# Construct an open set that contains the intersection of two closed sets and contained in a given open set

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 $$$$A_1 \cap A_2 \subset O \subset V$$$$ and $$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$$: $$$$V_1=intA_1 \cup (\cup_{x \in \partial A_1} B_{\varepsilon_1}(x)) \quad V_2=intA_2 \cup (\cup_{x \in \partial A_2} B_{\varepsilon_2}(x))$$$$ where $$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 $$O\neq V$$.

• What is wrong with $O=V$? – Kavi Rama Murthy Nov 20 '19 at 8:16
• 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. – DanielWainfleet Nov 20 '19 at 10:24
• @DanielWainfleet Thanks for reminding, here we are talking about euclidean space. – Huaixin Nov 20 '19 at 14:57

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.

• I see. In fact this question is part of a proof I'm doing, it seems I lost many details here... Thanks for reminding! – Huaixin Nov 20 '19 at 15:06

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