# Sequences $(U_n)$ of neighborhoods of $0$ in a LCA group with $m(U_n)\to 0$

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 $m(U_n)\to 0$?

• No, it is not always true, because you could have picked all $U_i$ equal to $G$! – Mariano Suárez-Álvarez Apr 20 '16 at 22:18
• 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. – M.González Apr 21 '16 at 13:45

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 $U_n\subset W_n$.