Let $(X,d)$ be a metric space and let $A,B$ be non-empty subsets of $X$.

  1. Prove that $$A^\circ\setminus \bar{B} \subseteq (A\setminus B)^\circ$$ and find an example for proper inclusion, where $A^{\circ}$ indicates the set of interior points of the set $ A $.

I can prove it but I can't construct an example.

  1. $ \overline {A\setminus B} \subseteq \bar A \setminus B^{\circ} $ find an example for proper inclusion

In this question, would $X=\mathbb R$, $A=[1,3]$, $B=[1,2]$ work?

Then $A\setminus B=(2,3]$, $\overline {A\setminus B} = [2,3]$, and $\bar A=[1,3]$, $B^{\circ}=(1,2)$ and $ \bar A \setminus B^{\circ} = \{1\}\cup[2,3]$

Is it correct for that part? Thanks


2 Answers 2


The sets are equal in the first part, in any topological space:

$$A^{\circ}\setminus \overline{B} = A^{\circ} \cap (X\setminus\overline{B}) = A^{\circ} \cap (X\setminus B)^{\circ} = (A \cap (X \setminus B))^{\circ} = (A \setminus B)^{\circ}$$

where we use that interior commutes with finite intersection and the standard complement relation $X \setminus \overline{B} = (X\setminus B)^{\circ}$.

So no example can be found for the first part.

Your example for the second one does seem to work.


I think that $A^{\circ}\setminus \bar{B} = (A\setminus B)^{\circ}.$ Here is a proof of the inclusion $(A\setminus B)^{\circ}\subset A^{\circ}\setminus \bar{B} .$

Let $x\in (A\setminus B)^{\circ}.$ Then there exists a neighbourhood $U$ containing $x$ of $A$ such that $U\cap B=\phi.$ Therefore $x\in A^{\circ}$ and $x$ cannot be a limit point of $B$. Hence $x\in A^{\circ}\setminus \bar{B} .$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .