Exercise 17.18

Proof. $\text{(⇒)}$ We show this by contrapositive. So suppose that $y\ne 1$ or $x$ is not a limit point of $A$ from above and that $y\ne 0$ or $x$ is not a limit point from below.

Case: $y\ne 0$ and $y\ne 1$. Clearly then $0<y<1$. If $y=b$ then the dictionary order interval $(x\times 0,x\times 1)$ is a basis element that contains $x\times y$ and that does not contain any other points of $B$, if indeed $x\in A$ so that $x\times y=x\times b$ is in $B$. If $y<b$ then the dictionary order interval $(x\times 0,x\times b)$ is a basis element with the same properties. Lastly, if $y>b$ then the dictionary order interval $(x\times b,x\times 1)$ is a basis element that contains $x\times y$ but no points of $B$.

Case: $y=0$ or $y=1$. If $y=0$ then we have

Case: $x=0$. Then, if $b=y=0$, we have that the dictionary order interval $[0\times 0,0\times 1)$ is a basis element containing $x\times y=0\times 0$ but no other points of $B$, if indeed $x=0\in A$ so that $x\times y\in B$. If $b\ne 0$ then $0<b$ so that the interval $[0\times 0,0\times b)$ is a basis element with the same properties.

Case: $x\ne 0$. Then $0<x$ and it has to be that $x$ is a not a limit point of $A$ from below. Thus there is an interval $(c,d)$ or $(c,1]$ (or $[0,d)$ in which case let $c=0$ in what follows) that contains $x$ but no other points $y\in A$ where $y<x$. If $b=y=0$ then it is easy to show that $(c\times 1,x\times 1)$ (or $(c\times 1,x\times 1]$ if $x=1$) is a basis element that contains $x\times y$ but no other points of $B$, if indeed $x\in A$ so that $x\times y\in B$. If $b\ne y=0$ then $0<b$ so that $(c\times 1,x\times b)$ is a basis element with the same property.

If $y=1$, then an analogous argument shows analogous results.

Thus in all cases and sub-cases it follows that $x\times y$ is not a limit point of $B$, which shows the desired result by contrapositive.

$\text{(⇐)}$ Now suppose that either $y=1$ and $x$ is a limit point of $A$ from above or $y=0$ and $x$ is a limit point of $A$ from below. In the first case consider any dictionary order interval $C=(a\times c,d\times e)$ that contains $x\times y$. Then it has to be that $x<d$ since otherwise it would have to be that $y=1<e$ since $x\times y\prec d\times e$, which is of course impossible. Then, since $x$ is a limit point of $A$ from above, it follows that the open set $[0,d)$ contains a point $z\in A$ where $x<z$ so that $x<z<d$. It then follows that the point $z\times b$ is in both $C$ and $B$, and is of course distinct from $x\times y$ since $x<z$. The same argument can be made if $C$ is a basis element in the form of $[0\times 0,d\times e)$ or $(a\times c,1\times 1]$. This suffices to show that $x\times y$ is a limit point of $B$ since $C$ was an arbitrary basis element.

An analogous argument can be made in the case when $y=0$ and $x$ is a limit point of $A$ from below, which shows the desired result. □

Proof. First, let $K=\left\{1∕n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}\subset [0,1]$ so that clearly $A=\left\{x\times 0\mid x\in K\right\}$. It is easy to show that $0$ is the only limit point of $K$ and it is a limit point from above only. It then follows from Lemma 1 that $0\times 1$ is the only limit point of $A$ so that $\overline{A}=A\cup \left\{0\times 1\right\}$ since the closure is the union of the set and the set of its limit points. □

Proof. This time let $L=\left\{1-1∕n\mid n\in {\mathbb{Z}}_{\text{+}}\right\}$ so that clearly $B=\left\{x\times \frac{1}{2}\mid x\in L\right\}$. It is trivial to show that $1$ is the only limit point of $L$ and that it is a limit point from below only. Hence $1\times 0$ is the only limit point of $B$ by Lemma 1 so that the result follows. □

Proof. First, we clearly have that $C=\left\{x\times 0\mid x\in (0,1)\right\}$. It is easy to show that every point of $(0,1)$ is a limit point both from above and below, that $0$ is a limit point from above only, and that $1$ is a limit point from below only. Thus it follows that the set of limit points of $C$ are then $\left\{x\times 0\mid 0<x\le 1\right\}\cup \left\{x\times 1\mid 0\le x<1\right\}$ by Lemma 1 . As many of these points are contained in $C$ itself, the result follows. □

Proof. The limit points of $D$ are the same as for $C$ above for the same reasons, i.e. $\left\{x\times 0\mid 0<x\le 1\right\}\cup \left\{x\times 1\mid 0\le x<1\right\}$. The result then follows. □

Proof. Let $F=\left\{\frac{1}{2}\right\}\times [0,1]$ so that we must show that $\overline{E}=F$.

$\text{(⊂)}$ Consider any $x\times y$ where $x\times y\notin F$ so that simply $x\ne \frac{1}{2}$ since it has to be that $y\in [0,1]$. If $x<\frac{1}{2}$ then the basis element $[0\times 0,\frac{1}{2}\times 0)$ clearly contains $x\times y$ but no elements of $E$. If $x>\frac{1}{2}$ then the basis element $(\frac{1}{2}\times 1,1\times 1]$ clearly contains $x\times y$ but no elements of $E$ either. This shows that $x\times y$ is a not in $\overline{E}$ by Theorem 17.5 part (b). Hence $\overline{E}\subset F$ by contrapositive.

$\text{(⊃)}$ Consider any $x\times y\in F$ so that $x=\frac{1}{2}$ and $y\in [0,1]$. If $y\in (0,1)$ then $x\times y\in E$ so that $x\times y\in \overline{E}$ since obviously $E\subset \overline{E}$. If $y=0$ then consider any dictionary order interval $F=(a\times c,b\times d)$ containing $x\times y=\frac{1}{2}\times 0$. In particular we have that $\frac{1}{2}\times 0\prec b\times d$ so that either $\frac{1}{2}<b$, or $b=\frac{1}{2}$ and $0<d$. In the first case we have that $\frac{1}{2}\times \frac{1}{2}$ is in both $F$ and $E$. In the second case let $z=d∕2$ so that we have $0<z<d\le 1$. Then clearly the point $\frac{1}{2}\times z$ is in $F$, but we also have that $\frac{1}{2}\times z$ is in $E$ since $0<z<1$. The same argument applies if the basis element $F$ is of the form $[0\times 0,b\times d)$ or $(a\times c,1\times 1]$. A similar argument shows an analogous result in the case when $y=1$. This shows by Theorem 17.5 part (b) that $x\times y\in \overline{E}$ since $F$ was an arbitrary basis element, which of course shows that $\overline{E}\supset F$ since $x\times y$ was arbitrary. □