0
$\begingroup$

Let $G$ denote an infinite compact abelian group with Haar measure $m$ (so $m(G)=1$).

Given a neighborhood $U_1$ of the unit $0$ in $G$ we can find a symmetric neighborhood $U_2$ of $0$ such that $U_2+U_2\subset U_1$. Proceeding in this way we construct a sequence $(U_n)$ of symmetric neighborhoods of $0$ with $U_{n+1}+U_{n+1}\subset U_n$.

Is it always true that $m(U_n)\to 0$?

Can we choose the sequence so that $m(U_n)\to 0$?

$\endgroup$
  • $\begingroup$ No, it is not always true, because you could have picked all $U_i$ equal to $G$! $\endgroup$ – Mariano Suárez-Álvarez Apr 20 '16 at 22:18
  • $\begingroup$ Thank you for your comment. So the answer to the first part of the question is no. However we can take $U_1$ with $m(U_1)<1/2$, and then it looks reasonable that $m(U_n)<2^{-n}$. But I do not know how to prove it. $\endgroup$ – M.González Apr 21 '16 at 13:45
0
$\begingroup$

Your questions are easy.

Is it always true that $m(U_n)\to 0$?

No. Let $G=\mathbb Z_2^\omega$ be a Tychonoff product of a discrete finite group $\mathbb Z_2=\mathbb Z/2\mathbb Z$. For each $n$ put $V_n=\{x=(x_i)\in G: x_i=0 $ for all $i\le n\}$. Then $\{V_n\}$ is a base at the zero of the group $G$ consisting of its subgroups. Thus for each $n$ you can put $U_i=V_i$ for $i\le n$ and $U_i=V_n$ for $i>n$.

Can we choose the sequence so that $m(U_n)\to 0$?

Yes (of course, provided the group $G$ is Hausdorff). For this purpose fix an arbitrary sequence $\{x_n\}$ of distinct points of the group $G$. Since the group $G$ is Hausdorff, for each $n$ there exists a neighborhood $W_n$ of the zero such that a family $\{x_iW_n:i\le n\}$ consists of mutially disjoint sets. Thus $m(W_n)\le 1/n$. Now it suffices to choose each $U_n\subset W_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.