How to verify this function is continuous? Recently, I'm reading a paper "Spaces with a regular Gδ-diagonal" of A.V.Arhangel’skii's. I can't understand the function $d$ in the example 9 is continuous. Could someone help me? Thanks ahead:)
 A: The space is $X=X_0\cup X_1\cup U$, where $X_0=\Bbb R\times\{0\}$, $X_1=\Bbb R\times\{-1\}$, and $U=\Bbb R\times(0,\to)$. 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
$$V_n(x)=\{x\}\cup\left\{\langle s,t\rangle\in U:t=s-a\text{ and }0<t<\frac1n\right\}$$ and 
$$V_n(x')=\{x\}\cup\left\{\langle s,t\rangle\in U:t=a-s\text{ and }0<t<\frac1n\right\}\;.$$ 
The topology $\tau$ on $X$ is obtained by isolating each point of $U$ and taking 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.

Added: In other words, $X$ is the closed upper half-plane with the $x$-axis doubled. Points in the open upper half-plane are isolated; basic open nbhds of points in one copy, $X_0$, of the $x$-axis are spikes diagonally up and to the right; and basic open nbhds of points in the other copy, $X_1$, of the $x$-axis are spikes diagonally up and to the left.

Now define $d:X\times X\to\Bbb R$ as follows.


*

*If $x\in X_0$ and $y=\langle s,t\rangle\in V_1(x)\setminus\{x\}$, let $d(x,y)=d(y,x)=t$.  

*If $x'\in X_1$ and $y=\langle s,t\rangle\in V_1(x')\setminus\{x'\}$, let $d(x',y)=d(y,x')=t$.  

*For distinct $y=\langle s,t\rangle,z=\langle u,v\rangle\in V_1(x)\setminus\{x\}$, let $d(y,z)=\max\{t,v\}$.  

*For distinct $y=\langle s,t\rangle,z=\langle u,v\rangle\in V_1(x')\setminus\{x'\}$, let $d(y,z)=\max\{t,v\}$.  

*For all other distinct $x,y\in X$ let $d(x,y)=1$, and of course $d(x,x)=0$ for all $x\in X$.


The claim in the paper is that $d$ is a continuous symmetric that generates $\tau$. That it’s a symmetric is obvious, and it’s not hard to check that it generates $\tau$.
$X$ is first countable, so it suffices to show that that if $\langle x_n:n\in\Bbb Z^+\rangle\to x$ and $\langle y_n:n\in\Bbb Z^+\rangle\to y$, then $\langle d(x_n,y_n):n\in\Bbb Z^+\rangle\to d(x,y)$. This is easily done by cases.


*

*If $x$ and $y$ are isolated points, we may assume without loss of generality that $x_n=x$ and $y_n=y$ for all $n\in\Bbb Z^+$, so that $d(x_n,y_n)=d(x,y)$ for all $n$, and the result is trivial.

*Suppose that $x\in X_0$ and $y=\langle s,t\rangle\in V_1(x)\setminus\{x\}$. Then $y$ is an isolated point in $X$, so we may assume that $y_n=y$ for all $n\in\Bbb N$. Either $\langle x_n:n\in\Bbb N\rangle$ is eventually constant at $x$, in which case the result is trivial, or we may assume (by passing to a subsequence if necessary) that $x_n=\langle s_n,t_n\rangle\in V_n(x)\setminus\{y\}$ for $n\in\Bbb Z^+$. But then $$d(x_n,y_n)=d(x_n,y)=\max\{t_n,t\}=t=d(x,y)$$ for all sufficiently large $n$, since $\langle t_n:n\in\Bbb Z^+\rangle\to 0$. The case of $x'\in X_1$ and $y\in V_1(x)\setminus\{x'\}$ is exactly similar.

*Suppose that $x$ and $y$ are distinct points of $X_0$. If both sequences are eventually constant, the result is trivial, so assume that the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ is non-trivial. As before, we may assume that $x_n=\langle s_n,t_n\rangle\in V_n(x)$. We may further assume that $y_n\in V_1(y)$ for all $n\in\Bbb Z^+$. Then for all $n\in\Bbb Z^+$ we have $y_n\notin V_1(x)$, so $d(x_n,y_n)=1=d(x,y)$ by clause (5) of the definition of $d$. The case of distinct points of $X_1$ is exactly similar.

*Suppose that $x\in X_0$ and $y=x'\in X_1$. Then for each $u\in V_1(x)$ and $v\in V_1(x')$ we have $d(u,v)=1$, since $u$ and $v$ don’t fall under any of the first four clauses in the definition of $d$, and the result is trivial.

*Finally, suppose that $x\in X_0$, $y\in X_1$, and $y\ne x'$. The sets $V_1(x)$ and $V_1(y)$ intersect in at most one point, and there is an $n_0\in\Bbb Z^+$ such that $V_{n_0}(x)\cap V_{n_0}(y)=\varnothing$. Thus, we may without loss of generality assume that $x_n\in V_{n_0}(x)$ and $y_n\in V_{n_0}(y)$ for all $n\in\Bbb Z^+$. But then, just as in Case 3, $d(x_n,y_n)=1=d(x,y)$ for all $n\in\Bbb Z^+$, and we’re done.
