Let $pr_1: \Bbb R^2 \to \Bbb R$ be the projection map, $pr_1(x,y)=x$. Show that $pr_1$ is open and not closed.
I approached this the following way. Let $V \subset \Bbb R^2$ be open and of the form $V=\{(x,y) : d(x,y)< r\}$, then $pr_1(V)=(-r,r)$ which is open in the standard topology of $\Bbb R$.
Now if $A \subset \Bbb R^2$ is closed and of the form $A=\{(x,y) : d(x,y) \le r \}$, then $pr_1(A)= [-r,r]$ which is closed in the standard topology.
The problem is that I feel like the sets $V$ and $A$ are not capturing all the possible subsets of $\Bbb R^2$ since a subset of $\Bbb R^2$ could also be of the form $[a,b] \times [c,d]$ which doesn't work with the open and closed disks I'm considering?
Edit: If I suppose that for an open set $V \in \Bbb R^2$ the image $pr_1(V)$ is not open then I must have that $\Bbb R \setminus pr_1(V)$ is open.
Now $pr_1(V)$ can be of 4 different forms $(a,b), [a,b), (a,b]$ or $[a,b]$ the first
$\Bbb R \setminus(a,b) = (-\infty,a) \cup(b, \infty)$ is open
$\Bbb R \setminus[a,b) = (-\infty,a] \cup (b, \infty)$ seems to be not open
$\Bbb R \setminus(a,b] = (-\infty, a) \cup[b, \infty)$ seems also not open
and $\Bbb R \setminus[a,b] = (-\infty, a] \cup [b, \infty)$ would also not be open?
This would have led to contradiction, if the first set would not be open?
