# Proof of the least upper bound or completeness axiom corollary (greatest lower bound)

First off, I don't think this is a duplicate. Sorry if it is! Second this is not a homework question, so completeness is much appreciated.

I am trying to prove the corollary of the least upper bound axiom: Every nonempty set of real numbers that is bounded below has a greatest lower bound.

The notes I took in class seem to suggest that to prove the corollary you need to prove the least upper bound axiom that every non empty set of reals that is bounded above has a least upper bound first.

Let $A \in \mathbb{R}$ with $A \neq \emptyset$ and suppose $\{x | x \in A\}$ is bounded below. Define $-A = \{-x | x \in A \}$. This is bounded above by the negative of any lower bound for A. (If $k = \text{lub}(A)$ then $x \geq k \implies x \leq -k \implies$ $-k$ is an upper bound for $-A$.) $A \neq \emptyset$ so $-A \neq \emptyset$. The least upper bound axiom implies $-A$ has a least upper bound, call it $\gamma$.

Now, the hint to complete the proof is to guess $-\gamma$ is a greatest lower bound of $A$ and use the theorem "Let $\beta$ be a lower bound for $A$. Then $\beta = \text{glb}(A)$ if and only if for any $v > \beta$, there is an $x$ in $A$ with $\beta ≤ x < v$."

Two things: (1) How to complete the proof. (2) Isn't this proof method, by using the least upper bound, saying that $A$ must have to have a least upper bound when the corollary only says that $A$ must be bounded below? Or does relying on $-A$ as opposed to $A$ mean that $A$ doesn't have to have a least upper bound for the proof to work?

## 2 Answers

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.

You could use aswell another version of the completeness axiom: Cantor's intersection theorem. In fact you can consider a sequence of innested intervals. Consider $a\in A$ and $b\in B$, where B is your nonempty set and A is the set of lower bounds of B. Now you can consider $[a_{n+1 n}, b_{n+1}]=[a_{n},\dfrac{a_{n}-b_{n}}{2}]$ if $\dfrac{a_{n}-b_{n}}{2} \in B$ or $[a_{n+1}, b_{n+1}]=[\dfrac{a_{n}-b_{n}}{2},b_{n}]$ if $\dfrac{a_{n}-b_{n}}{2} \in A$. As a corollary of the Cantor intersection theorem (the proof is not very difficult) there exists one and only one element in every such subset and this elemente is $inf(A)$.