mathbb{R}^2$. I want to make a selector function $f_\mathbb{P}$ such that when we write $f_\mathbb{P}(x, y) = 0$, the graph of the function is precisely $\mathbb{P}$. I have solved the problem for finite $\mathbb{P}$, containing $(x_1, y_1)$, $(x_2, y_2)$, ..., $(x_N, y_N)$: $$f_\mathbb{P}(x, y) = \prod_{n=1}^N (|x-x_n| + |y-y_n|) = 0.$$
Now, I tried to extend this in a straight-forward way to the countable infinite $\mathbb{P}$ case, giving: $$f_\mathbb{P}(x, y) = \prod_{n=1}^\infty (|x-x_n| + |y-y_n|) = 0.$$
But, this no longer works. Since we have situations like $\prod_{n=1}^\infty \frac{1}{10^n} = 0$ where infinite product of non-zero terms can still equal $0$, which was not possible before for the finite case.
I did make partial progress on this problem. Consider $$\prod_{n=1}^\infty A_n(|x-x_n| + |y-y_n|).$$ If $A_n$ is an increasing function of $n$, increasing at a large enough rate, then it can prevent some bad cases. But no matter what its rate of increase, we can find a large enough negative offset such that in overall effect, the problem remains unsolved.
Can someone help me solve this problem and/or the general problem of finding selector function $f_\mathbb{P}$ for any $\mathbb{P} \subseteq \mathbb{R}^n$.