Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:)

share|improve this question
    
The question was reposted at MO. –  Martin Sleziak Jun 20 '12 at 11:43
add comment

2 Answers

up vote 1 down vote accepted
+250

Suppose there is a continuous function $f:Z^2\to[0,1]$, with $f^{-1}(\{0\})=\{(z,z):z\in Z\}$.

  1. 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 $d_m(x)=0$.

  2. 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.

  3. 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.

share|improve this answer
add comment

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.

share|improve this answer
    
Thanks Brain;however, you haven't know clearly what i searched. I want to know this space whether has a zeroset diagonal.:) –  Paul Jun 22 '12 at 1:05
    
I even find this space has a regular $G_\delta$ diagonal. But whether having a zeroset is puzzling me. –  Paul Jun 22 '12 at 1:13
    
@John: Sorry: for some reason I read zeroset and thought $G_\delta$. I’ll think about it some more. –  Brian M. Scott Jun 22 '12 at 12:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.