1
$\begingroup$

The Harmonic series is well known and its divergence was proven back in the middle ages.

I've taken an introductory course in model theory so I know a bit about RCF and some properties of it. We did not explore it thoroughly though and haven't seen many interesting examples.

However, I do know that we can take some real closed field which is large enough (i.e. has cofinality $>\aleph_0$) and then the harmonic series will possibly converge.

My question if we take some $\mathcal{F}$ to be a model of RCF in which $\mathbb{R}$ is embedded and that the type $p(x) = \{ x > n | n\in\mathbb{N}\}$ is realized, $$x = \sum_{n \in \mathbb{N}^+} \frac{1}{n}$$ then $\forall y\in\mathbb{R}(x>y)$ then obviously $x$ is an upper-bound for the real numbers in the field we've chosen. However since $x$ is a non-Archimedean number, it is also clear that $x-1$ is an upper bound of the real numbers in $\mathcal{F}$.

This is the part where I get confused. What is $x$ and what is the conditions required for it to exist in the model?

Improve this question
$\endgroup$
3
  • $\begingroup$ What is {x > n|n in N} supposed to be? Is it {x : x > n and n in N}? Is it {n : x > n and n in N}? $\endgroup$
    – user5810
    Commented Sep 15, 2010 at 2:47
  • $\begingroup$ Ricky: A type is a set of (usually first-order) formulas. The type $p(x)$ has 1 free variable $x$. The formulas that form this type are all the statements $x>1$, $x>2$, $x>3$, etc. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 15, 2010 at 3:41
  • $\begingroup$ @Ricky: It is a collection of formulae with a free variable $x$, saying that $p(x)$ is realized meaning that there is some element in $\mathcal{F}$ for which all the formulae in $p(x)$ are true. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 15, 2010 at 3:44

1 Answer 1

Reset to default
1
$\begingroup$

Real closed fields are not complete (unless they are isomorphic to the reals), so the fact that some increasing sequence is bounded does not imply that it has a supremum.

If x is the sum of the harmonic series, then we seem to get x=1+ 1/3 + ...+ 1/2+1/4+...>1/2+1/4...+1/2+1/4+..=x/2+x/2 = x, suggesting that x does not exist in any real closed field.

Improve this answer
$\endgroup$
4
  • $\begingroup$ When you say complete, do you mean metrically? Because that'd be obvious since we define metrics using the reals. Or do you mean in the sense that it is a complete order (i.e. all the Dedekind cuts are realized)? $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 15, 2010 at 3:47
  • $\begingroup$ @AK: I think he means that the order is "bounded complete": every set which is bounded above has a least upper bound. $\endgroup$
    – Pete L. Clark
    Commented Sep 15, 2010 at 3:57
  • $\begingroup$ @Pete: So there are no real closed fields which are Dedekind-closed except the real numbers? $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 15, 2010 at 4:07
  • $\begingroup$ @AK: Yes, I believe so. (This is not really my area of core expertise.) See for instance emis.de/proceedings/TopoSym2001/20.pdf $\endgroup$
    – Pete L. Clark
    Commented Sep 15, 2010 at 5:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .