“The deepest and most important of the fundamental properties of the real numbers” according to Edmund Landau

In his book Differential and Integral Calculus, Edmund Landau gives an introductory chapter, and before starting it, he assumes as true some theorems. Among them, it is one he calls: "The deepest and most important of the fundan1ental properties of the real numbers"

Let there be given any division of all the real numbers into two classes, having the following properties:

$(a)$ Neither class is empty.

$(b)$ Every number of the first class is smaller than every number of the second class. (In other words, if $a<b$ and if $a$ lies in the second class, then $b$ lies in the second class.)

Then there exists a unique real number $\xi$ ; such that every $\eta < \xi$ belongs to the first class and every $\eta > \xi$ belongs to the second class.

What is this theorem called? Should it be a consequence of the least upper bound property of the real numbers?

If we change "real" by "rational" in the first sentence, this is the approach to Dedekind cuts and thus one of the possible constructions of the real numbers. The last sentence can be stated as "each of the division determines a one (two) unique set(s) of rational numbers" or something of the sort.

• Yes, it follows from the completeness axiom, by which the first class (since it is bounded above) has a least upper bound $\xi_1 \in \mathbb{R}$, and the second class (since it is bounded below) has a greatest lower bound $\xi_2 \in \mathbb{R}$. You can easily show that $\xi_1 = \xi_2$ (since $\xi_1 < \xi_2$ and $\xi_1 > \xi_2$ lead to contradictions), and that $\xi \equiv \xi_1$ satisfies the conclusions. Note that $\xi$ itself can belong to either class. – mjqxxxx Jul 2 '12 at 0:26
• One way to look at it is that if you do the Dedekind construction on $\mathbb R$ instead of $\mathbb Q$, you don't get any new elements out of it -- the completion-by-cuts is an idempotent construction. – Henning Makholm Jul 2 '12 at 1:29
• They wrote like that back then, portentous. Completeness is indeed useful. – André Nicolas Jul 2 '12 at 1:31
• @HenningMakholm Interesting! – Pedro Tamaroff Jul 2 '12 at 5:16
• Possibly related? math.stackexchange.com/questions/579013/… – Don Larynx Nov 24 '13 at 9:05