Counting to Infinity and Beyond

Anyone can learn to count in the ordinals, even a child, and so let us learn how to count to $\omega^2$, the first compound limit ordinal.

The large-format poster is available:

Some close-up views:

I would like to thank the many people who had made helpful suggestions concerning my poster, including Andrej Bauer and especially Saul Schleimer, who offered many detailed suggestions.

Infinite sets and Foundations—Interviewed on the Daniel Rubin Show

I was interviewed 26 August 2021 by mathematician Daniel Rubin on his show, and we had a lively, wideranging discussion spanning mathematics, infinity, and the philosophy of mathematics. Please enjoy!


0:00 Intro

2:11 Joel’s background. Interaction between math and philosophy

9:04 Joel’s work; infinite chess.

14:45 Infinite ordinals

22:27 The Cantor-Bendixson process

29:41 Uncountable ordinals

32:10 First order vs. second order theories

41:16 Non-standard analysis

46:57 The ZFC axioms and well-ordering of the reals

58:11 Showing independence of statements. Models and forcing.

1:04:38 Sets, classes, and categories

1:19:22 Is there one true set theory? Are projective sets Lebesgue measurable?

1:30:20 What does set theory look like if certain axioms are rejected?

1:36:06 How to judge philosophical positions about math

1:42:01 Concrete math where set theory becomes relevant. Tarski-Seidenberg on positive polynomials.

1:48:48 Goodstein sequences and the use of infinite ordinals

1:58:43 The state of set theory today

2:01:41 Joel’s recent books

Go check out the other episodes on Daniel’s channel!

Cantor’s Ice Cream Shoppe

Welcome to Cantor’s Ice Cream Shoppe! A huge choice of flavors—pile your cone high with as many scoops as you want!

Have two scoops, or three, four, or more! Why not infinitely many? Would you like $\omega$ many scoops, or $\omega\cdot2+5$ many scoops? You can have any countable ordinal number of scoops on your cone.

And furthermore, after ordering your scoops, you can order more scoops to be placed on top—all I ask is that you let me know how many such extra orders you plan to make. Let’s simply proceed transfinitely. You can announce any countable ordinal $\eta$, which will be the number of successive orders you will make; each order is a countable ordinal number of ice cream scoops to be placed on top of whatever cone is being assembled.

In fact, I’ll even let you change your mind about $\eta$ as we proceed, so as to give you more orders to make a taller cone.

So the process is:

  • You pick a countable ordinal $\eta$, which is the number of orders you will make.
  • For each order, you can pick any countable ordinal number of scoops to be added to the top of your ice-cream cone.
  • After making your order, you can freely increase $\eta$ to any larger countable ordinal, giving you the chance to make as many additional orders as you like.

At each limit stage of the ordering process, the ice cream cone you are assembling has all the scoops you’ve ordered so far, and we set the current $\eta$ value to the supremum of the values you had chosen so far.

If at any stage, you’ve used up your $\eta$ many orders, then the process has completed, and I serve you your ice cream cone. Enjoy!

Question. Can you arrange to achieve uncountably many scoops on your cone?

Although at each stage we place only countably many ice cream scoops onto the cone, nevertheless we can keep giving ourselves extra stages, as many as we want, simply by increasing $\eta$. Can you describe a systematic process of increasing the number of steps that will enable you to make uncountably many orders? This would achieve an unountable ice cream cone.

What is your solution? Give it some thought before proceeding. My solution appears below.

Alas, I claim that at Cantor’s Ice Cream Shoppe you cannot make an ice cream cone with uncountably many scoops. Specifically, I claim that there will inevitably come a countable ordinal stage at which you have used up all your orders.

Suppose that you begin by ordering $\beta_0$ many scoops, and setting a large value $\eta_0$ for the number of orders you will make. You subsequently order $\beta_1$ many additional scoops, and then $\beta_2$ many on top of that, and so on. At each stage, you may also have increased the value of $\eta_0$ to $\eta_1$ and then $\eta_2$ and so on. Probably all of these are enormous countable ordinals, making a huge ice cream cone.

At each stage $\alpha$, provided $\alpha<\eta_\alpha$, then you can make an order of $\beta_\alpha$ many scoops on top of your cone, and increase $\eta_\alpha$ to $\eta_{\alpha+1}$, if desired, or keep it the same.

At a limit stage $\lambda$, your cone has $\sum_{\alpha<\lambda}\beta_\alpha$ many scoops, and we update the $\eta$ value to the supremum of your earlier declarations $\eta_\lambda=\sup_{\alpha<\lambda}\eta_\alpha$.

What I claim now is that there will inevitably come a countable stage $\lambda$ for which $\lambda=\eta_\lambda$, meaning that you have used up all your orders with no possibility to further increase $\eta$. To see this, consider the sequence $$\eta_0\leq \eta_{\eta_0}\leq \eta_{\eta_{\eta_0}}\leq\cdots$$ We can define the sequence recursively by $\lambda_0=\eta_0$ and $\lambda_{n+1}=\eta_{\lambda_n}$. Let $\lambda=\sup_{n<\omega}\lambda_n$, the limit of this sequence. This is a countable supremum of countable ordinals and hence countable. But notice that $$\eta_\lambda=\sup_{n<\omega}\eta_{\lambda_n}=\sup_{n<\omega}\lambda_{n+1}=\lambda.$$ That is, $\eta_\lambda=\lambda$ itself, and so your orders have run out at $\lambda$, with no possibility to add more scoops or to increase $\eta$. So your order process completed at a countable stage, and you have therefore altogether only a countable ordinal number of scoops of ice cream. I’m truly very sorry at your pitiable impoverishment.