Confusion in least upper bound axiom

Question

Least upper bound axiom: Every non-empty subset of $\mathbb R$ that has an upper bound must have a least upper bound.

This sounds too obvious as it works for both closed and open subsets of $\mathbb R$

A newcomer naive person will say:

Every subset of $\mathbb R$ must have a least upper bound

Then he realizes the first limitation that the subset of $\mathbb R$ must be non-empty.

Then after some time he realizes the second limitation that subset of $\mathbb R$ must have an upper bound.

Only then he states the least upper bound axiom completely.

No person who have gone through this axiom have found other limitations. This doesn't mean that there are no other limitations.

So how can we say that we have now stated the least upper bound axiom completely? And how can we justify its applications in several areas of mathematics (as there is a probability that the lub axiom is incomplete)?

Answer 1

In my opinion, your question is: what is $\mathbb{R}$?

If you follow an axiomatic approach, you define $\mathbb{R}$ as an ordered field (being an ordered field means to fulfill some axioms) that is Dedekind-complete i.e. that satisfies the least upper bound axiom: every non-empty subset of $\mathbb{R}$ with an upper bound in $\mathbb{R}$ has a least upper bound in $\mathbb{R}$ (see here). Therefore, $\mathbb{R}$ is any object that fulfill these axioms. It can be proved that such an object is unique, up to isomorphism. If you change the axioms (in particular, if you replace the least upper bound axiom with something else), likely you are talking of another mathematical object, which is not $\mathbb{R}$.

It turns out that the set of axioms I sketched above is not the only way to define $\mathbb{R}$: there are several equivalent axiomatic definitions of $\mathbb{R}$, which means that there are several sets of axioms that describe the same object $\mathbb{R}$. For instance, you can define $\mathbb{R}$ as an ordered field such that every Cauchy sequence converges (Cauchy axiom). Then you can prove that, in a ordered field, the least upper bound axiom follows from the Cauchy axiom, and vice-versa. This corroborates the idea that the least upper bound axiom (or other equivalent axioms such as the Cauchy one) formalizes correctly our intuition about $\mathbb{R}$.

Answer 2

Because we can prove that that axiom holds after having constructed $\mathbb R$ by some method (Dedekind cuts, Cauchy sequences of rational numbers, …). And, since it is proved for every non-empty subset of $\mathbb R$ with an upper bound, there can be no other conditions missing.