In $\mathbb R^n$ with usual topology (euclidean metric), exists a bounded open set $A$ such that $A$ is different from the interior of its closure?

Question

I expect that if $A$ is open set in $\mathbb R^n$ with the usual topology, then $A$ is contained in the interior of the closure of $A$.

But the converse is not necessarily true, I think that this is a hint for the pursued set.

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The motivation of this claiming is because I want a bounded open set $A$ in which a bounded continuous real valued function defined all over $\mathbb R^n$ is Darboux integrable over the closure of a whilst it isn't over $A$ itself.

Again the motivation of the aforementioned motivation is that there is a theorem stating that if an integral of an function exists over a bounded set then it exists at the interior of this set and both are equal, so it seems rather absurd, but I think it is to show that $A$ is not the interior of closure of $A$.

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Edit: Please if the answer may follows the motivation it would be of great help! Thank you in advance.

Answer 1

Let $A$ = $(0,1) \cup (1,2)$, then $A$ is open, bounded and we have $Cl(A) = [0,2]$ but $Int(Cl(A)) = Int([0,2])= (0,2) \neq A$.

Answer 2

Re: Your first sentence: If $C\supset D$ then $Int(C)\supset Int (D).$ If $A$ is open then $A=Int(A).$ So if $A$ is open then $Int(Cl(A))\supset Int (A)=A.$

A set which is the interior of its closure is called regular-open. And a set which is the closure of its interior is called regular-closed.