# Clopen subsets of tdlc groups

Let $$G$$ be a totally disconnected locally compact group, $$U$$ a closed and open subset of $$G$$. Is there a neighbourhood $$V$$ of the identity such that for all $$g \in U$$ we have $$gV \subset U$$?

Equivalently (by Van Danzig's theorem), is every closed and open subset of $$G$$ a union of cosets of some compact open subgroup of $$G$$?

• If $V$ exists, then since $e \in V$, you get $g \in gV \subset U$ for all $g \in G$. Commented Dec 3, 2019 at 13:57
• @PaulFrost sorry I meant for all $g \in U$. It should be correct now Commented Dec 7, 2019 at 15:35

For example, let $$G$$ be your favorite non-discrete tdlc group such that there exists $$U_0\supseteq U_1\supseteq \dots$$ a decreasing sequence of clopen neighborhoods of the identity, with $$\bigcap_{n=0}^\infty U_n = \{e\}$$, but $$U_n\neq \{e\}$$ for all $$n$$. For example, you could take $$G$$ to be the automorphism group of the complete binary tree and $$U_n$$ to be the set of all automorphisms which fix the ball of radius $$n$$ around the root.
Now consider $$G\times \mathbb{Z}$$, which is also tdlc. Let $$U = \{(x,n)\mid n\geq 0, x\in U_n\}$$. This set is clopen in $$G\times \mathbb{Z}$$. But for any neighborhood $$V$$ of the identity $$(e,0)$$, there is some $$n$$ such that $$V\not\subseteq U_n$$. So letting $$g = (e,n)$$, we have $$gV\not\subseteq U$$.
• Note that it's not really important for such a decreasing sequence to exist--if it doesn't, you can just take a product with a larger discrete group than $\mathbb{Z}$ (as long as $G$ is not discrete so $\{e\}$ itself is not clopen). Commented Dec 7, 2019 at 16:09
No. For instance, let $$G=\mathbb{Z}_p\times\mathbb{Z}$$ and consider the set $$U=\bigcup_{n\in\mathbb{N}} p^n\mathbb{Z}_p\times \{n\}$$. More generally, given any non-discrete totally disconnected locally compact group $$H$$, you can take a product of $$H$$ with a discrete group and then take a union of shrinking clopen subsets of different cosets of $$H$$ in the product.