Let $\mathbb{P}_{ 2}(\mathbb{N})=\{A\subset \mathbb{N}:|A|= 2\}$, is there any explicit injective function $f:\mathbb{P}_{2}(\mathbb{N})\longrightarrow \mathbb{N}$, such that $f$ is a polynomial in two variables?
I have tried with $f(\{x,y\})= xy+y+x$, however this function is not injective:
$f(\{x,y\})=x+y+xy=f(\{0,x+y+xy\})$
The problem seems to be discussed in the beginning of this post:
Is There An Injective Cubic Polynomial $\mathbb Z^2 \rightarrow \mathbb Z$?
In particular, the function:
$f(x,y) = (x+y)^2 + y$
Is given as an example, along with some discussion of the same problem over integers.
If you are allowed to assume that $\{x,y\}$ is represented with $x<y$, then $\{x,y\}\mapsto x+\frac{y(y-1)}2$ is a bijection $\def\N{{\Bbb N}}\binom{\Bbb N}2\to\N$, where $\binom\N2$ is the set you denote as $\Bbb P_2(\Bbb N)$. This can be extended to $\{x_1,\ldots,x_n\}\mapsto\sum_{k=1}^n\binom{x_k}k$ which similarly gives a bijection $\binom\N n\to\N$ called the combinatorial number system.
The function
$$f(x, y) = (x + y)^3 + xy$$
is symmetric and injective on $\mathbb P_2(\mathbb N)$, because from $(x + y)^3 ≤ f(x, y) < (x + y + 1)^3$ we see that one can uniquely recover $x + y = \lfloor \sqrt[3]{f(x, y)}\rfloor$ and $xy = f(x, y) - \lfloor \sqrt[3]{f(x,y)}\rfloor^3$.
