2

I recently proved the following result:

(1) Let $M$ be a model of Neutral (Absolute) Geometry with set of points $\mathbb{P}$. Then $|\mathbb{P}|= \mathfrak{c}$.

However, the proof relies upon the following theorem:

(2) If $κ$ and $λ$ are cardinal numbers such that $κ$ is infinite and $0<λ≤κ$, the union of $λ$ sets of cardinality $κ$ has cardinality $κ$. In other words, the union of at least one and at most $κ$ sets of cardinality $κ$ has cardinality $κ$ (First answer here).

My question is this: I don't know which axioms are necessary to prove (2) but I'm certain they aren't precisely those of Neutral Geometry. Assuming my proof contains no errors, is (1) still a valid theorem? Loosely speaking, I think the answer is yes since (2) was, in the proof, never used to draw any geometric conclusions but I'm not sure.

EDIT: The axioms of Neutral Geometry can be found here

The following terms are undefined: Distance, point, line, half-plane, angle measure and area

Community
  • 1
M10687
  • 3,250
  • 2
    Since (2) is a standard result in set theory, it would help if you added something about the theory of Neutral Geometry (so that the notion of a model for that theory is clearer). Note the downward Löwenheim-Skolem theorem, associated with Skolem's paradox, that any first-order theory with an infinite model must have a countable model. – hardmath Feb 25 '16 at 01:19
  • Ok I'll add some information. – M10687 Feb 25 '16 at 01:22
  • Also, it's definitely certain that Neutral Geometry can't have a countable model. – M10687 Feb 25 '16 at 01:23
  • 1
    @hardmath Re: cardinality, the issue is the role of the reals. If we formulate neutral geometry as a first-order theory, then it has a countable model - unless we make the language of size continuum. One way to do this would be to have a symbol for each real. I think something like that is what's going on under the hood - in your axioms about distance, you're tacitly assuming that the "reals" being talked about are in fact the reals. If we formulate neutral geometry in a finite language, where the field of distances is just a real closed field, then it does have countable models. – Noah Schweber Feb 25 '16 at 02:20

1 Answers1

1

This answer refers to Hilbert axioms of neutral geometry, however maybe your theory is equivalent to Hilbert's.

It can be proved that neutral geometry has exactly two models up to isomorphism. Adding Euclid's fifth postulate or its negation we get a theory which is categorical, which means every two models are isomorphic. $\mathbb{R}^2$ and Klein model are the models of euclidian and hyperbolic geometry respectively and both of them are of cardinality $\mathfrak{c}$.

However you don't really need to know that this theory is categorical to prove that the set $P$ of all points has the cardinality $\mathfrak{c}$. All you need is some theorems which can be derived from the axioms. Take a line $L$ and a point $o\in L$. Let $A$ be one of the halflines contained in $L$ beginning in $o$ and $M$ be one of the halfplanes determined by $L$. For a given point $x\in P$ let $p(x)$ be an ortogonal projection on a line $L$ and define $$f(x)=\begin{cases}|xp(x)| & , x\in M \\ 0 & , x\in L \\ -|ap(x)| & , x\in M^*\end{cases}$$ $$g(x)=\begin{cases}|op(x)| & , p(x)\in A \\ 0 & , p(x)=o \\ -|op(x)| & , p(x)\in A^*\end{cases}$$

Then $x\mapsto (f(x),g(x))$ is a bijection between $P$ and $\mathbb{R}^2$.

Kulisty
  • 1,468