Consider following question: Let $X=\{(x,y) \in \mathbb{R}^2 : x\ge 0 \text{ or } y=0 \}$ and let $T$ be the subspace topology on $X$ induced by the usual topology on $\mathbb{R}^2$. Suppose $\mathbb{R}$ has usual topology and define $f: X\to \mathbb{R}$ by $f((x,y))=x$ for all $(x,y) \in X$. Then prove that $f$ is a quotient map, not open map, not closed map.
I have proved it quotient map, not open map, but unable to find a closed set whose image is not closed due to the fact the map $f((x,y))= x$ is defined ( defined on open sets).
So, Can you please tell how should I prove that this map is not closed.
$\{(x,y): x>0,y>0,xy=1\}$ is closed and its image is $(0,\infty)$.