I'm considering the following problem over uniform space which seems like it should hold, but I can't seem to show it.
Let $X$ be a connected uniform space with a uniformity $\mathcal{E}$. I denote for an entourage $U$, $$ U[x]:= \{ y\in X: (x,y)\in U \} \quad \text{and} \quad U[A]:= \cup_{a\in A} U[a]. $$ I assume that the uniformity is separated and that the space is locally totally bounded, in the sense that every point has a neighborhood which is totally bounded. I consider a compact set $A\subseteq X$ and a point $b\notin A$, such that $b\in U[A]$ for some $U\in \mathcal{E}$ symmetric.
I think that the following should hold:
There exists an $a\in \partial A$ such that $b\in U[a]$
but I have been unable to show it. I was also wondering if this is perhaps not true, and my intuition on this might be wrong.
My attempt so far:
Since $U$ is symmetric $U[b]\cap A\neq \emptyset$. I know that the space is regular, hence there exists a symmetric $U'\in \mathcal{E}$ such that $U'[b]\cap A=\emptyset$. I can assume without loss of generality that $U'[b]$ is $U$-small. Since $U[b]$ is totally bounded, there exists an $m$ such that $(b, a') \in (U')^m$, for some $a'\in A$. I would like to say that there exists an $a\in \partial A$, such that $(b,a)\in (U')^{m-1}$.
I would appreciate any hint, directions and counter-examples.