# zeroset-diagonal

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

what if X is additionally CCC?

• 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. – Joel David Hamkins Oct 18 '11 at 1:58
• 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.) – Todd Eisworth Oct 18 '11 at 13:49

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 $f=\sum_{n=1}^\infty 2^{-n}f_{(d_n,e_n)}$.
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.
• Compact spaces aren't going to work, as compact spaces with $G_\delta$-diagonal are metrizable. – Todd Eisworth Oct 18 '11 at 13:46