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I am trying to prove the following assertion: Let $B$ be a simply-connected set, and let $B' \subsetneq B$ be a proper open connected subset. Then, there exists a point $x \in \partial B'$ of the boundary such that there is a basis of neighborhoods of $x$ in $B$ whose intersections with $B'$ are connected.

This seems like it should be true. For example, in the one-dimensional case, a proper open subinterval of an interval always has an endpoint satisfying this property, whereas for a circle, one can take the complement of a point $x \in S^1$, and the boundary point $x$ does not have the property in question.

I am happy if one takes the special case of $B$ a manifold, or even a product of intervals.

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    $\begingroup$ False. Take B = C (complex plane) and B' = C minus [0,infty) and x = 1. $\endgroup$
    – mme
    Commented Feb 15, 2023 at 1:17
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    $\begingroup$ @mme: if I understand the question correctly, it asserts an existence of such an $x$ rather than claiming that it's true for any $x$ on the boundary, i.e. I don't think you are allowed to pick $x=1$ :-) $\endgroup$
    – M.G.
    Commented Feb 15, 2023 at 1:51
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    $\begingroup$ There is a 1-dim simply-connected continuum B that may serve as a counter-example. $\endgroup$
    – Wlod AA
    Commented Feb 15, 2023 at 3:20
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    $\begingroup$ Thanks for the correction! $\endgroup$
    – mme
    Commented Feb 15, 2023 at 3:38

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It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also simply connected)

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