Prove that there exists no set $A$ such that $A,Cl(A),Int(A),Cl(Int A)$ are pairwise distinct.

Question

Let $A\subseteq \Bbb R$ ,let $Cl(A)$= closure of $A$, $Int A=$ interior of $A$.

Prove that there exists no set $A$ such that $A,Cl(A),Int(A),Cl(Int A)$ are pairwise distinct.

Suppose that there exists a set $A$ such that $A,Cl(A),Int(A),Cl(Int A)$ are pairwise distinct.Then let us assume that $a\in A\setminus Int A$ and let $b\in Cl(A)\setminus A$.

But how to derive a contradiction from here.Please help.

A set $X$ satisfying $X=Cl(Int(X))$ is called a regular closed set. A set $Y $ satisfying $Y=Int(Cl(Y))$ is called a regular open set. — DanielWainfleet, Aug 07 '16 at 18:28

score 3 · Accepted Answer · answered Aug 06 '16 at 15:44

3

It's not true :

$A = ]0,1]\cup \{2\}$

$Cl(A) = [0,1] \cup \{2\}$

$Int(A) = ]0,1[$

$Cl(Int(A)) = [0,1]$

And these 4 sets are pairwise distincts

answered Aug 06 '16 at 15:44

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Prove that there exists no set $A$ such that $A,Cl(A),Int(A),Cl(Int A)$ are pairwise distinct.

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