Geometric definitions of infinity

Solution 1:

(i) finite vs. inifinite number of points

Coming from a background of projective geometry, I've dealt with both finite and infinite projective planes. Other people might be more familiar with affine planes. Both concepts generalize to higher dimensions. And at the core of both axiomatizations, you usually have a set of points which is either finite or infinite. So you could classify geometric spaces using the tools from set theory.

I guess that these days, mathematicians are very used to considering set theory as one of the most fundamental concepts and building stuff on that. So I'd say the above view is probably state of the art. I'm pretty sure, though, that there were times where set theory was far less established, and where people were more used to thinking about geometric concepts in a geometric language, distinct from set theory. I don't know whether anybody considered finite geometries at that time, though.

There are ways to translate between geometry and arithmetic, so one can re-establish the existence of an underlying number field for some (namely the Desarguesian) projective planes, treating geometric axioms as first class citizens and deriving field axioms from these. In the spirit of such a translation, you can construct a sequence $1 + 1 + \cdots$ using geometric constructions, and in an infinite space you can be assured that you will obtain a countably infinite sequence of distinct points, whereas in a finite space you are guaranteed to end up where you started, sooner or later.

Note that all of the above is geared towards finite geometric spaces in the most fundamental meaning of the word, namely finitely many points. If you want to distinguish between finite and infinite lengths, areas or volumes in some geometric space where even a finite volume would contain an infinite number of points, then you are in my opinion not so much asking about geometry but rather about measure theory.

Solution 2:

This question already has an answer, but I think it'd be worth expanding on a couple other fun conditions that will imply infinity in the convex $S\subseteq \mathbb{R}^n$ way described at the end of the edit. I find these interesting in the same way as sneaky order-based definitions of finite sets like "set has a well ordering $\le$ such that $\ge$ is also a well ordering".

Let $C=(|S|>1)\land(x,y\in S \land d(x,y)=d(x,z) \implies z\in S)$. That is, $C$ says $S$ has at least 2 points and $S$ is closed under drawing spheres around points in the set ($d$ is distance). The sneaky thing here is that any sphere in $\mathbb{R}^n$ is uncountably infinite and you can get to any other point in space by drawing spheres using points on other spheres. In the $n=2$ case this is equivalent to the connectedness of the so-called unit distance graph of the plane. We could also require $S$ be closed under drawing lines and get a similar result.

Let $C=(S\ne\emptyset)\land(\forall T\subseteq S, \exists s \in S,\forall s\in T, d(s,t)>1)$. That is, for any set of points in $S$ there's a point which isn't close to any of them. So they have to spread out infinitely. Since you also required that $S$ is convex, this also means that all the points between these infinitely spread out points are in $S$ so it has to have infinite measure as desired.

Let $C=(S$ contains an $n$-cube$)\land(\forall x,y\in S, y$ is a midpoint of $xz \implies z\in S)$. That is, $S$ is closed under the points playing leapfrog. Naturally, from the hypercube you can leapfrog your way to any point on the lattice that hypercube is a facet of. Then convexity requires that $S$ is the whole $\mathbb{R}^n$.

I think I'll stop there. The unifying trick across these is that if the set always has at least "one more" point in an unbounded way then you can prove it's infinite. Getting the measure to be infinite is a bit trickier, but since convexity was guaranteed this doesn't present an issue.