Exercise 17.6

Proof. Suppose that $A\subset B$ and consider any $x\in \overline{A}$. Consider any neighborhood $U$ of $x$ so that $U$ intersects $A$ by Theorem 17.5 part (a). Hence there is a point $y\in U\cap A$ so that $y\in U$ and $y\in A$. But then clearly $y\in B$ also since $A\subset B$. Therefore $y\in U\cap B$ so that $U$ intersects $B$. Since $U$ was an arbitrary neighborhood of $x$, this shows that $x\in \overline{B}$, again by Theorem 17.5 part (a). This of course shows that $\overline{A}\subset \overline{B}$ as desired since $x$ was arbitrary. □

Proof. $\text{(⊂)}$ We show this by contrapositive. So suppose that $x\notin \overline{A}\cup \overline{B}$. Then clearly $x\notin \overline{A}$ and $x\notin \overline{B}$. Thus, by Theorem 17.5 part (a), there is an open set $U_A$ such that $U_A$ does not intersect $A$, and likewise an open $U_B$ that does not intersect $B$. Let $U=U_A\cap U_B$, which is clearly open since $U_A$ and $U_B$ are. We also note that $U$ contains $x$ since both $U_A$ and $U_B$ do. Then it must be that $U$ does not intersect $A$ since, if it did, then $U_A$ would also intersect $A$ since $U\subset U_A$. Similarly, $U$ cannot intersect $B$. Thus, for all $y\in U$, $y\notin A$ and $y\notin B$. This is logically equivalent to saying that there is no $y\in U$ where $y\in A$ or $y\in B$, therefore there is no $y\in U$ where $y\in A\cup B$. Hence $U$ and $A\cup B$ do not intersect. Since $U$ is open and contains $x$, this shows that $x\notin \overline{A\cup B}$, again by Theorem 17.5 part (a). Therefore, by contrapositive, $x\in \overline{A\cup B}$ implies that $x\in \overline{A}\cup \overline{B}$ so that $\overline{A\cup B}\subset \overline{A}\cup \overline{B}$.

$\text{(⊃)}$ Consider any $x\in \overline{A}\cup \overline{B}$ and any neighborhood $U$ of $x$. If $x\in \overline{A}$ then $U$ intersects $A$ by Theorem 17.5 part (a). Hence there is a $y\in U\cap A$ so that $y\in U$ and $y\in A$. Then clearly $y\in A\cup B$ so that $y$ is also in $U\cap (A\cup B)$. Hence $U$ intersects $A\cup B$. An analogous argument shows that this is also true if $x\in \overline{B}$ instead. Since $U$ was an arbitrary neighborhood, this shows that $x\in \overline{A\cup B}$ by Theorem 17.5 part (a). Hence $\overline{A}\cup \overline{B}\subset \overline{A\cup B}$ since $x$ was arbitrary. □

Proof. Consider any $x\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->{\overline{A}}_{\alpha }$ so that there is a particular $\beta$ where $x\in {\overline{A}}_{\beta }$. Suppose that $U$ is any open set containing $x$ so that $U$ intersects $A_{\beta }$ by Theorem 17.5 part (a) since $x\in {\overline{A}}_{\beta }$. Then clearly $U$ also intersects $\bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ since $A_{\beta }\subset \bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$. Since $U$ was an arbitrary open set containing $x$, this shows that $x\in \overline{\bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }}$ by Theorem 17.5 part (a). This shows that $\bigcup <!--\; FUNCTION\; APPLICATION\; -->{\overline{A}}_{\alpha }\subset \overline{\bigcup A_{\alpha }}$ since $x$ was arbitrary, which is of course the desired result. □