(a) (b) The inclusion $\subset$ holds in both cases. Using 6(a), $\overline{\cap A_{\alpha}}\subset\overline{A}_{\alpha}$ for every $\alpha$ , and, therefore, $\overline{\cap A_{\alpha}}\subset\cap\overline{A}_{\alpha}$ . The equality does not hold in general, as $A_{\alpha}$ ’s might have some common limit points which are not in their intersection and not limit points of the intersection. For example, the intersection can be empty, while there is a common limit point: $(0,1)$ and $(1,2)$ is an example. Another example is $A=\mathbb{Q}\subset\mathbb{R}$ and $B=\mathbb{R}-\mathbb{Q}$ , where $\overline{A\cap B}=\emptyset$ and $\overline{A}\cap\overline{B}=\mathbb{R}$ .

(c) The inclusion $\supset$ holds, and, in fact, $\overline{A-B}-\overline{B}=\overline{A}-\overline{B}$ . Therefore, the inclusion is proper iff the closures of $A-B$ and $B$ share some points. The same examples illustrate the fact: if $A=(0,1)$ and $B=(1,2)$ , then $\overline{A}-\overline{B}=$ $\overline{A-B}-\overline{B}=$ $[0,1)\subset$ $\overline{A-B}=[0,1]$ , and if $A=\mathbb{Q}$ and $B=\mathbb{R}-\mathbb{Q}$ , then $\overline{A}-\overline{B}=$ $\overline{A-B}-\overline{B}=$ $\mathbb{R}-\mathbb{R}=\emptyset\subset$ $\overline{A-B}=\mathbb{R}$ .

Proof 1. Using 6(a,b), $\overline{A}-\overline{B}=$ $(\overline{A-B}\cup\overline{A\cap B})-\overline{B}=$ $\overline{A-B}-\overline{B}$ .

Proof 2. Assume that $x$ is in $\overline{A}-\overline{B}$ , i.e. every neighborhood of $x$ intersects $A$ but there is some neighborhood $V$ of $x$ that does not intersect $B$ . Suppose there is some neighborhood $U$ of $x$ that does not intersects $A-B$ , then the neighborhood $U\cap V$ of $x$ does not intersects $A-B$ and it does not intersects $B$ , i.e. it does not intersects $A$ at all, contradiction. So, $x\in\overline{A-B}$ but $x\notin\overline{B}$ . The other direction is immediate using 6(a).