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:)
How to verify this function is continuous?
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1The example to which you refer looks involved, but if you added a description of it here I think that would better your chances of getting an answer. – 2012-06-11
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0I've tried, however it is a little complex to descripte here indeed. – 2012-06-11
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1I've three books edited by Arkhangel'skii, none of which is the one you mention. If you don't write down that example the ammount of people understanding what you are talking about might be, I'm afraid, pretty reduced and, thus, the probability to get some help is going to be pretty tiny. – 2012-06-11
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1The paper is linked, and appears to be open-access (I wasn't asked for any credentials). – 2012-06-11
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0@DonAntonio: I included the description in my answer, which makes the whole thing self-contained. – 2012-06-11
1 Answers
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
$$V_n(x)=\{x\}\cup\left\{\langle s,t\rangle\in U:t=s-a\text{ and }0
$$V_n(x')=\{x\}\cup\left\{\langle s,t\rangle\in U:t=a-s\text{ and }0 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. 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.