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The Tangent Disc Space, Application of Baire Category Theorem

joypav

Active member
Mar 21, 2017
151
Showing that the Moore Plane is not Normal

I am working on constructing a $T_3$ space that is not $T_4$.
I've started with $X$, the upper half plane.
$L=\left\{(p,q):q=0\right\}$, the x-axis.
And the basis for the topology are open balls centered at points $(p,q), q>0$ intersected with $\left\{(x,y):y>0\right\}$,
and tangent discs centered at $(p,\epsilon)$ of radius $\epsilon$ union with $(p,0)$.

My professor gave some "steps" to follow. I've shown that (Step 1) the basis induces a topology on X, (Step 2) every subset of L is closed, and (Step 3) that the space is Hausdorff and regular. Now I need to show that it is not normal.

The next step says this...

Let $H=\left\{(x,y) \in X: y=0, x \in Q\right\} ; K=\left\{(x,y) \in X: y=0, x \in R-Q\right\}$.
We are going to show that these are two closed sets that cannot be separated by disjoint open sets (not normal).

He suggests to proceed by way of contradiction and to use the fact that in the usual topology of the reals that L is a Baire Space.

So...
I will suppose that $U$ and $V$ are disjoint open sets in the topology of X containing $H$ and $K$ respectively.
For each $P \in K$ let $\epsilon_P$ denote a number $\epsilon$ so that $B_\epsilon(P) \subset V$.

Step 4 is to show that...
if $n \in \Bbb{N}$ then the set $L_\epsilon\left\{P \in L\cap\left(R-Q\right) : \epsilon_P > 1/n\right\}$ is nowhere dense in the usual topology of the reals.

Say I approach Step 4 by way of contradiction. Meaning, I assume $\exists n \in \Bbb{N}$ so that $L_\epsilon$ is not nowhere dense.
$\implies$ there is an open interval contained in $\overline{L_\epsilon}$.

Now, choose a rational $q$ such that $q$ is in this open interval, say $I^0$.
Now I want to show that any open neighborhood of $q$ will intersect $U$. i.e. $U$ and $V$ are not disjoint. A contradiction.

How do I go about showing this?


(After this, I can say that this is a contradiction of the Baire Category Theorem $\implies X$ is not normal.)
 
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