Talk:Equidistance is Independent of Betweenness

I can't wait to see a proof of this; it appears that induction on the form of the formula containing $\mathsf B$ which is used to try to define $\equiv$ cannot possibly be a fruitful track to pursue. It is, however, for now the only way I know to prove things like this. --Lord_Farin 13:43, 25 January 2012 (EST)

It's too advanced for me to tackle by myself at this point, but I'll quote Givant (p. 200)

This can be proved by means of the well-known method of Padoa, indeed by exhibiting, for any given system, two relational structions $\langle S_1,\mathsf{B}_1,\equiv_1 \rangle$ and $\langle S_2,\mathsf{B}_2,\equiv_2 \rangle$ which are both models of the given system, and in which $S_1$ and $\mathsf{B}_1$ respectively coincide with $S_2$ and $\mathsf{B}_2$, while $\equiv_1$ and $\equiv_2$ do not coincide, i.e., for some $x,y,z,u$, one of the formulas $xy \equiv_1 zu$ and $xy \equiv_2 zu$ holds and the other fails. - Givant

Then he gives such models on pp. 201-202, which I don't feel comfortable putting up. Perhaps after I finish Linear Algebra (I just had my first class today). Feel free to get to it before I do. --GFauxPas 17:29, 25 January 2012 (EST)

That sounds rather crafted, but it is effective (as is obvious). Thanks for posting it. --Lord_Farin 17:49, 25 January 2012 (EST)

'if $\mathcal{G}$ isn't strong enough to create $\R^2$, how can we use it? There is something incorrect in my presentation, particularly since we're going to use $\cdot$ and $\le$-GFP'

I wouldn't say that $\mathcal G$ isn't strong enough; rather, I'd even say that $\mathcal G$ is rich enough to permit multiple metrics on $\R^2$ to be defined. What you think of as $\R^2$ is actually equivalent to $(\R^2,\mathsf B_1,\equiv_1)$, while the other example is just a slightly different metric space on the same base set. $\mathcal G$ is shown to be strong enough as we can specify $\mathsf B$ and $\equiv$ such that the model becomes (at least intuitively) $\R^2$ with the Euclidean metric. There is no problem with that; the theory allows for us to define what those two concepts mean (which is the deepest meaning of 'undefined term' anyway; that it be specified explicitly in order to speak of a model for the theory; in some sense the role $\in$ plays in axiomatic set theory). --Lord_Farin 14:37, 13 February 2012 (EST)

P.m. (or GFP, if you happen to know), on a side track: what is the conceptual difference between rôle and role, if any? --Lord_Farin 14:37, 13 February 2012 (EST)

None. The first is pedantically accurate, the second is pragmatic and easily typed on a conventional qwerty keyboard. --prime mover 17:08, 13 February 2012 (EST)