A non-trivial closed polynomial function must not be surjective? Suppose $f:\mathbb{R}^2\to\mathbb{R}$ is a nonconstant polynomial function.
If $f$ is a closed mapping, then $f$ must not be a surjection?
Thanks in advance.
 A: EDITED:
If there is a sequence of points $p_n$ with $|p_n| \to \infty$ and $f(p_n)$ bounded, then there is a (maybe different) sequence forming a closed set $S$ such that $f(S)$ is not closed.  Show that if there is no such sequence $p_n$, then $\lim_{|z| \to \infty} f(z)$ is either $+\infty$ or $-\infty$.
A: Let me flesh out (and slightly rephrase) the details in Robert Israel's beautiful answer a bit.  Suppose $f$ is surjective.  Let $D_n\subset\mathbb{R}$ denote the closed disk of radius $n$ centered at the origin.  Then for any $n$, there exist $x,y\in \mathbb{R}^2\setminus D_n$  such that $f(x)<-1$ and $f(y)>1$ (if, for instance, no such $x$ existed, then the image of $f$ would be bounded below since $f(D_n)$ is compact and outside of $D_n$, $f$ is always above $-1$).  Since $\mathbb{R}^2\setminus D_n$ is connected, we can find a point $z_n\in\mathbb{R}^2\setminus D_n$ such that $f(z_n)=1/n$.  Now let $S=\{z_n\}$.  Since $z_n\in \mathbb{R}^2\setminus D_n$ for all $n$, $S$ contains only finitely many points in any bounded set, so $S$ is closed.  But $f(S)=\{1,1/2,1/3,\dots\}$ is not closed.  Thus $f$ is not a closed map.
