Talk:Cantor-Dedekind Hypothesis

I'm kind of skeptical of this because the constructible numbers (and even algebraic numbers) are countable. Do you know a book or anything that mentions the statement here as a theorem (or says euclidean lines have to be uncountable)? I don't know nearly enough (even elementary) geometry, and I could easily be misunderstanding something, but at the moment I don't see why euclidean geometry guarantees enough points are on a line to get such a bijection. -- Qedetc 01:25, 10 June 2011 (CDT)


 * We have that $\pi$ is a transcendental real number and so an element of an uncountable set. $\pi$ is on the real number line, although admittedly Euclid did not prove that $\pi$ was transcendental. I don't know the details of how the proof of this works (I've seen it but haven't analysed it properly) but "surely" it's a truism that the points on a line are uncountable. Besides, who says that the points on an infinite straight line have to be constructible? There is nothing in Euclid's Elements that says every point on a line has to be "constructible", all you need to do is be able to draw the straight line and it is at that level "assumed" that the points on it "exist". But only certain points on it are proved to be identified by a specific "location" on that line. That's my understanding of this and how I reach a philosophical acceptance of its truth. --prime mover 01:54, 10 June 2011 (CDT)


 * Sorry in advance for the huge wall of text below:


 * Even if we need countably many transcendentals, that isn't necessarily enough to force an uncountable amount of elements. For example, if you adjoin $\pi^{1/n}$ for every $n\in\mathbb N$ to the field of algebraic numbers, you still get a countable field.


 * I'd have to suspect that uncountably many elements are forced to exist by euclidean geometry to clear this up for me. The problem for me is that I don't see why something would be forced to exist if it wasn't constructible.  I'm not sure it's obvious that any notion of line has to have uncountably many points.  The set of points  $\{(0,0) + t(1,1): t\in\mathbb{Q}\}$ in $\mathbb{Q}^2 \subseteq  \mathbb{R}^2$ certainly looks like kind of line-ish if you try to graph it.  Intuition on this matter isn't really relevant though.  The theorem here is purporting to establish that it is in fact uncountable.


 * To help clarify my thoughts here, I'm not suggesting that the points on lines in euclidean geometry have to be constructible or algebraic. We know that $\mathbb{R}^n$ is a model of euclidean geometry, and it definitely has nonconstructible points on its lines.  What I'm wondering is whether the axioms of euclidean geometry force there to be anything more than the points you can construct.  Do we know somehow that the axioms guarantee that if you have a line AB of length 4 at point A that there actually is a point on the line distance $\pi$ toward $B$?  We know there's infinitely many points between $A$ and $B$, but that's not enough.  This might involve knowing something about the line being a continuum rather than just dense.  It wasn't clear to me from the axioms I had seen from The Elements that this was necessary.


 * I just went and looked at some wikipedia pages and noticed Hilbert used an axiom ( http://en.wikipedia.org/wiki/Hilbert%27s_axioms#V._Continuity ) which seems like it might prevent countable models (or at least the ones I was thinking of, since they could be extended to the $\mathbb{R}$ based models). I don't know if Hilbert added this because it was an implicit assumption in The Elements or just because he felt like it (nothing like this seems mentioned in the axioms from The Elements, as far as I can tell, but Euclid might have used it somewhere).


 * Birkhoff ( http://en.wikipedia.org/wiki/Birkhoff%27s_axioms ) apparently took this bijection as an axiom.


 * Tarski ( http://en.wikipedia.org/wiki/Tarski%27s_axioms#Betweenness_axioms ) apparently has a weaker first-order version of euclidean geometry and includes an axiom that adds a sort of completeness property (but only for first-order definable subsets of lines, of which there are only countably many). I think we can prove there's a countable model for these axioms.  This probably isn't what people think of when they think of euclidean geometry though.


 * I guess I'm asking whether (constructible numbers)$^n$ is ever considered a model of usual euclidean geometry. I don't think this is ridiculous, since, if you imagine taking the real plane and throwing out all the non-algebraic points, it still "looks like" a big dense plane.  It's not clear to me that this isn't "enough".  There might be some argument in The Elements which prevents it, but it looks like Tarski's weaker geometry doesn't forbid it (as far as I can tell), and Hilbert and Birkhoff kind of threw in axioms for it for reasons I'm not sure of.


 * The reason this is at all relevant is that if the axioms don't force uncountable models, then I don't see how anyone is going to define this bijection, since the axioms won't guarantee the points exist. If you've seen this proof somewhere, then I guess they've got something that does it.  I'm expressing interest in seeing it (or knowing where you saw it :D ) and trying to fill in some gaps in my very incomplete knowledge.  Again, I don't really know much about euclidean geometry or how its normally formally handled.
 * Qedetc 04:05, 10 June 2011 (CDT)