5
$\begingroup$

The following question arises from trying to understand Lemma 1.3(ii) of arXiv:math/0405063. I believe a particular case of the proof (and in fact I think the proof is essentially equivalent to this claim) is:

Let $G$ be a locally compact group. Let $C,D$ be cosets (not assumed open, closed etc.) each of which has empty interior. Then $C\cup D$ also has empty interior.

This is not try in general topology, of course: let $C,D$ be the rational, respectively, irrationals, in $\mathbb R$. However, I cannot decide if being a coset rules out this sort of example. Is the claim true, and if so, what is a proof?

Improve this question
$\endgroup$
10
  • $\begingroup$ @NateEldredge I think Matt means Lemma 1.3(ii), which -- FWIW -- the preceding text claims is following/adapting a similar argument given in Rudin's book Fourier Analysis on Groups $\endgroup$
    – Yemon Choi
    Commented Feb 2, 2020 at 16:31
  • $\begingroup$ For a coset (=right coset, =left coset) in a topological group, having empty interior is equivalent to being non-open. $\endgroup$
    – YCor
    Commented Feb 2, 2020 at 16:58
  • 2
    $\begingroup$ This is false. Take $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index 2). Then $G=H\cup (G-H)$ and both $H,G-H$ have empty interior. $\endgroup$
    – YCor
    Commented Feb 2, 2020 at 17:02
  • 1
    $\begingroup$ @YCor: Great! That would do it. Okay, I guess I need to go back to Rudin and try to reconstruct the non-abelian version of his proof... $\endgroup$
    – Matthew Daws
    Commented Feb 2, 2020 at 17:05
  • 1
    $\begingroup$ @YCor Since the question is aimed at clarifying/correcting a gap in an argument in research-level harmonic analysis (and may even be detecting errors in a standard reference, namely Rudin's book) maybe you can post your comment as an answer. As I am somewhat familiar with the Ilie-Spronk paper and the literature that cites it, I should point out that Matt's question seems to be the result of digging rather deeper into the proofs than most people have done during the last 15 years $\endgroup$
    – Yemon Choi
    Commented Feb 3, 2020 at 15:31

2 Answers 2

Reset to default
5
$\begingroup$

This is false. Take the (compact abelian) group $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index $2$, and clearly a subgroup of index 2 is either closed or dense). Then $G=H\cup (G\smallsetminus H)$ and both $H$ and its coset $G\smallsetminus H$ have empty interior.

Improve this answer
$\endgroup$
0
$\begingroup$

With an additional assumption that there exists an interior point $x\in C\cup D$ such that $x\in C\cap D$, one can prove this fact for any topological group. It doesn't require local compactness.

Note that $e$ is in the interior of $x^{-1}C\cup x^{-1}D$ and both sets are subgroups. Thus, WLOG, $x=e$ and $C$, $D$ are subgroups.

Now pick an open neighbhourhood of $e$ such that $U\subseteq C\cup D$. Pick another open neighbhourhood of $e$ such that $V^2\subseteq U$. Now an easy algebraic argument shows that $V$ must by a subset of $C$ or $D$. Otherwise, pick $y\in V\setminus D$ and $z\in V\setminus C$. The product $yz$ must be in $U$, so in $C$ or $D$. Both lead to a contradiction.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ A basis of compact open subgroups?? I must be misunderstanding, because that doesn't seem to be true at all for say, $\mathbb{R}$, not even if you meant "relatively compact". $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2020 at 14:25
  • $\begingroup$ Yes, you are right. This is for totally disconnected only... $\endgroup$
    – Bugs Bunny
    Commented Feb 2, 2020 at 14:26
  • $\begingroup$ I will fix it now... $\endgroup$
    – Bugs Bunny
    Commented Feb 2, 2020 at 14:30
  • 1
    $\begingroup$ Setting aside the WLOG for a moment, since I think it's causing confusion: you seem to be claiming that the original cosets $C,D$ must have nonempty intersection. I don't see how that follows from the assumptions. $\endgroup$
    – Nate Eldredge
    Commented Feb 2, 2020 at 14:56
  • 1
    $\begingroup$ @NateEldredge seems to have identified a gap in the argument near the start. You start with x that lies in the union of $C \cup D$, and then you claim that $x^{-1}C$ and $x^{-1}D$ are subgroups. This only works if $x$ belongs to $C$ and $x$ belongs to $D$ $\endgroup$
    – Yemon Choi
    Commented Feb 2, 2020 at 16:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .