Trouble understanding why Moore's Plane is not normal

Question

This question was already asked here(Show that the Moore plane is not normal) but I couldn't understand that answer and also I have another question.

On page 13 of Dolciani Expository text in Topology by S.G Krantz author gives an outline why Moore's plane is not normal. But I am unable to deduce it.

Adding image Firstly, Author gives two sets $A= \{(x,0)\in P \mid x\in \mathbb{Q}\}$ and $B=\{(x,0)\in P\mid x\in \mathbb{R}\setminus \mathbb{Q}\}$, here $P$ is the Moore plane.

Author tells that both of these sets are closed. In my opinion Closure of both $A$, $B$ should be set $\{(x,0)\in P\mid x\in \mathbb{R}\}$.

Then it it clear that they are disjoint sets.

But can you please tell how to prove that they cannot be seperated by open sets?

I can see it intutively but I am afraid that I am wrong as my intution was wrong in question 1 asked above which was very similar.

I am asking for your help as I am bad in topology as course was marred by a bad instructor and wasn't aware of existence of MSE network then.

Answer 1

The set $L:=\{(x,0): x \in \Bbb R\}$ is discrete in its subspace topology with resp. to the Moore plane. This holds as for a basic subset of $(x,0)$, which is $\{(x,0)\}$ unioned with a tangent ball inside the upper half plane $P :=\{(x,r):x \in \Bbb R, r>0\}$ only intersects $L$ in $\{(x,0)\}$, making it an isolated point in the subspace topology on $L$.

Also, $L$ is closed in the Moore plane (its complement is $P$ which is open by definition of the topology on the Moore plane), so all subsets of $L$ are closed in $L$ (as $L$ is discrete) and so also closed in the Moore plane ("closed in closed is closed") as $L$ is closed.

The fact that we use these two sets specifically has other reasons that will become clear later in the proof.

I personally think that the Jones' lemma proof (see my answer to the linked question in your first line) is the nicer way to go with the proof.