One example of a subset of a group $G$ which has to be closed in any topology on $G$ compatible with the group operations is a centraliser. Are there any other interesting examples?
$\begingroup$
$\endgroup$
9
-
$\begingroup$ What is "topology compatible with the group operations" and why does the centralizer have to be closed? E.g., is the anti-discrete topology compatible? $\endgroup$– Alex DegtyarevCommented Jan 22, 2014 at 10:39
-
1$\begingroup$ First: this only holds if the group is Hausdorff. Second: any set which is described by quant or-free equations in the language of groups. Third: is this a research-related question? $\endgroup$– user1688Commented Jan 22, 2014 at 10:40
-
$\begingroup$ @Alex: compatible means that the group is a topological group, i.e., the group operations are continuous. $\endgroup$– user1688Commented Jan 22, 2014 at 10:43
-
6$\begingroup$ Voting to reopen in light of Anton's nice answer. $\endgroup$– Benjamin SteinbergCommented Jan 22, 2014 at 22:00
-
1$\begingroup$ Anton, yes, this is related to a research problem I am working on about the rigidity of the group topology on certain locally compact groups, and yes, you are right that I should have said Hausdorff. $\endgroup$– RupertCommented Jan 23, 2014 at 12:02
|
Show 4 more comments
2 Answers
$\begingroup$
$\endgroup$
1
Subsets of a group that are closed with respect to any Hausdorff group topology are called unconditionally closed.
Clearly, all algebraic sets are unconditionally closed, where a subset of a group $G$ is called algebraic if it is an intersection of finite unions of the sets of solutions to some equations with coefficient from $G$.
A.A.Markov proved that for countable groups the converse is also true: $$ \hbox{unconditionally closed = algebraic}. $$ For uncountable groups, this is not always the case as follows from works of S. Shelah (under CH) and G. Hesse.
-
$\begingroup$ Klyachkon: Is there any English text about the work of Markov? Is there similar works for other algebraic systems? $\endgroup$– Sh.M1972Commented Jan 23, 2014 at 4:03
$\begingroup$
$\endgroup$
This is not really an answer, rather a longer (and not very deep) remark about the question of the topology and the assumption that it should be Hausdorff.
Not every topology that makes a group a topological group is Hausdorff. But to any topological group we can associate a Hausdorff topological group in a canonical way. Let $G$ be a topological group and denote by $H$ the closure of $\{e\}$. Then $H$ is a normal subgroup in $G$, and the quotient group $G/H$ is Hausdorff with respect to the quotient topology. See Proposition 1-4 (vi) on page 6 of Ramakrishnan and Valenza, Fourier analysis on number field, 1999.
Based on this result, Ramakrishnan and Valenza write "Part (vi) shows that every topological group projects by a continuous homomorphism onto a topological group with Hausdorff topology. In this sense the assumption that a given group is Hausdorff is not too serious."
Nonetheless, the assumption plays an important role in Rupert's question. For, if we take the trivial topology $\mathcal{O}=\{\emptyset,G\}$, the $H=G$ and $G/H$ is the trivial group. Btw, the example of the trivial topology shows that the answer to the question (if we don't require Hausdorff) is easy: $G$ and the empty set are the only subsets that are closed in any topology on $G$ that makes $G$ a topological group.
