I'm working on a problem from Munkres about open and closed maps. Here's the problem:
"Let $\pi_1 : \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$ be projection onto the first coordinate. Let $X$ be the subspace $(0\times \mathbb{R})\cup (\mathbb{R} \times 0)$ [I think this is just the x and y axes] of $\mathbb{R}\times \mathbb{R}$, and let $g=\pi_1|_X$. Show that $g$ is a closed map but not an open map."
So I think I've finished showing it is a closed map, but I'm stumped on showing it is not an open map. It seems to me that if for an open set of $X$, we get something like $(a,b)\times (c,d)$ where $(a,b)$ is an open interval (or the union of some open intervals) on the x-axis and $(c,d)$ is an open interval on the y-axis, and when we project we're always going to get $(a,b) \subset \mathbb{R}$ and it will be open.
Can anyone provide some advice? Thanks!
For an open interval $(c,d)$ with $c,d>0$ the set $\{0\}\times(c,d)$ is open in $X$ as it can be expressed as $\Bbb R\times(c,d)\cap X$, but its projection is $\{0\}$ which is not open in $\Bbb R$.