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A friend asked me today why we need to bother with the Completeness Axiom in calculus. Even though I have taken a real analysis course quite a while ago, I could not answer him and I realized I had the same question.

The Completeness "Axiom" for $\mathbb{R}$, or equivalently, the least upper bound property, is introduced early in a course in real analysis. It is then shown that it can be used to prove the Archimedean property, is related to concept of Cauchy sequences and so on.

Let's consider a formal proof of the limit of a function, the simplest I can think of: To prove $\lim_{x\to1} 2x = 2$, take $\delta=\epsilon/2$. Then $|x-1|<\epsilon/2$ implies $|2x-2|<\epsilon$ and the proof is complete.

The "Completeness Axiom" must be implicitly assumed somewhere in this short proof - where? This also hopefully explains why we can't take this kind of limit in $\mathbb{Q}$.

EDIT: According to a comment, it seems this limit proof does not require the Completeness Axiom. Since the derivative and the Riemann integral can be defined using limits, that begs the question: Is it possible to have differential and integral calculus based only on $\mathbb{Q}$? If so, what are the most significant results that would no longer hold (the Intermediate Value Theorem for example, as mentioned in the Answers)?

TSJ
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  • In fact, this argument makes perfect sense in $ \mathbb{Q} $. – Berrick Caleb Fillmore Apr 20 '15 at 04:29
  • Thanks Berrick Fillmore - could you take a look at the new edit in the question? – TSJ May 04 '15 at 11:11
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    Hi TSJ. Suppose that we have a bounded function $ f: \Bbb{Q} \to \Bbb{Q} $. If we define its Riemann integral over $ [a,b] $, where $ a,b \in \Bbb{Q} $ and $ a < b $, to be the limit of some sort of sequence built using rational numbers only, then the convergence of the sequence is not guaranteed even if it is Cauchy. – Berrick Caleb Fillmore May 04 '15 at 16:32

1 Answers1

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The only difference between $\mathbb Q$ and $\mathbb R$ is the completeness axiom. So any property which is true in $\mathbb R$ but not in $\mathbb Q$ needs the completeness axiom. If you proof for example the existence of $\sqrt 2$ you will need the completeness axiom somewhere in your proof (otherwise you could prove $\sqrt 2 \in \mathbb Q$).

Now you can think of all theorems like the intermediate value theorem which hold for $\mathbb R$ as your basic number system but not for $\mathbb Q$. You will not have these theorems if you do not have the completeness axiom...

EDIT: Please have a look at these threads for your additional question:

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Stephan Kulla
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