Geometric definitions of infinity

Question

There are several definitions of "infinite set" that are common in set theory. For instance, a set $S$ is finite if there is a bijection between a natural number $n$ and $S$; it is infinite otherwise. Alternatively, one can define a set $S$ to be infinite if there is an injection from the natural numbers into it. Alternatively, there is the definition of "Dedekind infinite" sets, i.e., those for which there is an injection into a proper subset of themselves. All of these definitions can be shown to be equivalent under suitable assumptions (e.g., some require choice).

All of these definitions, however, are set-theoretic in character, in that they rely on the set-theoretic definition of function, injection, etc. They also characterize what it means for a set to contain infinitely many objects, rather than for a space to be infinitely divisible or infinite in extension. Two of the definitions require appeal to the natural numbers and so have a bit of an "arithmetic" flavor. Here's my question: are there well-known, precise "geometric" definitions of what it means for space to be infinite? Of course, it depends what one counts as "geometric", and so I realize this question is vague. But I am interested in whether there definitions that rely less on set theoretic properties of functions and the natural numbers, and more on properties of "space" in some sense.

Of course, there are order-theoretic conditions that might be used to characterize infinite spaces (e.g., density and unboundedness entail a space is infinite in a sense). Any other ideas?

Edit: MVG's suggestion is really helpful. It's true I wasn't clear whether I wanted a purely "geometric" definition of a space (i) containing infinitely many points, (ii) having infinite measure, or (iii) containing a subspace that extends infinitely in some direction (e.g., a ray in the plane). MVG suggests "geometric" definitions of a space that contains infinitely many points. So here's perhaps a more precise question that I can ask now about how to use geometric concepts to define subspaces of $\mathbb{R}^n$ having infinite measure, or that contain sets that extend infinitely in some direction.

For example, say a subset $S$ of a plane is "infinite" if it has the following properties: 1. $S$ contains either a circle or square. 2. Every circle in $S$ is circumscribed by a square lying in $S$. 3. Every square in $S$ is circumscribed by a circle lying in $S$. 4. The interior of any square in $S$ lies in $S$.

Then $S$ is the entire plane, I think, as you can get a nested sequence of circles of arbitrarily large radius. A similar definition should work in higher definitions. So it seems, perhaps by cheating, that we've characterized what it would mean for section of a plane to be unbounded or infinite in every direction using only figures that can be constructed via straight-edges and compass.

Of course, my definition wouldn't pick out the upper-right quadrant of the plane as being "infinite" (as a square with vertex at the origin could not be inscribed in a circle lying in the quadrant), but I think we can modify my definition slightly to fix it (e.g., by inscribing squares in quarter circles and vice versa). In general, does something similar work for all convex subsets of the plane?

So perhaps here's the more precise question.

Problem: Let $S$ be a convex subset of $\mathbb{R}^n$. Find a collection of conditions/sentences $C$ such that (i) $C$ mentions only figures that are constructible via by standard constraints (e.g., in the plane, straight-edge and compass in the plane) and (ii) $S$ satisfies $C$ if and only if the Lebesgue measure of $S$ is infinite, or $S$ contains some subset $T$ that has infinite Lebesgue measure of lower dimension.

Answer 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.

Answer 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.