Let $G$ be a group and $\mathcal T$ be a topology on $G$ and the function $$ \begin{align*} &f:G\times G\to G\\ &f(x,y)=xy^{-1} \end{align*} $$ be continuous at $(1,1)$.

Is $(G,\mathcal T)$ completely regular?

  • $\begingroup$ No: Take the trivial topology on G. $\endgroup$ – Moishe Kohan Sep 1 '14 at 22:22
  • $\begingroup$ The trivial topology is completely regular. With it the group is a topological group and any topological group is completely regular. $\endgroup$ – Minimus Heximus Sep 2 '14 at 2:13
  • $\begingroup$ Completely regular includes Hausdorff. $\endgroup$ – Moishe Kohan Sep 2 '14 at 3:01
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    $\begingroup$ I accept wikipedia's definition: en.wikipedia.org/wiki/Tychonoff_space. Otherwise I would say Tychonov. $\endgroup$ – Minimus Heximus Sep 2 '14 at 3:03

For $n\in\Bbb N$ let $B_n=\{k\in\Bbb Z:|k|\ge n\}$. (Note: $0\in\Bbb N$.) Let $G$ be $\Bbb Z$ under addition, and let $\mathscr{T}=\{\varnothing\}\cup\{B_n:n\in\Bbb N\}$. The only nbhd of $0$ is $B_0=\Bbb Z$ itself, so $$f:G\times G\to G:\langle x,y\rangle\mapsto x-y$$ is continuous at $\langle 0,0\rangle$.

$B_1$ is an open nbhd of $1$ whose complement is $\{0\}$; suppose that $\varphi:G\to[0,1]$ is such that $\varphi(1)=0$ and $\varphi(0)=1$. Let $U=(0,1]$, and note that $U$ is an open nbhd of $1$ in $[0,1]$. But $0\in\varphi^{-1}[U]$, and $1\notin\varphi^{-1}[U]$, so $\varphi^{-1}[U]\ne G$ is not a nbhd of $0$, and $\varphi$ is not continuous. Thus, $G$ is not completely regular.

  • $\begingroup$ @user138171: You’re welcome. I haven’t been around since last December, so I’m working my way back through the unanswered general topology questions to see if I can dispose of some of them. $\endgroup$ – Brian M. Scott Oct 5 '14 at 23:24

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