Exercise 17.17

Proof. First let $A=(a,b)$ and $C=[a,b)$ so that we must show that $\overline{A}=C$.

$\text{(⊃)}$ Consider any $x\in C$.

Case: $x=a$. Consider any basis element $B=[c,d)$ that contains $x=a$ so that $c\le x=a<d$. Let $e=\min <!--\; FUNCTION\; APPLICATION\; -->(b,d)$ so that $a<e$ since both $a<d$ and $a<b$. Then of course there is a real $y$ between $a$ and $e$ so that $a<y<e$. Thus we have $c\le a<y<e\le d$ so that $y\in B$. Also $a<y<e\le b$ so that $y\in A$. Hence $B$ intersects $A$ so that $x=a\in \overline{A}$ by Theorem 17.5 part (b) since $B$ was an arbitrary basis element.

Case: $x\ne a$. Then it has to be that $x\in (a,b)=A$ so that $x\in \overline{A}$ since obviously $A\subset \overline{A}$.

This shows that $C\subset \overline{A}$ since $x$ was arbitrary.

$\text{(⊂)}$ Now consider any real $x$ where $x\notin C$ so that either $x<a$ or $x\ge b$. If $x<a$ then the basis element $B=[x,a)$ clearly contains $x$ but does not intersect $A$. If $x\ge b$ then the basis element $B=[b,x+1)$ contains $x$ and does not intersect $A$. Either way, it follows from Theorem 17.5 part (b) that $x\notin \overline{A}$. Since $x$ was arbitrary, the contrapositive shows that $\overline{A}\subset C$. □

Proof. Let $A=(a,b)$. Consider any real $x$, and we shall consider an exhaustive list of cases that will show whether $x\in \overline{A}$ or $x\notin \overline{A}$.

Case: $x<a$. Obviously, there is a rational $p$ where $p<x$ since the rationals are unbounded below. Similarly, there is a rational $q$ where $x<q<a$ since the rationals are order-dense in the reals. The set $B=[p,q)$ is then clearly a basis element of $T_c$ that contains $x$. It is also trivial to show that $B$ does not intersect $A$ since $q<a$, which shows that $x\notin \overline{A}$ by Theorem 17.5 part (b) whether $b$ is rational or not.

Case: $x=a$. Consider any basis element $B=[p,q)$ (where $p$ and $q$ are rational) that contains $x=a$ so that $p\le x=a<q$. Let $d=\min <!--\; FUNCTION\; APPLICATION\; -->(b,q)$ so that $a<d$ since both $a<q$ and $a<b$. Then of course there is a real $y$ between $a$ and $d$ so that $a<y<d$. Thus we have $p\le a<y<d\le q$ so that $y\in B$. Also $a<y<d\le b$ so that $y\in A$. Hence $B$ intersects $A$ so that $x=a\in \overline{A}$ by Theorem 17.5 part (b) since $B$ was an arbitrary basis element. Note that this is true whether or not $b$ is rational.

Case: $a<x<b$. Then clearly $x\in (a,b)=A$ so that $x\in \overline{A}$ since obviously $A\subset \overline{A}$.

Case: $x=b$.

Case: $b$ is rational. Then there is another rational $q$ where $q>b$ since the rationals are unbounded above. Then clearly the set $B=[b,q)$ is a basis element of $T_c$ that contains $b$. Also clearly $B$ does not intersect $A$ since $y\in A$ implies that $y<b$ and hence $y\notin B$. This shows that $x=b\notin \overline{A}$ by Theorem 17.5 part (b).

Case: $b$ is irrational. Then consider any basis element $B=[p,q)$ containing $b$ so that $p$ and $q$ are rational. Thus $p\le b<q$, but since $p$ is rational but $b$ is not, it has to be that $p<b<q$. Let $c=\max <!--\; FUNCTION\; APPLICATION\; -->(p,a)$ so that $c<b$ since both $a<b$ and $p<b$. There is then a real $y$ where $c<y<b$ so that $a\le c<y<b$ and hence $y\in A$. Also $p\le c<y<b<q$ so that also $y\in B$. Therefore $B$ and $A$ intersect, which shows that $x=b\in \overline{A}$ by Theorem 17.5 part (b) since $B$ was arbitrary.

Case: $x>b$. Then there are clearly rationals $p$ and $q$ where $b<p<x$ and $x<q$. Then clearly the set $B=[p,q)$ is a basis element that contains $x$ and does not intersect $A$. This of course shows that $x\notin \overline{A}$ by Theorem 17.5 part (b) again, noting that this is true regardless of the rationality of $b$.

These cases taken together show the desired results. □