Does this space have a zeroset diagonal? Recently I constructed a special topology space which is modifying the example 2.17 of Arhangel'skii as follows :
The space is $ Z=X_0\cup X_1\cup X$, where $X_0=\Bbb R\times\{0\}$, $X_1=\Bbb R\times\{-1\}$, and $X=\Bbb R\times(0,\to)$. For $x=\langle a,0\rangle\in X_0$ let $x'=\langle a,-1\rangle\in X_1$. For $n\in\Bbb Z^+$ and $x=\langle a,0\rangle\in X_0$ let
$V_n(x)=\{x\}\cup$ a triangle which has the sides adjacent to the vertex $x$ of equal length and an angle at $x$ of measure $30^0$ with height equal to $\frac1n$ and height slope equal to $1$.
$V_n(x')=\{x'\}\cup$ a triangle which has the sides adjacent to the vertex $x$ of equal length and an angle at $x$ of measure $30^0$ with height equal to $\frac1n$ and height slope equal to $-1$.
The topology on $Z$ obtained by this: The topology $\tau$ on $X$ is the smallest topology obtained by $\tau_1\cup \tau_2$, where $\tau_1$ is Euclidean topology on $X$ and  $\tau_2$ is the topology of countable complements on $X$ (Much more details see 63# of counterexample in topology); and we take the families $\{V_n(x):n\in\Bbb Z^+\}$ and $\{V_n(x'):n\in\Bbb Z^+\}$ as local bases at $x\in X_0$ and $ x' \in X_1$, respectively. 
Does this space have a zeroset diagonal? Any idea will be appreciated:)
 A: Yes. For $n\in\Bbb Z^+$ let 
$$\mathscr{G}_n=\Big\{V_n(x):x\in X_0\Big\}\cup\Big\{V_n(x'):x;\in X_1\Big\}\cup\left\{B\left(x,\frac1n\right):x\in X\right\}\;,$$
where $B(x,\epsilon)$ is the open $\epsilon$-ball centred at $x$ in the Euclidean topology. Clearly each $\mathscr{G}_n$ is an open cover of $Z$, and for each $x\in Z$,
$$\bigcap_{n\in\Bbb Z^+}\operatorname{st}(x,\mathscr{G}_n)=\{x\}\;.$$
It’s well-known that the existence of a $G_\delta$-diagonal is equivalent to the existence of a countable family of open covers with this property, so $Z$ has a $G_\delta$-diagonal.
A: Suppose there is a continuous function $f:Z^2\to[0,1]$, with $f^{-1}(\{0\})=\{(z,z):z\in Z\}$.


*

*Define $d_m(x)$ for positive integer $m$ and $x\in X_0$ in the following way:


*

*If $\min (f(x,x'),f(x',x))>1/m$, then for $n$ large enough:
$$V_n(x)\times V_n(x)\ \cup\ V_n(x')\times V_n(x')\subseteq f^{-1}([0,1/m))\\
V_n(x)\times V_n(x')\ \cup\ V_n(x')\times V_n(x)\subseteq f^{-1}((1/m,1])$$
Let $d_m(x)=1/n$ for such an $n$.

*Otherwise, let $d_m(x)=0$.


*Consider the triangles $V_1((0,0))$ and $V_1((\varepsilon,0))$. They intersect for small enough $\varepsilon>0$, let $k=1/\varepsilon$.
Suppose $1/n=\min(d_m(x),d_m(y))\ge k\cdot|x-y|$ with $x>y$.
Then $V_n(x)$ and $V_n(y)$ intersect, as do $V_n(x')$ and $V_n(y)$. But
$$V_n(y)\times V_n(y)\subseteq f^{-1}([0,1/m))$$
while
$$V_n(x)\times V_n(x')\subseteq f^{-1}((1/m,1])$$
which is a contradiction.

*So we have $d_m(y)<k\cdot|x-y|$ whenever $k\cdot|x-y|\le d_m(x)$ and $x\ne y$. This means that there cannot be more than $1+t/k$ points where $d_m(x)\ge t$, so that there are only countably many elements in $$X(m)=\{x\in X_0: d_m(x)>0\}$$
Since $\bigcup_m X(m)$ is countable, there must be some $x$ such that $d_m(x)=0$ for all $m$: but then $f(x,x')=0$ or $f(x',x)=0$, which is impossible.
Therefore, the space does not have a zeroset diagonal.
