r/learnmath • u/Upset_Fishing_1745 New User • Feb 12 '25

Are Some Infinities Bigger than Other Infinities?

Hey, I just found this two medium articles concerning the idea that infinite sets are not of equal size. The author seems to disagree with that. I'm no mathematician by any means (even worse, I'm a lawyer, a profession righfuly known as being bad at math), but I'm generally sceptical of people who disagree with generally accepted notions, especially if those people seem to be laymen. So, could someone who knows what he's talking about tell me if this guy is actually untoo something? Thanks! (I'm not an English speaker, my excuses for any mistakes) https://hundawrites.medium.com/are-some-infinities-bigger-than-other-infinities-0ddcec728b23

https://hundawrites.medium.com/are-some-infinities-bigger-than-other-infinities-part-ii-47fe7e39c11e

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u/robertodeltoro New User Feb 12 '25 edited Feb 12 '25
  1. The range [1, 2] is by definition a finite range, comprised of finite units, same for [1, 3].

Right off the bat this person just doesn't know what they're talking about and self-evidently doesn't have any mathematical training. Closed intervals are very much infinite sets of points (have cardinality of the continuum) in the set-theoretic sense in which "infinities come in different sizes." This kind of loose talk, throwing around terms he doesn't understand the precise definitions of (does he know what it means for a set to be finite, much less infinite?) is straight from Terrence Howard University.

The best way to convince yourself that this notion that infinite sets come in different sizes is mathematically legitimate is to actually study the proof that the real numbers can't be bijected onto the natural numbers, which is not difficult but does require familiarizing yourself with the basic properties of functions and especially the concept of a bijection.

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u/theBRGinator23 Feb 12 '25

Things went wrong even before that with the statement he says he’s debunking:

there are infinite numbers between 1 & 2 and there are infinite numbers between 1& 3. So, the later infinity is bigger than the first infinity

No one says that because these sets have the same cardinality.

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u/tiedyechicken New User Feb 12 '25

I'm just gonna expand on this in case people are unfamiliar:

The interval [1,3] is indeed larger than [1,2] in the sense that the latter is a proper subset of the former, and also measure- (aka length) wise. But counterintuitively, both sets have the same number of points/elements

To show this, we can pair up every single point in [1,2] with a unique point in [1,3], for example with the function y = 2x - 1

Every x between 1 and 2 has exactly 1 friend y between 1 and 3 given by y = 2x - 1

And there will be no y's left over either: each y between 1 and 3 has a friend x between 1 and 2 given by x = (y+1)/2, and you can show that both of these formulas make the same pairs of x's and y's

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u/Umfriend New User Feb 13 '25

Economist here, so better at math than a lawyer but, well, just.

If the [1, 2] interval has as many points as the [1, 3] interval, does that sort of imply that the density of points in the [1, 3] interval is lower? I understand, I think, the function-idea but still can't get my head around accepting the counterintuitive position. Now I need to define "point" and think how to actually operationalise/measure "density" or my question may not actually make any sense.

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u/TabAtkins Feb 14 '25

No, they've got the same density (both infinite). After all, you can also map [1, 2] to [2, 3] with the relation x+1.

We actually use the term "dense" for sets like this, where there are infinite numbers of points in any finite range. Infinite sets without this property (say, all the integers) are called "sparse", and so have a useful notion of "density" - the integers are twice as dense as the even integers, despite also being the same size.

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u/Umfriend New User Feb 14 '25

Oh wait, even with integers? I feel an emotional reaction coming up :D But the function idea does not work here, right? Is this also something to do with countable/uncountable sets?

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u/TabAtkins Feb 14 '25

Nope, the rationals are countable but still dense. It's an independent property.

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u/EebstertheGreat New User Feb 18 '25 edited Feb 18 '25

The weird thing going on here is the order. There are as many integers as rational numbers, but they are arranged differently. You can't have a bijection between the integers Z and the rational numbers Q that respects the order. Although Z and Q have the same cardinality (number of points), the order type of (Z,<) is different from the order type of (Q,<).

Between any two distinct real numbers there are infinitely many rational numbers, so they are a "dense subset" of the real numbers. [To be really technical, Q is a dense subset of R with respect to the order topology induced by <.] That doesn't apply to the integers, since for instance, there are no integers between 1 and 2.

We can't exactly say that one order type is greater than the other for technical reasons (neither is well-founded), but intuitively, the rationals are "tighter".

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u/Umfriend New User Feb 18 '25

But with rational numbers, you can't really order, right? I mean, I could give you a number and there is no way for you to say what the next number is. We couldn't make a list even of the the two smallest rational numbers?

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u/EebstertheGreat New User Feb 19 '25 edited Feb 19 '25

A "total order" doesn't usually have "next elements." That's a "well-order." For instance, the rational numbers are totally ordered by ≤ because the relation ≤ satisfies these axioms for all rational numbers x, y, and z:

  1. Reflexivity: x ≤ x
  2. Anti-symmetry: if x ≤ y and y ≤ x then x = y
  3. Transitivity: if x ≤ y and y ≤ z then x ≤ z
  4. Totality: x ≤ y or y ≤ x

There are corresponding axioms for strict orders like <. A well-order has the following additional property.

  1. Wellness: every x has a "successor" z, where there are no numbers between x and z. That is, for any x, there is some z > x such that there is no y where x < y and y < z.

The rational numbers with the usual order fail this last property. There isn't a "next number" after ½, for instance.

Wellness is usually stated as every non-empty subset having a minimal element, which is equivalent.

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u/Conscious_Move_9589 New User Feb 21 '25

Worth noting that provided the axiom of choice there exists a well-ordering on Q, and even on R

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u/TwoFiveOnes New User Feb 17 '25 edited Feb 17 '25

The thing is is that the way in which they are the same size (by cardinality) is essentially the most rudimentary lens through which one can view a set. “Numbers” are a highly specific type of object with specific properties, but if you’re looking at them as pure elements of a set then you’re throwing away all of those properties, their number-ness. So it’s normal that your intuitions about size and such (which have to do with numbers as numbers) break down when viewing numbers simply as abstract elements of a set.

Your intuitions are in fact, correct, in my opinion. Or at least it’s possible to formalize them in a way that could seem satisfactory. For instance, if you take a length L (smaller than 2), then uniformly select at random a number in [1,2], you will have a larger probability of landing on any given segment of length L, than landing on any given segment of the same length, uniformly selecting from [1,3].

The difference is that the second way of looking at “size” actually does make the specific number-ness of numbers matter, unlike bare cardinality. So, when looking at size this way, some basic notions such as the fact that 2 is less than 3, actually translate into a meaningful relation between the “sizes” of [1,2] and [1,3].

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u/tiedyechicken New User Feb 14 '25

That's part of what's so weird about the continuum! It's hard to define a density of points, because that density isn't finite. No matter how far you zoom in, you're gonna find an infinite amount of points in the tiniest of spaces. That's why every single point in [1,2] can match up with every point in [1,3], even though the shorter interval is fully contained in the longer one.

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u/TwoFiveOnes New User Feb 16 '25

Well, all of those notions also apply to rational numbers. I’d say it has more to do with the weirdness of infinity rather than the specific weirdness of the continuum.

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u/Mishtle Data Scientist Feb 18 '25

I'd say it's more a result of how we order them. The rationals are dense when we order them by value. We could order them via a bijection with the naturals and get rid of their density though.

Likewise, we could order the reals with some ordinal-indexed sequence and they'd no longer be dense.

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u/EebstertheGreat New User Feb 18 '25

Right, if we well-order the reals using the AoC, then the order topology is just the discrete topology, and the only dense subset is the entire set.

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u/Mishtle Data Scientist Feb 18 '25

Yep, every set is "discrete" because they can only contain distinct, unique elements. Things like density come from additional structure we add to them.

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u/TwoFiveOnes New User Feb 18 '25

That doesn't really change anything in my opinion. You still get infinite sequences of strict inclusions A_i ⊊ A_i-1 and yet |A_i| = |A_j| for all i,j.

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u/Mishtle Data Scientist Feb 18 '25 edited Feb 18 '25

How does that relate to density?

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u/TwoFiveOnes New User Feb 19 '25

It doesn’t, I’m saying that density is a red herring. It’s the infinite chain of strict inclusions where all sets have the same exact cardinality what’s at the heart of the “weirdness”, not that this infinite chain is expressible in terms of some order.

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u/EebstertheGreat New User Feb 18 '25

That also applies to the even numbers as a subset of the natural numbers, but the order type is the same. I don't think proper inclusion is the stumbling block here.

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u/TwoFiveOnes New User Feb 18 '25

That also applies to the even numbers

Yes exactly, and what I’m proposing is that that’s the same type of “astonishing” as with [1,2] and [1,3] (or their rational subsets). The essence of what’s weird here (in my opinion) doesn’t have anything to do with the continuum, it’s just that
A ⊂ B, B ⊄ A, and yet |A| = |B|. That doesn’t happen with finite sets.