Infinity doesn’t make sense

The various notions of “infinity” appearing in mathematics exist primarily to make life easier: for example, it gives us a way of talking about unending processes or patterns, and can also help us count (e.g., how many numbers there are). However, despite the general convenience of infinities, they really don’t make that much sense.

Fair warning, this post is a lot more technical than my previous two in the “3 Levels” subcategory…

Over lunch

Infinity is often denoted by the “sideways eight” symbol $\infty$ , which perhaps reflects endlessness since you can trace over the infinity symbol with a pencil repeatedly without end. Wikipedia mentions a few possible reasons for why this symbol was chosen: for instance, perhaps it’s because the symbol resembles the Roman numerals CIƆ or CƆ—meaning $1\,000$ and could serve as a stand-in for “a very large amount”—or maybe it’s because it resembles $\omega$ , the final letter “omega” of the Greek alphabet.

In any case, denoting “infinity” with such a benign and small symbol as $\infty$ hides just how big infinity really is… It’s hard enough to understand how big some finite numbers are. I have around $500=5\times10^2$ friends (on Facebook), which is a surprising amount for a student of mathematics. This is not an exceedingly large number to imagine; for example, you might have seen this many people all at once if you take a first year course in calculus at a large university. Let’s use this as our starting point.

The University of Alberta only has roughly $40\,000=4\times10^4$ students—which is eighty times the number of friends I have—and even this number is hard (for me) to digest. The population of Edmonton as a whole is roughly $1\,000\,000=10^6$ —that is twenty five times the population of U of A. Going further, the population of Canada is about $37\,000\,000=3.7\times10^7$ —thirty seven times the population of Edmonton. Then, the population of North America is roughly $370\,000\,000=3.7\times10^8$ —ten times that of Canada. At about twenty times this, the world population is around $7\,900\,000\,000=7.9\times10^9$ . This number is already enormous, but even this pales in comparison to numbers beyond our world: there are $100\,000\,000\,000=10^{11}$ stars in the Milky Way. To populate the Milky Way with stars, every person in the entire world would have to make about thirteen stars each.

While the number $100\,000\,000\,000=10^{11}$ is extremely large, the individual steps I took to get to it might seem rather unremarkable: for example, needing only 25 copies of the U of A to match the population of Edmonton makes Edmonton seem very small. Having each person in the world create thirteen stars to populate the Milky Way actually makes the Milky Way seem disappointingly small compared to the population of Earth as well—one might expect the Milky Way to be much larger! The thing is, you can pretty much always pull tricks like this to make enormous numbers appear way smaller than they really are. This is the fate of being finite.

Infinity, on the other hand, cannot ever be reached using these finite tricks: the only way to go from “finite” to “infinite” is if you do something infinite. You could pretend every star of the Milky Way had a billion Earths, each with ten billion people, but if every person made a finite number of stars, then you would only have a finite number of stars in total. Having infinitely many of something is so incomprehensibly large, our small finite brains have no shot at all at really grasping it.

The disparity between the finite and the infinite is so great that very little intuition one has about finite numbers carries to the infinite setting. One of the most well-known examples of this is Hilbert’s Hotel. Suppose you owned a grand hotel with infinitely many rooms, accommodating one guest in each room. More precisely, you have countably infinitely many rooms, in the sense that every room is numbered 1, 2, 3, 4, … indefinitely. Every room in your hotel can only house one guest, so your hotel is at maximum capacity at the moment.

Now, a new guest enters the hotel to ask for a room. You’re at maximum capacity, so in a finite world, there would be no way to accommodate him; however, things are different in the infinite setting. Being the hotel manager, you instruct all existing guests to move one room down: the guest in the first room moves to room 2, the guest in room 2 moves to room 3, and so on. More precisely, you instruct the guest in room $R$ to move to room $R+1$ . Now, room 1 is empty, and all of your existing guests still have a room! This allows you to assign the brand new guest to room 1 with no fuss. In very rough mathematical symbols, this says $\infty+1=\infty$ .

If you had two new guests, you could just repeat this process twice (that is, tell the guest in room $R$ to go to room $R+2$ , which leaves rooms 1 and 2 vacant). Therefore, $\infty+2=\infty$ . In fact, you can keep adding finitely many new guests to your hotel with absolutely no problem! None of this would be possible if you had finitely many guests… no matter how big this finite number was. If you had ten billion rooms and you were at maximum capacity, you would be unable to accommodate even a single new guest.

What if you wanted to accommodate infinitely many guests? Surely this would be impossible—that’s as many new guests as you have rooms, and all the rooms are full! Nonetheless, it is possible. The trick is to move all your guests into rooms with even room numbers: send the guest in the first room to room 2, send the guest in room 2 to room 4, send the guest in room 3 to room 6, and so on. More precisely, send the guest in room $R$ to room $2\times R$ . Now, all of the odd-numbered rooms are vacant (and there are infinitely many odd numbers), so you are able to accommodate the infinitely many new guests.

This means that $\infty+\infty=\infty$ as well: you can double your hotel capacity easily if you have infinitely many rooms!

On the bus

At this point, you might get the impression that infinity—like the letter omega—is just somehow the “final number” that you just can’t get past: $\infty+1$ , $\infty+2$ , and even $\infty+\infty$ all seem to be equal to $\infty$ . However, this is still finite thinking, and is not true.

So how many new guests can Hilbert’s Hotel accommodate? We have already seen that the hotel can accommodate an additional (countably) infinite number of guests, and you might be able to see how we could take this a bit further. Suppose there were two limousines, each with infinitely many new guests. Can you accommodate all of them?

Indeed: just send the guest in room $R$ to room $3\times R$ . Now, send the guests in the first limousine to rooms 1, 4, 7, 10, … and send the guests in the second limousine to rooms 2, 5, 8, 11, …. You can do something similar to accommodate any finite number of limousines. What about infinitely many limousines?

Surprisingly, it’s still possible. There are several ways to go about it. First, let’s deal with the guests in the hotel: take the guest’s room number (in decimal) and send them to the room with zeroes after each digit. This means the guest in the first room goes to room 10, the guest in room 12 goes to room 1020, and the guest in room 301 goes to room 300010. Now, for the new guest $N$ in limousine $L$ , write $N$ and $L$ out in decimal, and add zeroes on the left until they are the same length. The room number we will send this guest to will have as its digits the result of alternating between the digits of $N$ and $L$ (starting with $N$ ). For example, guest #66 in limousine #33 will go to room 6363; guest #45 in limousine #2 will go to room 4052; guest #17 in limousine #820 will go to room 81270. You can try to convince yourself that this assignment really works (hint: reverse engineer from the final room number where each guest must have come from). This means $(\infty\times\infty)+\infty=\infty$ .

What if there were infinitely many new guests in each limousine, infinitely many limousines being carried in each truck, and infinitely many trucks? It’s still possible! (Hint: try to extend the above example… for guests in the hotel, take the guest’s room number and send them to the room with two zeroes after each digit.) How about infinitely many new guests per infinitely many limousines per infinitely many trucks per infinitely many larger trucks? Yes! How about infinitely many guests per infinitely many limousines per infinitely many trucks per infinitely many larger trucks per infinitely many even larger trucks? Still yes!

How about infinitely many guests per infinitely many limousines per infinitely many trucks per infinitely many larger trucks per infinitely many even larger trucks per … (indefinitely)??

This is finally when there are too many guests. No matter what you might try, it is actually impossible to accommodate this many new guests. To prove this, we will use Cantor’s diagonal argument. For simplicity, let’s even pretend that your hotel started out empty, so we only need to deal with new guests. Every new guest can be identified with an infinite sequence of numbers $G=(g_1,g_2,g_3,g_4,\dots)$ : the new guest is guest # $g_1$ in limousine # $g_2$ in truck # $g_3$ in larger truck # $g_4$ in ….

Suppose there is a way to accommodate every guest to a room in your hotel. Then, we can list out our guests $G_1, G_2, G_3, G_4, \dots$ in order based on their room number: the guest in the first room is $G_1=(g_{1,1},g_{1,2},g_{1,3},g_{1,4},\dots)$ , the guest in the second room is $G_2=(g_{2,1},g_{2,2},g_{2,3},g_{2,4},\dots)$ , and so on. I will prove that this accommodation misses someone (in fact, it misses many people). One such angry guest—let’s call him $\check G$ —is guest # $g_{1,1}+1$ in limousine # $g_{2,2}+1$ in truck # $g_{3,3}+1$ in larger truck # $g_{4,4}+1$ in….

Example of $\check G$ for some arbitrary room assignment $G_1, G_2, G_3, G_4, \dotsc$ of the guests. Note that $\check G$ disagrees with each $G_R$ at the $R$ th column (indicated in red)_{[partial code]}

Indeed, it’s impossible to find $\check G$ in your hotel. If $\check G$ were in room $R$ , then this would mean $\check G = G_R$ . Look at the $R$ th number of the sequence for $\check G$ : by definition, it is $g_{R,R}+1$ . On the other hand, the $R$ th number of $G_R$ is $g_{R,R}$ . These are not equal, so $\check G\neq G_R$ —a contradiction! Therefore, any assignment of guests to rooms will have missed someone, and therefore there are too many guests to accommodate.

Cantor’s diagonal argument generalises to prove that you can always find a bigger infinity. Therefore, we give these infinities different names (in particular, we don’t use $\infty$ to denote infinity). For instance, the infinity I have been using thus far (“countable infinity”) is more appropriately denoted by the symbol $\aleph_0$ (the zero denotes that this is the smallest infinity). With Hilbert’s Hotel, we have proved that $\aleph_0+1=\aleph_0$ , that $\aleph_0+\aleph_0=\aleph_0$ , and moreover that $\aleph_0\times\aleph_0=\aleph_0$ . Cantor’s diagonal argument proved, on the other hand, that $\aleph_0^{\aleph_0} \gneqq \aleph_0$ , so $\aleph_0^{\aleph_0}$ is a bigger infinity than $\aleph_0$ .

What do these identities mean in mathematics? Precisely speaking, $\aleph_0$ represents the size of the set $\mathbb{N} = \{1, 2, 3, 4, \dots\}$ . The fact that $\aleph_0+1=\aleph_0$ means that $\mathbb{N}_0 = \{0, 1, 2, 3, 4, \dots\}$ is not bigger than $\mathbb{N}$ . As for $\aleph_0+\aleph_0=\aleph_0$ , this implies that the set $\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$ of all integers is also not bigger than $\mathbb{N}$ . Moreover, $\aleph_0\times\aleph_0=\aleph_0$ implies that the set $\mathbb{Q} = \{\frac nm \mid n, m \in \mathbb{Z}\}$ of all rational numbers is still not bigger than $\mathbb{N}$ !

Finally, Cantor’s diagonal argument that $\aleph_0^{\aleph_0}\gneqq\aleph_0$ implies that the set $\mathbb{R}$ of all real numbers is finally bigger than $\mathbb{N}$ . Since $\mathbb{R}$ is a continuum, the size of this set is sometimes denoted $\mathfrak{c}$ and called the cardinality of the continuum. One can prove that $\mathfrak{c} = \aleph_0^{\aleph_0}$ , which demonstrates how $\mathfrak{c} \gneqq\aleph_0$ . By adapting Cantor’s diagonal argument, you can then prove that the set of functions on $\mathbb{R}$ is even bigger than $\mathfrak{c}$ (that is, $\mathfrak{c}^{\mathfrak{c}} \gneqq\mathfrak{c}$ ), and so on—there are infinitely many infinities!

Pretty crazy, right? It actually gets crazier. Think about the gap between $\aleph_0$ and $\mathfrak c$ . Are there any infinities that lie in this gap? Cantor spent a lot of time thinking about this problem in the 1870’s, and was convinced that the answer should be no: the smallest infinity that is larger than $\aleph_0$ should be $\mathfrak c$ . This statement is known as the Continuum Hypothesis. Cantor invested a lot of effort in trying to prove this claim, but never succeeded.

Around half a century later, Gödel proved that the usual foundations of mathematics were incapable of proving the Continuum Hypothesis to be true. In 1963, Cohen complemented this result and proved that the usual foundations of mathematics were also incapable of proving the Continuum Hypothesis to be false! Indeed, it turns out that the existence of an infinity that sits between $\aleph_0$ and $\mathfrak c$ cannot be proven nor disproven!

In the lounge

The independence of the Continuum Hypothesis, at least to a relative non-expert like myself, still strikes me as an extremely surprisingly “down-to-Earth” example for the incompleteness of ZFC. I mean, counterintuitive examples in math are usually unworldly, if not somewhat snarky. Mathematically enthused undergrads usually meet their first otherworldly counterexamples in analysis (e.g., continuous but nowhere differentiable functions like the Weierstrass function or the blancmange curve), but it often seems like most “intuitive but false” results are still true in practice, and you have to go out of your way to encounter problems.

Even the proof of the (in)famous Gödel’s Incompleteness Theorems is done by constructing sentences that indirectly refer to the underlying logical system and provability within that system. This is somehow a cheap shot since it doesn’t really seem related to the actual mathematics of the system per se: it’s true that ZFC cannot be proved consistent within its own system (unless ZFC is inconsistent), but this statement says more about ZFC than the mathematics we hope to encapsulate with ZFC. On the other hand, the Continuum Hypothesis is way more “mathematical” since it can be spun as a statement about the real numbers. More precisely, the Continuum Hypothesis asserts that any uncountable subset of real numbers is in bijection with the real numbers… how such a statement cannot be proven in ZFC is much more baffling.

