Exercise 17.8

Proof. Consider any $x\in \overline{A\cap B}$ and any open set $U$ containing $x$. Then by, Theorem 17.5 part (a), $U$ intersects $A\cap B$, from which it immediately follows that $U$ intersects both $A$ and $B$. However, since $U$ was an arbitrary neighborhood of $x$, it follows from Theorem 17.5 part (a) again that $x$ is in both $\overline{A}$ and $\overline{B}$. Hence $x\in \overline{A}\cap \overline{B}$, which shows that $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$ since $x$ was arbitrary.

Now consider the standard topology on $ℝ$ with $A=[-1,0)$ and $B=(0,1]$. As these are clearly disjoint, we have that $A\cap B=\varnothing$ so that $\overline{A\cap B}=\varnothing$ also. However, since we also clearly have that $\overline{A}=[-1,0]$ and $\overline{B}=[0,1]$, it follows that $\overline{A}\cap \overline{B}=\left\{0\right\}$. Thus clearly $\overline{A\cap B}=\varnothing \ne \left\{0\right\}=\overline{A}-\overline{B}$ as desired. □

Proof. Consider any $x\in \overline{\bigcap <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }}$ and any open set $U$ of $x$. Then, by Theorem 17.5 part (a), $U$ intersects $\bigcap <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ so that, for any particular $A_{\beta }$, $U$ intersects $A_{\beta }$. This shows that $x\in {\overline{A}}_{\beta }$ by Theorem 17.5 part (a) so that $x\in {\overline{A}}_{\alpha }$ for every $\alpha$ since $\beta$ was arbitrary. Hence $x\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->{\overline{A}}_{\alpha }$, which shows that $\overline{\bigcap <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }}\subset \bigcap <!--\; FUNCTION\; APPLICATION\; -->{\overline{A}}_{\alpha }$ since $x$ was arbitrary.

As in part (a), equality fails if we have $A_1=[-1,0)$ and $A_2=(0,1]$ in the standard topology on $ℝ$. By the same argument as in part (a) it follows that $\overline{{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^2{A}_n}=\varnothing \ne \left\{0\right\}={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^2{\overline{A}}_n$. □

Proof. Consider any $x\in \overline{A}-\overline{B}$ and any open set $U$ containing $x$. Then $x\in \overline{A}$ so that every open set containing $x$ intersects $A$ by Theorem 17.5 part (a). Also $x\notin \overline{B}$ so that there is an open set $V$ containing $x$ that does not intersect $B$, also by Theorem 17.5 part (a). Let $W=U\cap V$ so that $W$ contains $x$ since both $x\in U$ and $x\in V$. Now, since $W$ is also an open set containing $x$, $W$ intersects $A$ so that there is a $y\in W$ where also $y\in A$. It also cannot be that $y\in B$ since we have $y\in W\subset V$ so that then $V$ would intersect $B$. Therefore $y\in A-B$. Also, we have $y\in W\subset U$ so that also $y\in U$. Hence $U$ intersects $A-B$, which shows that $x\in \overline{A-B}$ by Theorem 17.5 part (a) since $U$ was an arbitrary neighborhood of $x$. Therefore $\overline{A-B}\supset \overline{A}-\overline{B}$ as desired since $x$ was arbitrary.

As a counterexample to equality, consider the standard topology on $ℝ$ with $A=[0,2]$ and $B=(1,3]$. Then clearly $\overline{A}=A=[0,2]$ and $\overline{B}=[1,3]$, from which it is easily shown that $\overline{A}-\overline{B}=[0,1)$. But we also have $A-B=[0,1]$ so that obviously $\overline{A-B}=[0,1]$ as well. Therefore $\overline{A-B}=[0,1]\ne [0,1)=\overline{A}-\overline{B}$ as desired. □