MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The real numbers can be constructed as the set of Cauchy sequences (e.g. 3, 3.1, 3.14, 3.141, 3.1415,...) under the equivalence relation that their difference tends to 0.

  • Real numbers are equivalence classes of real number sequences, $a = (a_1, a_2, a_3, \dots)$
  • These sequences must be Cauchy:
    • you give me an $\epsilon > 0$
    • I can give you can $N > 0$ such that
    • if you give me any $n > N $
    • we have $|a_n - a_m| < \epsilon$
  • Equivalence relation: $a \equiv b$ if for any $\epsilon > 0$ , there exist $N > 0$ with $|a_n - b_n|< \epsilon$ for all $n > N$.

Then we have to check these have the properties of the real numbers we like: $(a+b)+c = a + (b+c)$, least upper bound property, etc.

Oh, I tried writing the definition of Cauchy sequence as a "game". Any leads on that are welcome in the comments.

I had been reading about the divergent sequences 1 - 2 + 3 - 4 + ... and how to add them. The partial sums are 1, -1, 2, -2, 3, -3, ... which diverge. We can try averaging the partial sums get 1, 0, 2/3, 0, 3/5, 0, 5/7, 0,... which still doesn't converge

Maybe instead use the simpler example 1 -1 + 1 -1+... whose partial sums are 1,0,1,0,1,0 and we "claim" the sum should be "$\frac{1}{2}$". This is the Cesaro sum of the alternating sequence. What's iffy is that we can rearrange the sequence to -1+1-1+1-... which should tend to $-\frac{1}{2}$. But that rearrangement is "infinite" in some sense.

I would like to know about the collection infinite sequences of (say) rational numbers admitting Cesaro averages. We could say $a \equiv b$ if $a_n - b_n$ has Cesaro mean tending to $0$, I think this is another construction of the real numbers.

What if we just ask $a_n - b_n$ be Cauchy? What is the set of equivalence classes of rational numbers with Cesaro averages modulo the Cauchy sequences. It's probably not even Hausdorff.

Loosely related, there is a construction of the reals due to A'Campo of real numbers as "slopes" or "almost homomorphisms".

  • A "slope" is a sequence of number $ \lambda: \mathbb{Z} \to \mathbb{Z}$ with $\{ \lambda(m)+\lambda(n)-\lambda(m+n): m, n \in \mathbb{Z} \}$ a finite set.
  • Two slopes are "equivalent" if $\{\lambda(n) - \lambda(n') : n \in \mathbb{Z} \}$ is a finite set.
  • The "positive" slopes have finite $\{ n : \lambda(n) \leq 0\}$.
  • ${\bf 1}$ is the identity map ${\bf 1} : n \mapsto n$.

The sum is $\lambda(n) + \lambda'(n)$ and the product is $\lambda(\lambda'(n))$ with $n \in \mathbb{Z}$.

This idea is connected with rotation number of oriented homomorphisms of the circle.

share|cite|improve this question
Thanks for sharing this "slopes" construction, it's interesting. – Zsbán Ambrus Jan 19 '12 at 14:49
The construction is originally due to Schanuel, not A’Campo. See for some historical remarks. – Emil Jeřábek Jan 19 '12 at 17:03
They were independent... and the emphases are different. Street/Schnauel are concerned with $\mathbb{R}$ as a logical construction. With A'Campo it seems to come out of geometry & topology. In any case, the integer sequence $\{n \mapsto [αn]\}$ maps to the real number $\alpha \in \mathbb{R}$. – john mangual Jan 19 '12 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.