What is your intuitive understanding of infinity? [duplicate]

What is your intuitive understanding of infinity?

Mine is the following, I prepared it as image:

Those were the main points I got to after thinking by myself about what infinity is, without looking at definitions and textbooks.

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marked as duplicate by egreg, Jack D'Aurizio, Hans Engler, Eric Stucky, CookieJun 22 '14 at 0:40

I don't think that (4) is an equivalent definition to the accepted definition (in bijection with a natural number ordinal), as you take the ultraproduct of the paths of length $n$ to produce a graph with exactly the properties you say but infinite in the traditional sense. However, I think you're trying to get at the fact that a set is finite if and only if there exists a well-ordering of the set whose opposite order is also a well-ordering (i.e. every non-empty subset has both a minimum and a maximum). Just a note – Hayden Jun 21 '14 at 14:41
@Hayden: If you look closely, then you'll notice that this is a way to mathematically strip down the definition of a finite well-order. It is an order with a maximum and a minimum, where every non-endpoint has a unique successor and a unique predecessor. – Asaf Karagila Jun 21 '14 at 14:50
@AsafKaragila Is the reason the ultraproduct I mentioned does not fulfill your requirements for the fact that the two edges guaranteed to exist (for the vertices that aren't the end-points, of course), because you can't unambiguously say (for vertices in the bi-infinite center) which corresponds to the successor and which corresponds to the predecessor? This was my assumption, and the OP only mentioned that every node that wasn't one of the end-points connected to two other nodes, but did not indicate any idea of order except that one end-node was called 'start' and the other 'end'. – Hayden Jun 21 '14 at 14:58
@Hayden: I'm not quite sure what you mean by ultraproduct. If you take any ultrapower of a finite structure, the result is isomorphic to that finite structure. If you take an ultraproduct of increasingly large structures, then you effectively show that there is no first-order statement in the language of these structure which is true if and only if the structure is finite; but that's fine, nobody claims that there is. Within set theory we can define what is finite internally to that universe. – Asaf Karagila Jun 21 '14 at 15:01
@AsafKaragila I was only using the ultraproduct to show that there existed an infinite path with all the properties the OP mentioned (which, strictly speaking, did not mention any inherent ordering). But what you say is right, my first encounter with this example of an ultraproduct was as the ultraproduct of paths of finite length that are structures of first-order graph theory, and is typically used to show that there is no first-order formula for two vertices being connected. Using it here was just to guarantee it had all the properties stated. – Hayden Jun 21 '14 at 15:17

My intuition about infinity is ineffable. It has been developed, and is developing, through my journey through mathematics. Being privileged to work in set theory, I have dabbled a lot with infinity, and with even stranger infinities than most, as a consequence of work related to the axiom of choice.

But what is infinity?

Well, yes. Infinite is not finite. The fourth point is in fact the definition of being finite in modern set theory. It means that you can be put in bijection with a finite ordinal. The third point is known as being Dedekind-infinite, and its equivalence to infiniteness requires using the axiom of choice. But those are two valid (and in a context where choice is assumed, equivalent) ways to characterize finite and infinite.

But the second point is problematic. If you want to talk about infinite quantities, then they are just as numbers as $6,42$ and $31337$. Cardinal numbers, finite and infinite, are mathematical formulations of the quantity of "how many elements are there". So infinity is not "simpler" or "more basic" than finite numbers. Both have equal rights to be called numbers and both are equally simple or complicated.

On the other hand, if you only want to talk about infinity in the context of calculus, or measure theory, or other contexts where $\infty$ denotes a point which is larger than all the elements of the domain (or something along these lines), then indeed infinity becomes "simpler", but the finite numbers of your domain become much more fundamental. If you define $\infty$ as something larger than all the real numbers, then you have to have the real numbers before you can talk about infinity.

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I would have given +5 for this if I could, thanks! – user3111311 Jun 21 '14 at 14:56
Well, not to be strict about it, but a 50 points bounty is +5. Then again, there's really no need for that here. Just glad I could help. – Asaf Karagila Jun 21 '14 at 14:57
"The fourth point is in fact the definition of being finite in modern set theory. It means that you can be put in bijection with a finite ordinal" -- sort of. I think the questioner's wording is consistent with the ordering of the integers in Charles's answer, which arguably is a "chain" per the questioner's definition "two nodes with one link and all other nodes with two links". It's intuitively obvious that the questioner intends the chain to be bijective with a natural number, but I suspect the only way to formalise that is to say it, in which case you don't need to talk about the links ;-) – Steve Jessop Jun 21 '14 at 18:37
@Steve: As I discussed with Hayden in the comments. Yes, there are mathematical inaccuracies. But the idea is there. Recall that when we draw a partial order we don't draw all the edges, just the immediate edges. One could mistake that as a graph. But if you think about that chain as a linear order instead, then it's fine. – Asaf Karagila Jun 21 '14 at 18:46

My first actual precise introduction to the geometric concept of infinity came from reading about inversive geometry as a kid.

An inversion about a circle of radius $r$ consists of moving every point in the following way: you draw the ray from the center of the circle through the point and let $d$ be the distance from the point to the center. You then move the point to the place on the ray where it is a distance $r^2 / d$ from the center.

Henceforth, I will call the center of the circle we are inverting about the "origin".

This geometric transformation has some pretty neat properties (and is closely related to linear fractional transformations of the complex plane): circles that don't go through the origin get transformed into circles. Circles that do go through the origin get transformed into lines. Lines get transformed into circles through the origin.

Of course, the transformation isn't defined at the center. So the book said that we'll add another point to the plane called "infinity", and an inversion swaps the point at infinity with the center of the circle.

This naturally fills in some edge conditions: e.g. when I said a line gets transformed into a circle, that's not quite true: the circle would actually be missing a point (the origin). But after adding the point at infinity (and decreeing that it lies on every line), the inversion of a line truly is a circle. And similarly, the inversion of a circle through the origin is well defined at all of its points and it becomes a line (complete with its point at infinity).

Through studying inversive geometry, it became pretty easy to intuit the idea of the point at infinity in terms of actually being a point; there is no need for any of these nebulous heuristics like trying to think about limiting processes or ill-defined quantities that somehow keep growing in size.

I'm not sure how useful this will be to you, though, since the geometric concept of infinity has very little in common with the infinite quantities that appear in set theory.

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One of my favorite intuitive understandings of the infinite is the Riemann sphere. It doesn't work in all contexts but it's great in complex analysis and helps get at the basis of projective space, too.

(4) is not a good understanding of a finite set because you can order the nonzero integers 1, 2, 3, ..., -3, -2, -1 such that 1 and -1 are the start and end and every other element $n$ is between $n-1$ and $n+1$ (in either order).

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$f(x)=1-e^{-x}$ is also constantly increasing as $x$ increases but its limit is just 1. So that is not a well defined way to view infinity IMO. – GinKin Jun 21 '14 at 15:02
Well, you'd have to say that the second derivative of your constantly increasing amount is always >= 0. But then we're getting into how Calculus defines infinity, which is only one of the many ways you can. – Andrew Larsson Jun 21 '14 at 21:50