In the second to last section of this page -- with the puzzle, a claim is made that assuming the axiom of choice and the continuum hypothesis, you can order the real numbers in [0, 1] so that each number only has a countable number of predecessors.

https://www.scottaaronson.com/democritus/lec2.html

This definitely doesn't seem reasonable to me. How can that be possible?

## Ordering a real interval with countable predecessors - possible?

**Moderators:** gmalivuk, Moderators General, Prelates

- gmalivuk
- GNU Terry Pratchett
**Posts:**26356**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

### Re: Ordering a real interval with countable predecessors - possible?

If you take all the ordinals with finitely many predecessors, you end up with ℵ

If you take all the ordinals with at most ℵ

If you assume CH, then you also have ℵ

_{0}of them. (That is, all integers have finitely many predecessors, but there are infinitely many integers.)If you take all the ordinals with at most ℵ

_{0}predecessors, you have ℵ_{1}of them (the next larger cardinal).If you assume CH, then you also have ℵ

_{1}real numbers, and if you assume AC you can well-order them, and since there's as many of them as there are ordinals with at most ℵ_{0}predecessors, you can well-order the reals (or the reals on [0,1]) so each only has at most countably many predecessors.### Re: Ordering a real interval with countable predecessors - possible?

gmalivuk wrote:If you take all the ordinals with at most ℵ_{0}predecessors, you have ℵ_{1}of them (the next larger cardinal).

This was the unintuitive part for me. Here's a proof I tried to put together, although I've never worked with ordinal numbers before.

Suppose the cardinality of the set X of all ordinals with at most ℵ

_{0}predecessors was just ℵ

_{0}instead of ℵ

_{1}. But then, X would be itself an ordinal with ℵ

_{0}predecessors, which means X contains itself. But since ordinals have to be strictly well ordered, we can't have X < X, so we have a contradiction.

The part I'm not sure about is whether the set of all ordinals with at most ℵ

_{0}predecessors must be an ordinal itself.

- gmalivuk
- GNU Terry Pratchett
**Posts:**26356**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

### Re: Ordering a real interval with countable predecessors - possible?

That's pretty much the proof given on the page you linked to, yes.>-) wrote:↶gmalivuk wrote:If you take all the ordinals with at most ℵ_{0}predecessors, you have ℵ_{1}of them (the next larger cardinal).

This was the unintuitive part for me. Here's a proof I tried to put together, although I've never worked with ordinal numbers before.

Suppose the cardinality of the set X of all ordinals with at most ℵ_{0}predecessors was just ℵ_{0}instead of ℵ_{1}. But then, X would be itself an ordinal with ℵ_{0}predecessors, which means X contains itself. But since ordinals have to be strictly well ordered, we can't have X < X, so we have a contradiction.

How are ordinals defined, set-wise?The part I'm not sure about is whether the set of all ordinals with at most ℵ_{0}predecessors must be an ordinal itself.

### Who is online

Users browsing this forum: No registered users and 4 guests