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Let $G$ be a topological group that is Hausdorff, that is, for every pair $(g,h)$ of distinct elements of $G$, there exist disjoint open sets $U_g$ and $U_h$ such that $g \in U_g$ and $h \in U_h$.

Say $g$ and $h$ are invariantly separated if $U_g$ and $U_h$ can be chosen to be invariant under conjugation in $G$.

Now let $K$ be the set of elements that are not invariantly separated from the identity. Equivalently, $K$ is the intersection of all closed invariant identity neighbourhoods in $G$.

It often happens that $K$ is non-trivial: for instance in the $ax+b$ group over $\mathbb{R}$, $K$ consists of all the translations.

Does it ever happen that $K \not= \{1\}$, but $K \cap U = \{1\}$ for some identity neighbourhood $U$? Having $1$ isolated in $K$ ensures that the group is distal, i.e. there are no conjugacy classes that approach the identity, but perhaps there is some other way for invariant separation to fail. A locally compact example would be especially interesting.

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