Is the function $\,f(x, y) = x-y\,$ closed? Is the function $\,\,f: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$, defined as $$f(x,y)=x-y,$$ closed?

The function $f$ is not closed:
Take
$$
F=\left\{(x,y):y\ge x+\frac{1}{1+x^2}\right\}.
$$
Then $y>x$, for all $(x,y)\in F$, and hence $0\not\in f(F)$. But $0\in\overline{f(F)}$,
as
$$
\left(n,n+\frac{1}{1+n^2}\right)\in F,
$$
and thus
$$
x-y=\frac{1}{1+n^2}\in f(F).
$$
