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Let $A,B\subseteq\mathbb R^d$ with $A$ closed and $B$ open and such that $A\cap B\neq\emptyset$. Assume that there exists a sequence $(x_k)_{k\in\mathbb N}\subseteq\mathbb R^d$ conveging in Euclidean norm to some $x\in A\cap B$. Do we have $x_k\in A\cap B$ for some $k\in\mathbb N$? If $x$ is an interior point of $A$ then of course this is true, since $B$ is an open set and $A$ then contains a ball around $x$ of size $\varepsilon>0$. So the question becomes problematic if $x\in\partial A$ (boundary of $A$).

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up vote 1 down vote accepted

Let $x_i = (1 + \frac{1}{i},0)$, let $A$ be the closed unit ball and $B$ some small open ball around $(1,0)$. $(x_i)_{i\in\mathbb{N}}$ the converges to $(1,0) \in A \cap B$, yet none of the $x_i$ lie in $A$ and hence not in $A \cap B$. The answer is therefore no.

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No. Take $A=\{0\}$ and $B=\mathbb{R}^n$, $x_k=(1/k,0,\ldots,0)$.

More generally, let $x$ be any non-interior point in $A$. For each $n$, pick $x_n$ with $|x_n-x|<1/n$, $x_n\notin A$, and $x_n\in B$ (the latter is automatic when $n$ is large enough).

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For a counterexample take $A=[-1,0], \ B=(-1,1) \subseteq \mathbb{R}$ and $x_n=\frac{1}{n}$.

This sequence $(x_n)\to 0 \in A \cap B$ but $x_n \not \in (-1,0]=A \cap B, \ \forall n \in \mathbb{N} $

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