Proof of Real Number Property In his introduction to Calculus, Apostol gives a foundation for the properties
 of real numbers. After laying down the field and order axioms but before stating the least-upper-bound axiom, the author uses the set $T = \{x \ : \ 0 \leq x < 1\}$ as an example of a set which has no maximum element, yet has a least-upper-bound, namely, 1.
Is the author relying on intuition to make these assertions, or can we really prove these statements even before introducing the least-upper-bound axiom?
 A: The least upper bound axiom is not needed to prove that this specific, concretely-given set of real numbers has a least upper bound: all that’s needed is to prove that $x<1$ for each $x\in T$, which is a matter of definition, and that if $u<1$, then there is an $x\in T$ such that $u<x$. The latter can be done by taking $x=\frac12(u+1)$: that $u<x<1$ can be proved from the field and order axioms.
A: Certainly $T$ has an upper bound, namely $1$. Now to show that $1$ is the least upper bound, let $a$ be any upper bound of $T$. Can $a$ be less than $1$? No because $T$ contains elements arbitrarily close to $1$, i.e. for any $x < 1$ there is an element $y \in T$ such that $x < y$ which is to say that $x$ is not an upper bound for $T$. Therefore the upper bound $a$ is greater than or equal to the upper bound $1$, and since $a$ was arbitrary we conclude that $1$ is the least upper bound of $T$.
Note that the exact same argument works for $T' = \{x \in \mathbb Q \mathrel| 0 \leq x < 1\}$ even though $\mathbb Q$ doesn't have the least upper bound property.
The least upper bound property of $\mathbb R$ is there to guarantee least upper bounds of "weird" sets like $T_\xi = \{r \in \mathbb Q \mathrel| r < \xi\}$. Those sets don't have least upper bounds in the rationals when $\xi$ is irrational, but they do in the reals.
