I was trying to show that the Moore Plane is a regular but not normal space. I’m not sure of my proof of the latter property, and felt like asking if it looked alright.
We show that the disjoint closed sets $A=\Bbb{Q} \times \{0 \}$ and $B = (\Bbb{R - Q}) \times \{0 \}$ can not be separated by disjoint open sets.
Suppose that $U, V$ are open disjoint sets such that $A\subseteq U$ and $B \subseteq V$. Then for each $x\in \Bbb{Q}$ and $y\in \Bbb{R-Q}$ there exist “tangent disks” $B_x$ and $B_y$ such that $(x,0)\in B_x \subseteq U$ and $(y,0)\in B_y \subseteq V$.
Suppose the radii of $B_x$ and $B_y$ are $r_x$ and $r_y$ respectively, and choose $q_x, q_y \in \Bbb{Q}$ such that $0<q_x<r_x$ and $0<q_y<r_y$.
We’ve thus constructed a function $f : \Bbb{R} \to \Bbb{Q}$, $f(u)= q_u$.
Now, since $\Bbb{R} = \bigcup_{q\in \Bbb{Q}} f^{-1}(\{q \} )$ and $\{f^{-1}(\{ q\})\}_{q\in \Bbb{Q}}$ is a countable family of subsets of the real line, it follows from the Baire Category Theorem that at least one of the $f^{-1}(\{q\} )$ has non-empty interior. Thus let $(a,b)\subseteq f^{-1}(\{q_0\})$ for some $a<b$ and $q_0\in \Bbb{Q}$.
Now, let $(x_n)_n \subset (a,b) \cap \Bbb{Q}$ and $(y_n)_n \subset (a,b) \cap (\Bbb{R-Q})$ be two sequences in $A$ and $B$ respectively such that $x_n,y_n \to \frac{a+b}{2}$ as $n\to \infty$ (in the Euclidean topology).
Since the radii of each of the $B_{x_n}$ and $B_{y_n}$ are strictly greater than $q_0 > 0$, it must be (“geometrically”) that they eventually intersect, for $n$ large enough.
I’m not sure about that last part in particular. Thank you in advance for your help.
You can avoid the slight handwaving at the end by changing your choices of tangent disks slightly, and you can simplify the last bit slightly as well. When you choose $B_x$ such that $\langle x,0\rangle\in B_x\subseteq U$, choose it to have a rational radius $q_x$ right from the start, and similar for the tangent disks at points of $B$. Then at the end fix any $x\in(a,b)\cap\Bbb Q$ and choose $y\in(a,b)\setminus\Bbb Q$ such that $|x-y|<2q_0$; then $\left\langle\frac{x+y}2,q_0\right\rangle\in B_x\cap B_y\subseteq U\cap V$.