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

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

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?

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

If you take all the ordinals with finitely many predecessors, you end up with ℵ0 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 ℵ0 predecessors, you have ℵ1 of them (the next larger cardinal).

If you assume CH, then you also have ℵ1 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 ℵ0 predecessors, you can well-order the reals (or the reals on [0,1]) so each only has at most countably many predecessors.
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### Re: Ordering a real interval with countable predecessors - possible?

gmalivuk wrote:If you take all the ordinals with at most ℵ0 predecessors, you have ℵ1 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 ℵ0 predecessors was just ℵ0 instead of ℵ1. But then, X would be itself an ordinal with ℵ0 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 ℵ0 predecessors must be an ordinal itself.

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

>-) wrote:
gmalivuk wrote:If you take all the ordinals with at most ℵ0 predecessors, you have ℵ1 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 ℵ0 predecessors was just ℵ0 instead of ℵ1. But then, X would be itself an ordinal with ℵ0 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.
That's pretty much the proof given on the page you linked to, yes.

The part I'm not sure about is whether the set of all ordinals with at most ℵ0 predecessors must be an ordinal itself.
How are ordinals defined, set-wise?
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