Name of an axiom

Question

What's the name of the following axiom: Let $A$ and $B$ be sets of real numbers, and $(\forall a\in A\text{ and }\forall b\in B)a \leq b$. Then $\exists c\in\mathbb R$ such that $(\forall a\in A\text{ and }\forall b\in B) a \leq c \leq b$

I thought it was the completness axiom, but that talks about least upper bounds.

Answer 1

This is the completeness axiom (modulo the missing requirement that $A$ and $B$ are both non-empty), but here completeness refers to the completeness of the order, rather than metric completeness defined with Cauchy sequences. This is sometimes called "least-upper bound property" or Dedekind completeness.

Answer 2

First, note that the statement as you gave it is not true if either $A$ or $B$ is empty. For example, let $A=\mathbb R$ and $B=\{\}$. Add the condition that both $A$ and $B$ are non-empty and the statement becomes true in the real numbers.

This is called the completeness axiom. There are several variations on it that are equivalent, and you hit on one of the variations. There are too many variations to name each one (though of course some groups of axioms do have names as a group).

Answer 3

Let me show how this axiom is equivalent to the least upper-bound property. Let $A$ be a nonempty set which is bounded above by a number $M$. Let $$ B = \left\{x : x \geq y\text{ for all $y$ in $A$}\right\} $$ So $B$ is nonempty too. Then by your axiom, there exists a point $c$ such that $$ a \leq c \leq b \quad\text{for all $a\in A$ and $b\in B$} \tag{1} $$

I claim $c$ is the least upper-bound of $A$. Clearly by the first half of (1) $c$ is an upper bound for $A$.
Suppose $c'$ were an upper bound for $A$ less than $c$. Then $c'$ would be in $B$, but less than $c$, contradicting the second half of (2). Therefore $c \leq c'$, and $c$ is the least upper bound of $A$.

To show why the least upper-bound property implies your axiom, let us be given two nonempty sets $A$ and $B$ with every element of $A$ less or equal to every element of $B$. So every element of $B$ is an upper bound for $A$. Let $c$ be the least upper bound of $A$. Then $c \geq a$ for each $a$ in $A$. If there were an element of $B$ less than $c$, it would be an upper bound of $A$ less than the least upper bound of $A$, a contradiction. Therefore $b \geq c$ for all $b$ in $B$.