Moreover, the independence of the Continuum Hypothesis leads to so many reasonable-sounding statements that are independent of ZFC. That link is a pleasure to read, but allow me to summarise some of my favourites:

[cf] The projective dimension of the $\mathbb{C}[x,y,z]$ -module $\mathbb{C}(x,y,z)$ is two or three depending on whether the Continuum Hypothesis is true or false, respectively. The cited answer reveals the culprit to be how projective dimension relies on decompositions into countably-generated submodules.
[cf] Suppose $f:[0,1]\times[0,1]\to[0,1]$ , where $f(x,-)$ and $f(-,x)$ are measurable for almost every $x\in[0,1]$ . If the Continuum Hypothesis is true, then it is possible to find such an $f$ where $\int_0^1\int_0^1f(x,y)\mathrm dx\mathrm dy \neq \int_0^1\int_0^1f(x,y)\mathrm dy\mathrm dx$ ; however, the unconditional statement is independent of ZFC.
[cf] This is concerned with Wetzel’s Problem: let be a set of holomorphic functions , and denote by the set of complex numbers obtained by evaluating every function in at .
- Note that if $\mathcal{F}(z)$ is finite for every $z\in\mathbb{C}$ , then $\mathcal{F}$ is also finite, as proven here. Wetzel asked if $\mathcal{F}$ were necessarily countable if all $\mathcal{F}(z)$ were countable.
- Erdős proved that Wetzel’s problem is equivalent to the negation of the Continuum Hypothesis: if the Continuum Hy;othesis were true, then there exists an uncountable family of holomorphic functions that take on countably many values at every complex number; if the Continuum Hypothesis were false, then no such family can exist.
[cf] Assuming the Continuum Hypothesis, you can prove that every function $f:\mathbb{R}^2\to\mathbb{R}$ decomposes as $f(x,y) = \sum_{n=1}^\infty g_n(x)h_n(y)$ . The unconditional statement, however, is independent of ZFC. This might sound more believable if you remember that $f$ is an arbitrary function, but nonetheless…

I have somewhat sidetracked to just talking about the Continuum Hypothesis, but I think this really highlights how difficult it is to wrap our heads around infinity. If you review the axioms of ZFC, only one of them directly addresses infinity (namely, the axiom of… infinity). The other axioms are stated sufficiently broadly to account for sets of any size, but clearly had finite intuition (in fact, the axiom of pairing is very finite, though this may be a consequence of only being able to quantify over finitely many variables in an axiom): we could assert an axiom of finiteness—claiming that all sets were finite—and the remaining axioms would still make sense; in fact, the axiom of choice becomes simply true.

On one hand, it’s rather incredible that the axiom of infinity, which makes just the innocent assertion that $\mathbb{N}$ is a set, is enough to blow ZFC into an enormous universe of infinities. On the other hand, how surprising can it be that a lot can’t be proven nor disproven about significantly larger sets (like $\mathbb{R}$ )? By no means am I an expert on this sort of thing, but our only real handle on infinities in ZFC is the C—the axiom of choice—and even then, it gives us a very weak handle.

A popular paraphrasing of the axiom of choice is something like this: given infinitely many nonempty boxes, we may choose an item from every box. However, we’re not really the ones doing the choosing; if anything, we’re just allowed to be given an item from every box. This is further highlighted by the axiom of global choice, as it basically says that we have been given (by whom? God?) some “chosen one” element from every single nonempty set in existence. (This is part of a conservative extension of ZFC, so this doesn’t weaken my point.) This is all we have, and even then there is a lot of interest in studying mathematics without the axiom of choice at all.

Without the axiom of choice, we are suddenly faced with even more independence results. Does the vector space of all real-valued sequences have a (Hamel) basis?^[cf] Between any two sets, can we always inject one into another? Granted, there are still some things that we can prove without the axiom of choice regarding infinite sets (the first one that comes to mind being the Cantor-Schröder-Bernstein Theorem, or maybe I’ll also count transfinite induction), and we can usually work around dependencies on this axiom, but it’s quite limiting for the ambient theory to lack the axiom of choice. If we were more capable of doing “infinitely many things in finite time” then perhaps we wouldn’t be faced with so many independence results (perhaps we would also be able to prove the axiom of choice, too), but we have ultimately been bounded by our finite minds.