The most popular music video of last year is, arguably, *Gangnam Style* by the famous Korean artist, Ψ. One of the noteworthy features is the presence of ‘accumulation points’ in the music; these occur at 1:07 and 2:29 in the video:

[youtube=http://www.youtube.com/watch?v=9bZkp7q19f0]

Also, the notes in the song are *well-ordered*, meaning that every set of notes has an earliest element. Indeed, since there are two limit points, the song has order type 2ω + *n*, where *n* is the (finite) number of notes after 2:29 in the video.

The night before last, this inspired me to produce music corresponding to a larger infinite ordinal. ω^2 limit points actually sound effective, but when you approach larger limit ordinals such as ω^3 and ω^4, it degenerates into an absolute cacophany. As a proof of concept, I decided to produce the sound of ω^4.

The scale above represents the entire duration (3:39) of the music, with a linear scale showing certain ordinals marked on it. The first clap you can hear is ‘1’, followed shortly by ‘2’, ‘3’, and all of the other positive integers before reaching a limit point at ω. The entire process is repeated again and again, resulting in limits of limits, such as ω^2 (0:10 in the sound file) and ω^3 (0:46). The music ends at ω^4, although it is possible in principle to produce ordinal music corresponding to any countable ordinal. Most pieces of popular music merely have finite ordinals, with *Gangnam Style* being a rare exception.

### Supertasks

Performing an infinite number of operations in a finite amount of time (such as producing music corresponding to ω) is a transfinite process known as a *supertask*. For instance, a computer capable of doing an elementary operation in 1/2 of a second, and the next in 1/4 of a second, and the third in 1/8 of a second, can effectively solve the halting problem in 1 second. Such a computer is known as a Zeno machine, since it resembles Zeno’s paradox.

Small mistake in second paragraph – Gangnam Style ordinal isn’t 2ω + n, because from non-associativity 2ω=ω. There should be ω*2+n in there

Ordinal multiplication is still associative; it’s the non-commutativity which is important here. Yes, strictly speaking it should be ω*2, as it’s distinct from 2*ω = sup{2, 4, 6, 8, …} = ω. When writing ordinals in Cantor normal form, it should be unambiguous that powers of ω are being post-multiplied by integers, rather than pre-multiplied.

I believe that although $2*\omega = \omega$, most people doing set theory will normally write $2\omega$ as a shorthand for $\omega * 2$, distinguishing $2\omega$ and $2*\omega$

As Adam said. How do you get maths in comments?

I don’t think that WordPress supports the insertion of formulae into comments, although you can copy and paste the omegas from the blog post. Anyway, your LaTeX in the previous post is sufficiently legible.

For future reference, use @latex x^3@ (or whatever) where @ represents $.

What’s wrong with this counterargument to ‘ordinal train’ problem?

Suppose we have a valid journey where the train becomes empty at ω_1. Then I shall consider an identical journey, except where I add a driver to the train at the first stop. The driver can be considered to be a passenger who never gets off the train. Thus I have a valid journey where the train is not empty at the end.

The ordinal train becomes empty many times before ω_1 (although precisely when this happens depends on the order that the passengers board/disembark). In your modified situation, the driver would be forced to leave at one of those stops.

Oops, I forgot the bit that said “if it’s empty”, and just assumed it never was.

I think that the article proving the OTP is a little confusing in its definition of f. Basically, what it means is that among the people on the train at stop alpha, some of them will be on at omega_1 and some won’t. f just kicks of the people who won’t be on at stop omega_1, and of course anyone who will be on at omega_1, we don’t care about.

Then again, there’s a much simpler way to guarantee the existence of a stop where the train is empty. Supposing otherwise, that the train is never empty before omega_1, we note that someone must get off at each countable ordinal. Hence, we can describe every countable ordinal with a finite sequence of ordinals xi, where x1= 1 and someone gets on at xj and off at x(j+1), as there’s no infinite decreasing sequence. Since there are countably many possible choices for each ordinal, we have shown that there are countably many ordinals, contradiction.

(That should say “countably many countable ordinals”)

I want to hear a music that represented by Church-Kleene ordinal.

Also, is it possible to reach w_1 from music?

No, we can reach any ordinal below omega_1 (i.e. any countable ordinal), but not omega_1 itself.

For a proof of the latter, suppose we have an order-preserving injection from omega_1 to R. Then, for each ordinal x, we can find a rational q between x and SUCC(x), thereby giving an injection into Q. Hence there are only countably many ordinals indexed by omega_1. Contradiction.

For a proof of the former, note that any successor ordinal can be expressed by appending an extra note, and any limit ordinal can be expressed by concatenating a countable sequence of pieces of ordinal music, and applying a homeomorphism from [0, infinity) to [0, 1) to map it into a finite interval.

Such music is possible, but no algorithm is capable of producing it.