We assume that we know the Least Upper Bound Principle: Every non-empty set $A$ which is bounded above has a least upper bound.
We want to show that every non-empty set $B$ which is bounded below has a greatest lower bound.
Let $A=-B$. So $A$ is the set of all numbers $-b$, where $b$ ranges over $B$.
First we show that if $B$ has a lower bound, then $A$ has an upper bound. For let $w$ be a lower bound for $B$. We show that $-w$ is an upper bound for $A$.
Because $w$ is a lower bound for $B$, we have $w\le b$ for any $b\in B$. Multiply through by $-1$. This reverses the inequality, and we conclude that $-w\ge -b$ for any $b\in B$. The numbers $-b$ are precisely the elements of $A$, so $-w\ge a$ for any $a\in A$.
Since $A$ is bounded above, it has a least upper bound $m$. We show that $-m$ is a greatest lower bound of $B$.
As usual, the proof consists of two parts (i) $-m$ is a lower bound for $B$ and (ii) nothing bigger than $-m$ is a lower bound for $B$.
The proofs again use the fact that multiplying by $-1$ reverses inequalities. To prove (i), suppose that $-m$ is not a lower bound for $B$. Then there is a $b\in B$ such that $b\lt -m$. But then $-b\gt m$. Since $-b\in A$, this contradicts the fact that $m$ is an upper bound of $A$.
To prove (ii), one uses the same strategy.
Suppose you have the axiom of completeness and assume you have $A$ and $B$ as in the statement of the cut property. Then, as $B$ is nonempty, $A$ has an upper bound. Let $c$ be the least upper bound for $A$.
- For $a\in A$, $a\le c$, because $c$ is an upper bound for $A$;
- For $b\in B$, $c\le b$, because $b$ is an upper bound for $A$ and $c$ is the least upper bound
Suppose you know the cut property. Consider a nonempty set $C$ with an upper bound. Then let
$$
A=\{x\in\mathbb{R}: x<c\text{, for some $c\in C$}\}
$$
and let $B$ be the complement of $A$. Since $C$ has an upper bound $b$, we have $b\notin A$, so $B$ is nonempty as well as $A$. The union of $A$ and $B$ is $\mathbb{R}$ by construction. Suppose $a\in A$ and $b\in B$. If $b\le a$, we have $b<c$ for some $c\in C$, so $b\in A$: a contradiction.
Now, the cut property provides $d$ so $a\le d$, for every $a\in A$, and $d\le b$, for every $b\in B$. Can you prove that $d$ is the least upper bound of $C$?
Best Answer
I will show about the strategy of this proof. We want to show that
The existence of a least upper bound on bounded sets by above imply that...
...if we have have two sets $X$ and $Y$ such that $x\le y$ for all $x\in X$ and all $y\in Y$ then exists a $c$ such that $x\le c\le y$ for all $x\in X$ and all $y\in Y$
Then we want show that for $S:=\sup(X)$ holds
$$x\le S\le y,\quad\forall x\in X,\forall y\in Y\tag{1}$$ provided that $x\le y$ for all $x\in X$ and all $y\in Y$.
Because $S$ is the supremum of $X$ then the LHS of (1) is clear, that is, $x\le S$ for all $x\in X$ by the definition of supremum. Then to conclude the proof we must show that $S\le y$ for all $y\in Y$.
Then we show that if $S\le y,\forall y\in Y$ is not true then it must be the case that exists some $y_0\in Y$ such that $y_0<S$. But then this would imply that $S$ is not the supremum of $X$, that is, $x\le y_0<S$ for all $x\in X$, a contradiction.
Hence if $S$ is the supremum of $X$ it must be the case that $S$ is a lower bound of $Y$, that is, that $x\le S\le y$ for all $x\in X$ and for all $y\in Y$.$\Box$