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Is it true that a topology space X with a zeroset diagonal is first countable?

what if X is additionally CCC?

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    $\begingroup$ Your question is interesting, but in order to attract more attentive answers, I would encourage you to edit your question to provide a bit more explanation. For example, a topological space $X$ has a zeroset diagonal when there is a continuous function $f:X\times X\to [0,1]$ with $\Delta=f^{-1}(0)$, where $\Delta=\{(x,x)\mid x\in X\}$ is the diagonal. First countable= every point has a countable local basis. CCC = countable chain condition = every family of disjoint open sets is countable. $\endgroup$
    – Joel David Hamkins
    Commented Oct 18, 2011 at 1:58
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    $\begingroup$ I would supplement Joel's statement by also remarking that without any information about motivation, no one can tell if this is a question arising in research, or simply a question arising from a problem set in a topology course. (I'm sensitive to this --- a few years ago, we had a student who posted their problem sets to the 'Ask a topologist' website.) $\endgroup$
    – Todd Eisworth
    Commented Oct 18, 2011 at 13:49

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All it takes is a countable, not first-countable, Tychonoff space, say a countable dense subset, $D$, of the Cantor cube $2^{\mathfrak{c}}$. For every point $(d,e)$ off the diagonal there is a continuous function $f_{(d,e)}$ on $D^2$ that is zero on the diagonal and has value $1$ at $(d,e)$. Now enumerate the complement of the diagonal as $\lbrace(d_n,e_n):n\in\mathbb{N}\rbrace$ and define $f=\sum_{n=1}^\infty 2^{-n}f_{(d_n,e_n)}$.

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  • $\begingroup$ Thanks. I'm very appreciate your idea of constructing the counterexample for me. It's very useful and simple! $\endgroup$
    – Paul
    Commented Oct 20, 2011 at 1:03
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I don't think this is true. You just need to find a space $X$ with $X\times X$ $T_6$ (a.k.a. perfectly normal) but not first countable. A space is $T_6$ if it's normal and every closed set is a $G_\delta$ set. These spaces have the property that for any two disjoint closed sets $A$ and $B$ you can find a map $f:X\rightarrow [0,1]$ with $f(A)=0$, $f(B)=1$, and $0 < f(x) < 1$ for $x\not \in A\cup B$. So such a space would have an $f$ taking the diagonal to $0$ and everything else to a positive number.

According to Counterexamples in Topology, Appert Space is an example of a space which is $T_6$ but not first countable (there are many other examples, see the table in the back). The problem is that products do not preserve the $T_6$ property, so you can't say anything about $X\times X$ for $X$ being Appert space. I'm too tired now to figure out whether or not $X\times X$ is $T_6$, but if it's not, I suggest looking at Fort Space and Arens Space. A useful paper on the interplay between $T_6$ and countability axioms can be found here.

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    $\begingroup$ I thought about this some more and was trying to come up with a space like I describe above via an uncountable discrete topology. This won't work as stated because the discrete topology is metrizable, hence first countable. However, a 1-point compactification might work. I think the problem now has been reduced to looking in Counterexamples in Topology $\endgroup$
    – David White
    Commented Oct 18, 2011 at 12:02
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    $\begingroup$ Compact spaces aren't going to work, as compact spaces with $G_\delta$-diagonal are metrizable. $\endgroup$
    – Todd Eisworth
    Commented Oct 18, 2011 at 13:46
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    $\begingroup$ There's a whole family of results in general topology of this sort, and it's an area of current research (Raushan Buzyakova at UNC Greensboro has been especially active of late). But the OPs questions have had so little context that I don't feel good about answering them, as they look like the kind of HW questions one gets in a good graduate course on general topology. $\endgroup$
    – Todd Eisworth
    Commented Oct 18, 2011 at 15:09
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    $\begingroup$ I see what you mean about his former questions. I didn't realize that when I answered. Just got an idea and decided to go for it. If this continues to get no upvotes, perhaps I'll delete it and see if we can get the question closed. $\endgroup$
    – David White
    Commented Oct 18, 2011 at 19:58

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