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Suppose $\mathcal T$ and $\mathcal S$ are two compatible Hausdorff topologies on a group $G$ and $\mathcal R$ is a maximum compatible topology on $G$ with $\mathcal R \subseteq \mathcal T\cap \mathcal S$.

Is $(G,\mathcal R)$ Hausdorff?

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Is there some reason for the strange spelling in the title? "I am moft ferioufly difpleafed". –  Todd Trimble Mar 22 '14 at 6:43
@ToddTrimble: there'ff ffome. –  user47958 Mar 22 '14 at 6:55
What does it mean for two topologies to be compatible? (That for both topologies $G$ is a topological group?) –  jmc Mar 22 '14 at 7:41
Is there some reason for the strange spelling in the title? “I om most seriously displeosed”. –  jmc Mar 22 '14 at 7:42
Topolgies are compatible with the group, making it a topological group. –  user47958 Mar 22 '14 at 7:49

2 Answers 2

up vote 3 down vote accepted

It seems the answer is no. Basically, the $p$-adic and Archimedean topologies on $\mathbb{Q}$ are incompatible enough that the maximal compatible topology contained in both of them is indiscrete. Here is an outline of the proof (I'm heading to bed so I haven't filled in the details):

$\bullet$ Any subset of $\mathbb{Q}$ which is open for both topologies must be dense for both topologies. Therefore, all elements of $\mathcal{R}$ are dense.

$\bullet$ If $\mathcal{R}$ is compatible with the group structure, the complement of the antidiagonal in $\mathbb{Q}\times \mathbb{Q}$ must be open. Since this complement contains e.g. $(1,1)$, it must contain a set of the form $U\times U$ for some $U\in \mathcal{R}$.

$\bullet$ This is impossible since $U\cap U^{-1}\neq \emptyset$ by density of $U$.

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What I remember from my notes on uniform spaces, is that $p$-adic uniformity on $\Bbb Q$ is the uniformity generated by $\{W_n\mid n\in \Bbb N \}$ where $W_n=\{(r,s)\in \Bbb Q^2\mid p^n|r-s\}$. Please give me definitions of Archimedean and $p$-adic topologies on $\Bbb Q$. –  user47958 Mar 22 '14 at 14:14
I think he meant to write "euclidean" instead of "archimedean". –  Johannes Hahn Mar 22 '14 at 18:38
Yes, by archimedean I meant the restriction of the usual topology on $\mathbb{R}$. For the definition of the $p$-adic topology, see –  Kevin Ventullo Mar 22 '14 at 21:50

Edit: This answer is completely wrong. Please ignore it, and see Kevin's answer.

A $T_{0}$-topological group $G$ is Hausdorff ($T_{2}$). Recall that $T_{0}$ means that for all $x,y \in G$, there is an open subset $U$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$.

Consequently (by homogeneity), a topological group is Hausdorff if and only if the one element subgroup of the unit element is closed.

Thus, for your question it suffices to show that $\{1\} \subset G$ is closed for the $\mathcal{R}$-topology. This is clear if we show that $\mathcal{R} = \mathcal{T} \cap \mathcal{S}$. Let $U$ be an open subset for $\mathcal{T}$ and $\mathcal{S}$. For every element $g \in G$, the set $gU$ is open for $\mathcal{T}$ and $\mathcal{S}$, because both were compatible. For the same reason $U^{-1}$ is open for both $\mathcal{T}$ and $\mathcal{S}$. Hence $\mathcal{R} = \mathcal{T} \cap \mathcal{S}$ is compatible.

Consequently $U = G - \{1\}$ is an open subset for $\mathcal{R}$, and hence $(G,\mathcal{R})$ is Hausdorff.

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For showing compatibility, why is it enough to show that $gU$ is open for all $g$ and $U^{-1}$ is open? For instance, if I take an abstract infinite group and put the cofinite topology on it, then both of those conditions are satsified, but the cofinite topology is not compatible. –  Kevin Ventullo Mar 22 '14 at 8:43
Let me be seriously displeosed too! There are examples which show the intersection of two compatible topologies is not necessarily compatible. you have proved inverse, and left (right) multiplication are continuous. While continuity of the 2-variable multiplication function needs a serious proof. –  user47958 Mar 22 '14 at 8:50
Hmm, you are very right! I don't know what made me think this is enough. Excuse my stupidity. –  jmc Mar 22 '14 at 9:00
You're answer is completely wrong, but if you add local compactness to $\mathcal R$, It may be completely correct according to a theorem due to Ellis. Good point for me, why I +1. –  user47958 Mar 22 '14 at 12:13
@MinimusHeximus — That is very generous. Thank you. –  jmc Mar 22 '14 at 12:19

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