Exercise 3.15

Proof. Let $A=[0,1]$ (or $A=[0,1)$) and consider any nonempty subset $A_0$ of $A$ that is bounded above in $A$. So suppose that $b$ is an upper bound of $A_0$ in $A$ so that $0\le b\le 1$ (or $0\le b<1$). Now, of course, $A_0$ has a least upper bound $c$ in $ℝ$ since it is also a nonempty subset of $ℝ$ that is bounded above. Obviously, $c\le b$ since it is the least upper bound. For any $x\in A_0$ we of course have that $x\in A$ so that $0\le x\le 1$ (or $0\le x<1$), and clearly $x\le c$ since it is an upper bound of $A_0$. Thus we have $0\le x\le c\le b\le 1$ (or $0\le x\le c\le b<1$) so that $c\in A$. Hence $A_0$ has a least upper bound in $A$ as desired. □

Proof. Let $X=[0,1]\times [0,1]$ (or $X=[0,1)\times [0,1]$). Suppose that $A$ is a nonempty subset of $X$ that is bounded above and that $x_1\times y_1$ is an upper bound of $A$ in $X$. Then of course we have $x_1\in [0,1]$ (or $x_1\in [0,1)$). Define $A_x=\left\{x\mid x_{}\times y_{}\in A\right\}$ so that clearly $A_x\subset [0,1]$ (or $A_x\subset [0,1)$). It then follows that $x\le x_1$ for any $x\in A_x$ since $x_1\times y_1$ is an upper bound of $A$ in the dictionary order. Also clearly $A_x$ is nonempty since $A$ is. Thus $A_x$ is a nonempty subset of $[0,1]$ (or $[0,1)$) that is bounded above (by $x_1$) so that it has a least upper bound $a$ by what was shown in part (a).

Now, if $a\notin A_x$ then set $b=0$. Otherwise, define $A_y=\left\{y\mid a\times y\in A\right\}$. Then, since $a\in A_x$, there is a $y\in [0,1]$ where $a\times y\in A$, which shows that $y\in A_y$ and hence $A_y\ne \varnothing$. We also clearly have that $A_y\subset [0,1]$ so that $A_y$ is bounded above by $1$. Then $A_y$ has a least upper bound $b$, again by what was shown in part (a). In either case, we assert that $a\times b$ is the least upper bound of $A$ in $X$ in the dictionary order.

First, it is obvious that $a\times b\in X$ is based on how $a$ and $b$ were defined. Consider any $x_{}\times y_{}\in A$ so that $x\in A_x$, and hence $x\le a$ since $a$ is the least upper bound of $A_x$. If $x<a$ then of course $x_{}\times y_{}\le a\times b$, so assume that $x=a$. Then $a=x\in A_x$ so that $b$ was defined as the least upper bound of $A_y$. Then we have that $y\in A_y$ since $x_{}\times y_{}=a\times y\in A$, and thus $y\le b$ since $b$ is the least upper bound of $A_y$. This shows that $x_{}\times y_{}\le a\times b$ so that we have shown that $a\times b$ is an upper bound of $A$ since $x_{}\times y_{}$ was arbitrary.

To show that it is the least upper bound consider any $x_0\times y_0\in X$ where $x_0\times y_0<a\times b$. If $x_0<a$ then there must be an $x\in A_x$ where $x_0<x$ since $x_0$ cannot be an upper bound of $A_x$ since $x_0<a$ and $a$ is the least upper bound of $A_x$. Then, since $x\in A_x$, there is a $y\in [0,1]$ where $x_{}\times y_{}\in A$. Hence $x_0\times y_0<x_{}\times y_{}$ since $x_0<x$ so that $x_0\times y_0$ is not an upper bound of $A$. On the other hand, if $x_0=a$ we must have $y_0<b$ so that it cannot be that $b=0$, and hence it must be that $a\in A_x$. Then there must be a $y\in A_y$ where $y_0<y$ since $y_0$ cannot be an upper bound of $A_y$ since $y_0<b$ and $b$ is the least upper bound of $A_y$. Hence $a\times y\in A$, $x_0=a$, and $y_0<y$ so that $x_0\times y_0<a\times y\in A$, which shows that $x_0\times y_0$ is not an upper bound of $A$. Thus in either case $x_0\times y_0$ is not an upper bound of $A$, which shows that $a\times b$ is the least upper bound since $x_0\times y_0<a\times b$ was arbitrary.

This of course completes the proof that $A$ has a least upper bound, which shows that $X$ has the least upper bound property since $A$ was an arbitrary nonempty subset. □

Proof. Let $X=[0,1]\times [0,1)$. Consider the set $A=\left\{0\right\}\times [0,1)$, which is clearly a nonempty subset of $X$. This subset also obviously has an upper bound in $X$ in the dictionary order, for example, the point $1\times 0$. So let $x_1\times y_1$ be any upper bound of $A$ in $X$ and suppose for the moment that $x_1=0$. Then $y_1\in [0,1)$ so that $0\le y_1<1$, but then there is a $y_0\in ℝ$ such that $0\le y_1<y_0<1$. Then $y_0\in [0,1)$ so that $x_1\times y_0=0\times y_0\in A$, but also $x_1\times y_1<x_1\times y_0$ so that $x_1\times y_1$ cannot be an upper bound of $A$. So it must be that in fact $x_1\ne 0$ and hence $x_1>0$. Now let $x_0=x_1∕2>0$ and $y_0=0$ so that clearly $x_{}\times y_{}<x_0\times y_0<x_1\times y_1$ for any $x_{}\times y_{}\in A$ since $x=0<x_1∕2=x_0<x_1$. This shows that $x_0\times y_0$ is still an upper bound of $A$ but that $x_0\times y_0<x_1\times y_1$. Since $x_1\times y_1$ was an arbitrary upper bound of $A$, this proves that $A$ can have no least upper bound! □