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For two closed set $A_1,A_2$ with $A_1 \cap A_2 \neq \varnothing $ and the intersection of $A_1$ and $A_2$ is contained in an given open set $V$, I want to construct an open set $O$ such that \begin{equation} A_1 \cap A_2 \subset O \subset V \end{equation} and $O \neq V $. My idea is to use the open cover of the boundary of $A_1$ and $A_2$. Specifically, let $\partial A_1$ and $\partial A_2$ to denote the boundary and $int A_1$, $int A_2$ to denote the interior of $A_1$ and $A_2$ respectively, then we have two open sets $V_1$ and $V_2$: \begin{equation} V_1=intA_1 \cup (\cup_{x \in \partial A_1} B_{\varepsilon_1}(x)) \quad V_2=intA_2 \cup (\cup_{x \in \partial A_2} B_{\varepsilon_2}(x)) \end{equation} where $B_{\varepsilon}(x)$ denotes the open ball centred at $x$ with radius $\varepsilon$. Then we have $A_1 \subset V_1$ and $A_2 \subset V_2$. I want to construct $O= V_1 \cap V_2$ by choosing appropriate $\varepsilon_1$ and $\varepsilon_2$.

Can this idea work? I'll appretiate it for any hints!

EDIT 1: We are in Euclidean space.

EDIT 2: Require $O\neq V$.

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    $\begingroup$ What is wrong with $O=V$? $\endgroup$ – Kavi Rama Murthy Nov 20 '19 at 8:16
  • $\begingroup$ Presumably you want $O\subsetneqq V.$ The notation $X\subset Y$ does not mean that $X$ must be a proper subset of $Y$.... What kind of space are we in ? You cannot get $O\subsetneqq V$ for every type of space. $\endgroup$ – DanielWainfleet Nov 20 '19 at 10:24
  • $\begingroup$ @DanielWainfleet Thanks for reminding, here we are talking about euclidean space. $\endgroup$ – Huaixin Nov 20 '19 at 14:57
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If O is a subset of V, then trivially O = V will suffice.
If O is a proper subset of V, then it cannot always be done.
Let the space S have only one point and both A's be S.
Notice that S is a metric space,
a concern you were reluctant to disclosed.

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  • $\begingroup$ I see. In fact this question is part of a proof I'm doing, it seems I lost many details here... Thanks for reminding! $\endgroup$ – Huaixin Nov 20 '19 at 15:06
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(1). A metric space is a normal ($T_4$) space.

(2). Euclidean space $\Bbb R^n$ is a connected space.

(3). In $\Bbb R^n,$ let $A=A_1\cap A_2$ be closed and let $A\subset V$ where $V$ is open, with $\emptyset\ne A\ne \Bbb R^n.$

The closed sets $A,\,\Bbb R^n\setminus V$ are disjoint, so by (1) there is a disjoint pair $O,\,O'$ of open sets with $A\subset O$ and $\Bbb R^n\setminus V\subset O'.$

The open set $O'$ is not empty and is not $\Bbb R^n$ so by (2), $O'\ne \Bbb R^n\setminus V.$ So $O'\supsetneqq \Bbb R^n\setminus V.$

Therefore $O\subset \Bbb R^n\setminus O'\subsetneqq \Bbb R^n\setminus (\Bbb R^n\setminus V)=V.$

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