Exercise 1.3.3

Proof.

a)

Since $A$ is bounded below, $B\ne \varnothing$. Moreover, $A$ is not empty, so there is an element $a\in A$. Since every element $b\in B$ is a lower bound for $A$, we have in particular $b\le a$, and this is true for every element of $B$, so $a$ is an upper bound for $B$. The Axiom of Completeness applied to $B$ asserts then the existence of an upper least upper bound $c$ of $B$: $$c=\sup <!--\; FUNCTION\; APPLICATION\; -->B.$$

Now we show that $c=\inf <!--\; FUNCTION\; APPLICATION\; -->A$.

$\text{∙}$

Reasoning by contradiction, suppose that $c$ is not a lower bound of $A$. Then there is some $a\in A$ such that $a<c$. But $a$ is an upper bound of $B$, as every element of $A$. Since $c$ is the least upper bound of $B$, we obtain $c\le a$. This is a contradiction, so $c$ is a lower bound of $A$.

$\text{∙}$

Let $m$ be any lower bound of $A$. By definition of $B$, $m\in B$. Therefore $m\le c$, because $c$ is an upper bound of $B$. This shows that $c$ is the greatest lower bound of $A$, so $$c=\inf <!--\; FUNCTION\; APPLICATION\; -->A.$$

b)

Let $A$ be non empty and bounded below. Then part (a) builds an element $c$ which is the greatest lower bound of $A$, using only the Axiom of Completeness for the existence of least upper bound. So there is no need to assert that greatest lower bounds exist as part of the Axiom of Completeness.