Consider the topological space (R,U) with usual topology U. Since there is no basic neigbourhood (1−ϵ,1+ϵ) such that (1−ϵ,1+ϵ)⊂A and ϵ>0, we conclude that 1∈/A∘. By analogy, 2∈/A∘. Therefore, the interior A∘ of A is emptyset. Since R∖A=(−∞,1)∪(1,2)∪(2,+∞) is an open set in usual topology, A is a closed set, and thus A=A. So, ext(A)=R∖A=(−∞,1)∪(1,2)∪(2,+∞) and the boundary b(A)=A∖A∘=A∖∅=A.
Since for each k∈Z there is no basic neigbourhood (k−ϵ,k+ϵ) such that (k−ϵ,k+ϵ)⊂Z and ϵ>0, we conclude that k∈/Z∘. Therefore, the interior Z∘ of Z is emptyset. Since R∖Z=⋃k∈Z(k,k+1) is an open set in usual topology, Z is a closed set, and thus \overline \mathbb Z=\mathbb Z. So, ext(\mathbb Z)=\mathbb R\setminus\overline \mathbb Z=\bigcup_{k\in\mathbb Z}(k,k+1) and the boundary b(\mathbb Z)=\overline \mathbb Z\setminus \mathbb Z^\circ=\mathbb Z\setminus\emptyset=\mathbb Z.
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