# [SOLVED]Upper Half Disc Topology, Hausdorff not Regular

#### joypav

##### Active member
Problem:
Let X be the upper half plane: $X = \left\{(x, y) \in E^2 : y \geq 0 \right\}$.
Define a basis for the topology as follows. If $P = (p, q)$ and $q > 0$ then if $\epsilon > 0$ then the set
$\left\{(x, y) \in E^2 : \sqrt{(x - p)^2 + (y - q)^2} < \epsilon\right\}\cap\left\{(x, y) : y > 0\right\}$
is a basis element; if $P = (p, q)$ and $q = 0$ then the set
$\left\{(x, y) \in E^2 : \sqrt{(x - p)^2 + (y - q)^2} < \epsilon\right\}\cap\left\{(x, y) : y > 0\right\}\cup\left\{(x, y)\right\}$
is a basis element.
Prove that X is Hausdorff but not regular.

Proof:
"Hausdorff"
Take $P_1, P_2 \in X, P_1 \ne P_2, P_1 = (p_1, q_1), P_2 = (p_2, q_2)$.
Define the distance between any two points $X = (x_1, y_1), Y = (x_2, y_2)$ as follows,
$d(X,Y) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$
Then, $P_1 \neq P_2 \implies d(P_1, P_2) > 0$. Let $d(P_1, P_2) = l > 0$, some $l$.

Choose open sets,
U = $\left\{(x, y) : \sqrt{(x - p_1)^2 + (y - q_1)^2} < l/3\right\}\cap\left\{(x, y) : y > 0\right\}$
V = $\left\{(x, y) : \sqrt{(x - p_2)^2 + (y - q_2)^2} < l/3\right\}\cap\left\{(x, y) : y > 0\right\}$

Obviously,
$P_1 \in U$ and $P_2 \in V$.

Claim: $U \cap V = \emptyset$

BWOC, assume $\exists P_0 \in X$ such that $P_0 \in U \cap V$.
$\implies P_0 \in U$ and $P_0 \in V$
$\implies d(P_0, P_1) < l/3$ and $d(P_0, P_2) < l/3$
But,
$l = d(P_1, P_2) \leq d(P_1, P_0) + d(P_0, P_2) < l/3 + l/3 = 2l/3$, a contradiction.

$\implies U \cap V = \emptyset$
$\implies$ X is Hausdorff.

"Not Regular"
All I need is the existence of one point and one closed set that cannot be separated using open sets.
I've yet to get a good idea for this part. Will I use one of the points on the x-axis?

#### GJA

##### Well-known member
MHB Math Scholar
Hi joypav,

Yes, your intuition is correct. You want to consider a point on the $x$-axis to show that your space $X$ with the given topology is not regular. From here you will want to consider why the basis elements as you've defined them will not allow this to happen.

I do not know if it was a typo in the given problem statement or when you posted to MHB, but I suspect when you define the basis elements in the case where $q=0$, you want to take $(x,0)$ and not $(x,y)$ as part of the intersection.