Answer to Question #144076 in Differential Geometry | Topology for ibrahim road

Question #144076
In (R, U) usual topology, find
A°,Z°, ext (A), ext (Z), b(A), b (Z)
Where A = {1,2} and Z set of all integer numbers
1
Expert's answer
2020-11-13T14:59:58-0500

Consider the topological space (R,U)(\mathbb R, U) with usual topology UU. Since there is no basic neigbourhood (1ϵ,1+ϵ)(1-\epsilon, 1+\epsilon) such that (1ϵ,1+ϵ)A(1-\epsilon, 1+\epsilon) \subset A and ϵ>0,\epsilon>0, we conclude that 1A1\notin A^\circ. By analogy, 2A2\notin A^\circ. Therefore, the interior AA^\circ of AA is emptyset. Since RA=(,1)(1,2)(2,+)\mathbb R\setminus A=(-\infty,1)\cup (1,2)\cup(2,+\infty) is an open set in usual topology, AA is a closed set, and thus A=A\overline A=A. So, ext(A)=RA=(,1)(1,2)(2,+)ext(A)=\mathbb R\setminus\overline A=(-\infty,1)\cup (1,2)\cup(2,+\infty) and the boundary b(A)=AA=A=A.b(A)=\overline A\setminus A^\circ=A\setminus\emptyset=A.


Since for each kZk\in\mathbb Z there is no basic neigbourhood (kϵ,k+ϵ)(k-\epsilon, k+\epsilon) such that (kϵ,k+ϵ)Z(k-\epsilon, k+\epsilon) \subset \mathbb Z and ϵ>0,\epsilon>0, we conclude that kZk\notin \mathbb Z^\circ. Therefore, the interior Z\mathbb Z^\circ of Z\mathbb Z is emptyset. Since RZ=kZ(k,k+1)\mathbb R\setminus \mathbb Z=\bigcup_{k\in\mathbb Z}(k,k+1) is an open set in usual topology, Z\mathbb 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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