I'm stuck with this one:
Show that $f:\mathbb{R^2}\to\mathbb{R}$, with $f(x,y)=x,$ is open and continuous, but not closed.
Here's what I've done so far:
$i)$ Open: Obviously the projection on the real axis of any open disk gives an open interval, so that's true.
$ii)$ Not closed: We can look at the graph of $g(x)=1/x$. It's a subset of $\mathbb{R^2}$, and closed with the usual topology in there, but it's projected into an open interval on $\mathbb{R}$.
What I cannot figure out is how to prove that this projection is continuous. That is, finding that open sets in $\mathbb{R^2}$ are mapped through $f^{-1}$ to open sets.
I'd appreciate any hint.
Thanks for your time.
Best Answer
You need to show that if $U$ is open in $\mathbb R$, then $f^{-1}(U)$ is open in $\mathbb R\times \mathbb R$.
But what is $f^{-1}(U)$? It's just $U\times \mathbb R$, which is clearly open in $\mathbb R\times \mathbb R$ (it is a basic open set in the product topology).