Show that the Niemytzki's tangent disc topology is completely regular.

Question

I need to show that the Niemytzki's tangent disc topology is completely regular. In by attempt I could show that there is a Urysohn function for a point $b$ on the upper half plane $\{(x,y): x\in\mathbb{R},y>0\}$ and a closed set $F$ that does not contain $b$. However I am stuck in the case where $b$ is a point on the line $y=0$. Could someone please help? Thanks.

Answer 1

As it is a Hausdorff space (why?) it suffice to show that it is functionally regular,
Let $F$ be a closed non-empty subset of $(Y,\mathcal{T}_N)$ (being the upper half place with the Niemytzki's topology) and $p=(x,y)$ a point not in $F$, we want to find a continuous function which completely separates $F$ and $p$, we consider the two cases;

• If $p\in Y'$ (denote by Y' the open upper half plane and D the x-axis line), recall that $Y\setminus F$ is open, so there exist an open $U$ in the canonical topology such that $p\in U \subseteq Y\setminus F$, so $Y\setminus U$ is closed in both the Moore topology and the canonical topology, following regularity of the latter there exists a Urysohn function $f$ such that $f(Y\setminus U)= 0,\ and\ f(p)= 1$, $f$ is continuous relatively to the canonical topology, constant on $D$ as $f(D)=0$ and vary only in $U\subset Y'$ hence continuous with respect to $\mathcal{T}_N$ (why?), as $F\subseteq Y\setminus U$, the same function is a Urysohn function for $F$ and $p$.
• If $p\in D$, we can have an element from the basis that contains $p$ and does not intersect $F$ (since $F$ is closed), say $B_p= B((x_0,\epsilon), \epsilon)\cup \{p\}$, let us define a function $f$ that satisfy $f(p)= 0$, $f(q)= 1\ if\ q\notin B_p$, and at any point $(x,y)$ inside $B_p$\ ; $f(x,y)= \frac{1}{2 y\epsilon}\big((x-x_0)^2+ y^2\big) $.
  We claim that $f$ is continuous, to see this, notice that $f^{-1}([0,\alpha))= \{p\}\cup B\big((x_0,\alpha\epsilon), \alpha\epsilon\big)$ (why?) is open, and $f^{-1}((\alpha,1])= Y\setminus \overline{B\big((x_0,\alpha\epsilon), \alpha\epsilon\big)}$ is also open,
  we conclude that $f$ is a Urysohn function for $p$ and $F$.