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 $O\neq V$.

  • 6
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.