# How someone got the idea for the completeness axiom?

I don't understand what was the Motivation for the completeness Axiom and why Analysis and calculus would not work without it: • Do you mean the completeness axiom? – rwbogl Feb 10 at 18:17
• I think these things usually come from trying to prove something, realising you're making an assumption, trying to prove the assumption is true, then finally realising that the assumption can't be proved or disproved and is an underlying axiom. But I'll await more informed answers. – timtfj Feb 10 at 18:18
• @rwbogl the title is now changed – New2Math Feb 10 at 18:28
• Related discussion can be found here and here. – Pedro Feb 10 at 18:51

## 3 Answers

I don't have any good historical information, but I do have some intuition for you.

Analysis, and therefore calculus, is centered around the idea of limits. What happens as we approach a certain point? What is the result of repeating a process indefinitely? The Greeks used this idea in their method of exhaustion to calculate the volume of solids. Newton and Leibniz used this idea with their "infinitesimals" and "fluxions." None of these were rigorous methods, but they got the job done.

When we finally started to introduce rigor into analysis, we needed to really understand what we meant by a limit. This gave rise to the $$\epsilon$$-$$\delta$$ definition of the limit and various other formalizations of ideas that were only intuitive up to that point.

The notion of limit and supremum (and infimum) are intimately connected. Roughly, we want to be able to say that, in as many cases as we reasonably can, a process that we carry out ought to have a limit. The completeness axiom (though it can be proven with some set theory) tells us that this always works in the context of bounded sets. Given a set bounded from above, we can inch our way closer and closer to the upper bound, and are guaranteed to approach some limit, namely the supremum.

This idea turns out to be crucial in defining and proving various things that calculus students often take for granted in analysis. As a basic example, how do you define $$x^p$$ for arbitrary real $$p$$? Calculus students are well aware that $$\frac{dx^p}{dx} = p x^{p - 1},$$ but how do we prove that? It turns out that one way to define this quantity is as $$x^p = \sup_{\substack{a/b \in \mathbb{Q} \\ a/b \leq p}} x^{a/b},$$ where $$x^{a/b} = \left(\sqrt[b]{x}\right)^a.$$ To even make this definition we must assume that such an upper bound exists for every suitable $$x$$.

So really, the completeness axiom is about just what is says: completeness. We want to be justified in taking "limits" - or things that look like limits - as often as possible; that is, we want our system to be as complete as possible.

• Okey this was really helpfull to fuel my Intuition About this Topic. I am just rephrasing what you were saying but if I take the above set which is bounded from below and not empty I can theoretically Always look at a Point between the Bound and the Point that I know that exists and test if it is in the set or not. If not then this Point is a lower Bound otherwise if I would say if it is not a lower Bound then there must be a Point that is smaller than the Point in Question which is called a lower Bound but such apoint cannot exist because the Point in Question is not in the set. – New2Math Feb 10 at 18:52
• If it is in the set then I go down, if it is not in the set I go up. Rinse and repeat this is a Limit process. – New2Math Feb 10 at 18:53
• That's the rough idea! The general result is this: Every bounded set contains a sequence that approaches its infimum (or supremum). – rwbogl Feb 10 at 19:32

Core theorems in analysis rely on this. Examples:

• Heine-Borel
• Continuous functions on closed intervals attain maxima/minima
• intermediate value theorem.

These theorems are essential in developping calculus (differentiation and integration).

Consider sequence or rational numbers $$a_n = (1+\frac 1 n)^n$$. Does it converge? If we're working with real numbers, then yes, $$\lim_{x\to\infty}a_n = e$$. If we were working with rational numbers (which don't satisfy the completness axiom) then no, it doesn't. Completness axiom ensures that sequences which "ought" to converge, actually do.