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

Question

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

Answer 1

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

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