Section 17: Problem 19 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

If $A\subset X$ , we define the boundary of $A$ by the equation $\mbox{Bd}A=\overline{A}\cap\overline{(X-A)}.$

(a) Show that $\mbox{Int}A$ and $\mbox{Bd}A$ are disjoint, and $\overline{A}=\mbox{Int}A\cup\mbox{Bd}A$ .

(b) Show that $\mbox{Bd}A=\emptyset$ $\Leftrightarrow$ $A$ is both open and closed.

(d) If $U$ is open, is it true that $U=\mbox{Int}(\overline{U})$ ? Justify your answer.

(a) $x\in\overline{A}$ iff for every open $U$ ,

implies $U\cap A\neq\emptyset$ iff there is an open $V$ such that

or for every open $U$ ,

implies $U\cap A\neq\emptyset\neq U\cap(X-A)$ iff $x\in Int(A)$ or $x\in Bd(A)$ . Also, the two cases are disjoint (either there is $V$ that does not intersect $X-A$ or every $U$ intersects $X-A$ ), so that $Int(A)\cap Bd(A)=\emptyset$ .

(b) $Bd(A)=\emptyset$ iff $\forall x$ there is an open set $U$ s.t. $x\in U\subset A$ or $x\in U\subset(X-A)$ iff $A,(X-A)\in\mathcal{T}$ .

(d) No, $U\subset\overline{U}$ is open, therefore, $U\subset Int(\overline{U})$ , but, for example, $\mathbb{R}-\{0\}\subsetneq Int(\overline{\mathbb{R}-\{0\}})=\mathbb{R}$ .

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Section 17: Problem 19 Solution

Section 17: Problem 19 Solution