Let $(G,\mathcal T)$ be a completely regular topological space. Is there a group structure on $G$ such that the function $$f:G\times G\to G$$ $$f(x,y)=xy^{1}$$ is continuous at $(1,1)$?

$\begingroup$ completely regular spaces need not be Hausdorff. $\endgroup$ – Minimus Heximus Sep 1 '14 at 18:15

$\begingroup$ The way you've phrased the question seems odd. Do you mean instead: "if G is a group equipped with a completely regular topology for which $f$ is continuous, what can we say about $G$?" Or are you really asking whether arbitrary completely regular spaces admit some kind of group structure, possibly a stupid or unnatural one, for which $f$ is continuous? $\endgroup$ – Yemon Choi Sep 1 '14 at 18:41

2$\begingroup$ @YemonChoi $G$ is a set. $\mathcal T$ is a completely regular topology on it. I'm looking for the stupid or unnatural structure. I asked your first question in math.se: math.stackexchange.com/questions/914517/… $\endgroup$ – Minimus Heximus Sep 1 '14 at 18:48

15$\begingroup$ Not if the identity 1 is specified in advance: the set $\{1\} \cup \{ 1 + \frac{1}{n}: n \in {\bf N}\}$ with the usual topology is a counterexample. (The problem here is that all the neighbourhoods of the identity are both infinite and cofinite, and in an infinite group the product of two cofinite sets is necessarily the whole group.) To extend this example to the case where 1 is not specified in advance, one needs a space where every point has a neighbourhood in which all smaller neighbourhoods have complements of strictly smaller cardinality. $\endgroup$ – Terry Tao Sep 1 '14 at 22:19

$\begingroup$ Depending on the interpretation of "all smaller neighborhoods have complements of strictly smaller cardinality", this may sound like the space is not Hausdorff. Actually, if all closed sets, different from the whole infinite space $G$, have their cardinalities smaller that $G$ then $G$ is not a Hausdorff space. (Just a trivial remark). $\endgroup$ – Włodzimierz Holsztyński Sep 2 '14 at 3:23
Here is a partial positive answer, going in the opposite direction of Terry´s comment.
Let $X$ be an infinite topological space. Suppose $X$ is first countable at $p \in X$ and for every open neighborhood $U$ of $p$ there is a smaller neighborhood $V \subseteq U$ such that $U \setminus V=U$. Then there is a group structure on $X$ with identity $p$ such that the operation $(x,y) \mapsto xy^{1}$ is continuous at $(p,p)$.
Just note that (starting with a $U_0$ of minimal cardinality) we can find $\{U_n : n \in \omega\}$ an open local base at $p$ such that for every $n \in \omega$ we have $U_{n+1} \subset U_n$ and $U_n \setminus U_{n+1}=U_0$. Now find (e.g. using compactness of firstorder logic) a group $G$ containing subgroups $H_0 \supset H_1 \supset H_2 \supset \cdots$, such that $G=X$ and $H_n \setminus H_{n+1}=H_0=U_0$ for every $n \in \omega$, and then transfer the group structure to $X$ in the obvious way.
Note that we can also change the hypothesis to: there is an open neighborhood of $p$ with no smaller neighborhoods (e.g. $p$ is isolated) and the conclusion still holds; just call $U_0$ such neighborhood, find $G$ and $H_0$ as before and ignore the rest.
Edit (Some details as to why such $G$ exists): Let $\kappa=X$ and $\mu=U_0$. Consider the language of first order logic consisting of a binary function symbol (for the group operation), countably many unary predicate symbols $\{P_n : n \in \omega\}$ and constant symbols $\{c^n_\alpha : \alpha \in \mu, n \in \omega\}$. In this language consider the first order theory that includes the group axioms, for each $n$ the axioms saying that $P_n$ is a subgroup and $P_{n+1} \subseteq P_n$, axioms saying that all the $c^n_\alpha$'s are distinct and $c^n_\alpha \in P_{n} \setminus P_{n+1}$. This theory is consistent because each finite fragment of it can be satisfied, e.g. using $\mathbb{Z}$ and finitely many of its subgroups. Since the language has size $\mu$, there is (by LowenheimSkolem) a model for this theory of size $\mu$. Inside this model we can find our $H_n$´s as the interpretations of the $P_n$'s. Finally we let $G=H_0 \times K$ where $K$ is any group of size $\kappa$.


1$\begingroup$ @unser47958, I added some details about that. $\endgroup$ – Ramiro de la Vega Sep 2 '14 at 19:08

2$\begingroup$ Instead of the compactness argument, you can e.g. take for $H_0$ the direct sum of $\omega$ copies of a group of size $\mu$, and for $H_n$ its subgroup omitting the first $n$ copies. $\endgroup$ – Emil Jeřábek Sep 2 '14 at 19:33

$\begingroup$ That´s much simpler @Emil, thank you. $\endgroup$ – Ramiro de la Vega Sep 2 '14 at 19:36