Timeline for Group structure on an arbitrary completely regular topological space that makes $(x,y)\mapsto xy^{-1}$ continuous at $(1,1)$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2014 at 12:31 | comment | added | YCor | @WlodzimierzHolsztynski: I first understood as you do, but Tao's statement is "if every point $x$ has a neighborhood $V$ such that every neighborhood $W$ of $x$ with $W\subset V$, we have $\#(W\smallsetminus V)<\#(G)$. Still I don't see immediately why this would answer the question. | |
| Sep 2, 2014 at 16:45 | answer | added | Ramiro de la Vega | timeline score: 3 | |
| Sep 2, 2014 at 3:23 | comment | added | Włodzimierz Holsztyński | 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). | |
| Sep 2, 2014 at 2:10 | history | edited | Minimus Heximus | CC BY-SA 3.0 |
edited title
|
| Sep 1, 2014 at 23:41 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
more accurate title
|
| Sep 1, 2014 at 22:30 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
improved title
|
| Sep 1, 2014 at 22:19 | comment | added | Terry Tao | 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. | |
| Sep 1, 2014 at 19:08 | review | Close votes | |||
| Sep 2, 2014 at 11:43 | |||||
| Sep 1, 2014 at 18:48 | comment | added | Minimus Heximus | @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/… | |
| Sep 1, 2014 at 18:41 | comment | added | Yemon Choi | 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? | |
| Sep 1, 2014 at 18:21 | history | edited | Minimus Heximus | CC BY-SA 3.0 |
edited title
|
| Sep 1, 2014 at 18:15 | comment | added | Minimus Heximus | completely regular spaces need not be Hausdorff. | |
| Sep 1, 2014 at 18:13 | history | asked | Minimus Heximus | CC BY-SA 3.0 |