A Regular Space that isn't Tychonoff A known example of a regular space that isn't Tychonoff is the "Mysior Plane" $X=X_0\cup\{p\}$, where
$$X_0=\{(x,y)\in\mathbb{R}^2 \ |\ y\geq 0\}$$
and $p\notin X_0$. The topology here is defined by declaring every point $(x,y)$ with $y>0$ to be isolated, an open set around a point $(x,0)$ be on the form $U_x\setminus F$, where
$$U_x=\{(x,y)\ |\ 0\leq y\leq 2\}\cup\{(x+y,y)\ |\ 0\leq y\leq 2\}$$
and $F$ is a finite set, and a neighborhood of $p$ to be on the form $U_n=\{p\}\cup\{(x,0)\ |\ x\geq n\}$ for $n\in\mathbb{N}$.
However, I'm having some trouble showing this space in not Tychonoff:
Let $H_j=\{(x,0)\ |\ j-1\leq x \leq j\}$, $H$ be the $x$-axis and let $f:X\rightarrow\mathbb{R}$ be a continuous function such that $f(H_1)=0$. It can be shown that the sets $K_j=f^{-1}(0)\cap H_j$ are infinite for each $j\in\mathbb{R}$. Why does that implies that $f(p)=0$?
 A: You have a typo in the definition of $K_j$: $K_j=f^{-1}[\{0\}]\cap H_j$. The proof that $K_j$ is infinite for each $j\in\Bbb Z^+$ is by induction on $j$. Clearly $K_1=H_1$ to get the induction started. For the induction step it will be convenient to let
$$U_x^-=\{\langle x,y\rangle:0\le y\le 2\}$$
and
$$U_x^+=\{\langle x+y,y\rangle:0\le y\le 2\}$$
for each $x\in\Bbb R$.
Now assume that $K_n$ is infinite. Then there is a countably infinite $C\subseteq K_n$. Let $\langle c,0\rangle\in C$. Suppose that $G$ is a $G_\delta$-set containing $\langle c,0\rangle$, i.e., $G=\bigcap_{n\in\Bbb Z^+}G_n$ for some open nbhds $G_n$ of $\langle c,0\rangle$. Then each $G_n$ contains all but finitely many points of $U_c^+$, so $G$ contains all but countably many points of $U_c^+$. Now
$$\begin{align*}
U_c^+\setminus f^{-1}[\{0\}]&=\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left[\frac1n,\to\right)\right]\right)\\
&\qquad\cup\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left(\leftarrow,-\frac1n\right]\right]\right)
\end{align*}$$
is a subset of $U_c^+$ that is a countable union of closed sets and does not contain $\langle c,0\rangle$, so its complement is a $G_\delta$-set containing $\langle c,0\rangle$. We’ve just seen that this complement contains all but countably many points of $U_c^+$, so $U_c^+\setminus f^{-1}[\{0\}]$ must be countable. $C$ is countable, so $B=\bigcup\limits_{\langle c,0\rangle\in C}\big(U_c^+\setminus f^{-1}[\{0\}]\big)$ is countable, and so is
$$P=\{\langle x,0\rangle:\langle x,y\rangle\in B\}\,.$$
Let $H=H_{n+1}\setminus P$; clearly $H$ is infinite. (In fact $|H|=|\Bbb R|$.)
Now let $\langle x,0\rangle\in H$. Then
$$U_x\cap U_c^+\cap f^{-1}[\{0\}]\supseteq U_x^-\cap U_c^+\cap f^{-1}[\{0\}]\ne\varnothing$$
for each $\langle c,0\rangle\in C$. That is, every open nbhd of $\langle x,0\rangle$ meets $f^{-1}[\{0\}]$, and $f^{-1}[\{0\}]$ is closed, so $\langle x,0\rangle\in f^{-1}[\{0\}]$. Thus, $H\subseteq f^{-1}[\{0\}]$, and therefore $K_{n+1}=H_{n+1}\cap f^{-1}[\{0\}]$ is infinite. This completes the induction step.
We now know that $K_n$ is infinite for each $n\in\Bbb Z^+$. In particular,
$$U_n\cap f^{-1}[\{0\}]\ne\varnothing$$
for each $n\in\Bbb Z^+$, so $p\in\operatorname{cl}f^{-1}[\{0\}]=f^{-1}[\{0\}]$, i.e., $f(p)=0$.
You may also find the answers to this question of some interest.
