Is the Half Disc Topology normal? Recall:
A metric space is Hausdorff, normal and regular.
Here normal implies regular since the singletons are closed.
Now consider the Half disc topology in $\mathbb R^2$:
https://en.m.wikipedia.org/wiki/Half-disk_topology
In the upper half plane P the standard metric on $\mathbb{R^2}$ induces the topology. The line $L=\{(x,0)|x \in \mathbb{R}\}$ is added with open balls defined differently.
This space is possibly not a metrizable (i.e. metric) space anymore.
1)The Half Disc topology is Hausdorff.
Would like to prove or disprove:
2)The Half Disc topology is normal
3)The Half Disc topology is regular.
If I can show that it is normal, since singletons are closed in a top. Hausdorff space, then $2)\rightarrow 3)$.
Any hints how to tackle $2)$ or $3)$?
Thank you.
Appended:
Alessandro gives a hint how to show that the Half Disc topology is NOT regular.
Attempt:
Consider an open ball $P$ centered at $x$ on $L$ ($x-$axis). The complement $P^c$ is closed and comprises the points on the circumference of the semi-circle and $L$ \ {$x $} , i.e. the $x-$axis excluding the point $x$.
Let $P^c \subset O$, open, and $x \in P_x$ where $P_x$ is a semi-circle ball centered at $x$ with radius $r>0.$
Choose a point $z\not =x$ on $L$ with $|z-x| <r$.
$z \in P^c \subset O$.
$z$ is an interior point of $O$, There is a semi-circle ball $P_z$ around $z$ s.t. $P_z \subset O$.
But then $P_z\cap P_x \not = \emptyset$, and we are done.
Not regular implies not normal.
 A: Unfortunately what you would like to show is false, so I cannot give any hints in that direction. I can give an hint in the opposite direction:
Hint: Consider $x\in L$ and let $P$ be an half-disk neighbourhood of $x$, so that $P$ consists of an open half disk in the upper plane plus the singleton $x$. Note that the closure of $P$ contains all points on the diameter of the open disk. Consider now the complement of $P$, which is a closed set. Can it be separated from $x$ by open neighbourhoods?
A: Allessandro has given a correct hint as to why $X$ is not regular (nor $T_3$) but there is a nice related space, the Niemytzki plane or Moore plane (not sure who has priority here) from which we can use the proof of non-normality: $L$ is a closed and discrete subspace of size $\mathfrak{c}=2^{\aleph_0}$, while the whole space $X$ is has a dense subset of size $\aleph_0$. Jones' lemma does the rest: $X$ cannot be normal.
But of course the non-regularity argument is already enough for that. But the Niemytzki plane is a nicer space overall: it's even completely regular and a Moore space.
