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

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?

• I don't understand your comment Normal Human. I don't have any formulae in the question. I understand math jax is for latex formulae? – a student Nov 25 '15 at 2:25
• Never mind the comment, it was an erroneous one. – user147263 Nov 25 '15 at 2:28
• There is a crucial difference between (**) and (*), which is that in (*) the least element is required to be an element of the set, whereas in (**) the infimum is not required to be an element of the set. If you somehow define the natural numbers in terms of the real numbers, you will be able to prove (*) from (**) (and the other axioms of the real numbers), but the proof will take some work. – Eric Wofsey Nov 25 '15 at 2:36
• @EricWofsey I think the fact that the least element is in the set follows from the fact that the sets in (*) consist of integers. Of course the $\inf$ of a set of integers is in the set. Or am I missing something? – a student Nov 26 '15 at 0:29
• Intuitively it is obvious that the inf of a set of natural numbers is in the set, but it is something you have to prove. To emphasize why you should be careful here, you're trying to prove the well-ordering principle, so you must make sure you aren't using any facts (no matter how basic or familiar) that can themselves only be proven using the well-ordering principle. Instead, you have to rely on whatever your definition of the natural numbers as a subset of the real numbers is, and properties of the real numbers that you are assuming as axioms or have otherwise proven. – Eric Wofsey Nov 26 '15 at 0:40

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.
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 $S$.