$f: R^2 \rightarrow R$ defined by f(x,y) = x+y is open map but not closed map

Question

Define $f: R^2 \rightarrow R$ by $f(x,y) = x+y$. I want to prove that 1. f is open, 2) f is not closed.

My try: We can see that f is sum of the first and second projection maps which are open. But I don't to how to conclude f is open. Similarly, I couldn't construct example for the second question. Any hint will be good. Thank you.

Answer 1

Let $U\subseteq\Bbb R^2$ be open. You want to show that $f[U]$ is open. So assume $z\in f[U]$. Then there is $(x,y)\in U$ with $x+y=z$. As $U$ is open, there exists $r>0$ such that $(x-r,x+r)\times(y-r,y+r)\subseteq U$. Conclude that $(z-2r,z+2r)\subseteq f[U]$.

Let $A=\Bbb Z\times\sqrt2\,\Bbb Z$. Then $A$ is closed. Show that $f[A]$ is dense in $\Bbb R$ (but it is still countable).