Upper half plane with irrational slope topology. Let $X$ be the upper half plane $\{(p,q);p,q \in \mathbb Q, q\geq 0 \}$ with the Irrational slope topology. Can we say that the set $$\{(p,q);p,q \in \mathbb Q, q> 0 \}$$
is a closed set of $X$.
I think this is true since an open neighborhood of points of the form $\{(p,0);p \in \mathbb Q\}$ is an open segment on the $X$ axis. 
Am I right?
Thank you!
 A: Yes, the subset $\{ \langle x,y \rangle \in X : y > 0 \}$ is closed in the irrational slope topology (with fixed irrational $\theta$).
Recall that for $\langle x , y \rangle \in X$ the basic open neighbourhoods of $\langle x,y \rangle$ are of the form $$N_{\epsilon} ( \langle x,y \rangle ) = \{ \langle x,y \rangle \} \cup B_\epsilon ( x + \tfrac{y}{\theta} ) \cup B_\epsilon ( x - \tfrac{y}{\theta} )$$ where for a real number $\zeta$ and $\epsilon > 0$ we set $$B_\epsilon ( \zeta ) = \{ \langle r , 0 \rangle \in X : | r - \zeta | < \epsilon \}.$$
It follows that the basic open neighbourhoods of $\langle x , 0 \rangle \in X$ must be subsets of the "$x$-axis" (since for arbitrary $\langle x,0 \rangle$, the only point of $N_\epsilon ( \langle x,y \rangle )$ which might not lie on the "$x$-axis" is $\langle x,y \rangle$ itself).  More concretely, a simple calculation shows that $N_\epsilon ( \langle x,0 \rangle ) = B_\epsilon ( x )$ for all $\epsilon > 0$. Therefore the "$x$-axis" is open in the space.
