On the relation of Completeness Axiom of real numbers and Well Ordering Axiom

Question

In my abstract algebra book one of the first facts stated is the Well Ordering Principle:

(*) Every non-empty set of positive integers has a smallest member.

In real analysis on the other hand one of the first things introduced are the real numbers and their Completeness Axiom:

Every nonempty set of real numbers having an upper bound must have a least upper bound.

Which is equivalent to:

(**) Every nonempty set of real numbers having a lower bound must have a biggest lower bound (infimum).

It has never been mentioned in any book I've read and I don't know if they have anything to do with each other but (*) and (**) seem to me to be such that (**) implies (*).

Is the Well Ordering Principle a consequence of the Completeness of the real numbers? Or do they have nothing to do with each other? How should I think of them in terms of how they relate to each other?

Is it okay to see one as a consequence of the other?

Answer 1

If you define $\Bbb{R}$ using Dedekind cuts over $\mathbb{Q}$, then $(**)$ can be proven without using $(*)$. (the Dedekind completion of a dense linear order without endpoints always has the least upper bound property).

It is tricky to say that $(*)$ follows from $(**)$ because $(*)$ is a defining caracteristic of $\mathbb{N}$ so in a way you summon $(*)$ as soon as you talk about $\mathbb{N}$. I am not sure you can define $\mathbb{N}$ knowing only that $\mathbb{R}$ is an ordered field with property $(**)$ in a way that would make the proof $(**) \rightarrow (*)$ possible.

Answer 2

I see the completeness axiom of $\mathbb{R}$ as a generalisation of the well ordering principle of $\mathbb{N}$. As every subset $S\subset\mathbb{Z}$ is bounded below, this condition need to be imposed on $S\subset\mathbb{R}$ to make the extension possible even though the bound doesn't have to be in $S$.