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Is this a theorem of ZF or a theorem of ZFC?

My suspicion is that it is a theorem of ZFC (the proof I've seen in P. L. Clark's online notes on countable ordinals requires selecting an element from each set of a countable collection of sets).

Does the situation change if the ordinals in the collection are themselves countable?

Dave L. Renfro
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Dacian Bonta
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  • For any nonempty set of (von Neumann) ordinals $A$, $\bigcap A$ is its least element. – BrianO Nov 20 '15 at 01:49
  • Also note that the axiom of choice is not about choosing an element from a collection of sets. It is about choosing one element from each set in an (infinite) collection of non-empty sets. – Eike Schulte Nov 20 '15 at 04:49
  • An infinite choice was made in the proof. I edited for clarity. – Dacian Bonta Nov 20 '15 at 09:14
  • edited to clarify that issue. the proof was creating an infinite descending chain, with choice from each set in the collection. – Dacian Bonta Nov 20 '15 at 09:29

2 Answers2

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This theorem, in fact, is provable in ZF. To see this, let $\alpha_i,i\in I$ be a set of ordinals. If we take union $\bigcup_{i\in I}(\alpha_i+1)$, we will get an ordinal $\beta$ greater than all of $\alpha_i$. But now $\{\alpha_i\}_{i\in I}$ is a subset of $\beta$, and $\beta$ is by definition well-ordered. Finally, every subset of a well-ordered set is well-ordered, and we can finish here.

Every step in this proof is valid without choice.

Wojowu
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  • Since you want each $\alpha_i$ to be an element of $\beta$ (not just a subset of $\beta$), you should define $\beta$ to be $(\bigcup_{i\in I}\alpha_i)+1$ or at least $\bigcup_{i\in I}(\alpha_i+1)$. – Andreas Blass Nov 19 '15 at 21:19
  • @AndreasBlass That's right, thank you. – Wojowu Nov 19 '15 at 21:20
  • Does the well-ordering of $\beta$ follows from its construction, rather than by definition? – Dacian Bonta Nov 19 '15 at 22:31
  • @DacianBonta: By the von Neumann definition of an ordinal, ordinals are indeed well-ordered by definition (and the well-order is the restriction of the set membership relation). It must be proved that a union of ordinals (the so-called supremum) is an ordinal, so the "big claim" of this answer, is that it's provable in ZF. – Steve Jessop Nov 19 '15 at 23:16
  • then, of course, the question would be whether AoC is hiding in one of the constructs used in von Neumann approach to ordinals. – Dacian Bonta Nov 20 '15 at 09:27
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It is not hard to show directly from the definitions:

  1. If $(A,<)$ is a well-ordered set, then for every $B\subseteq A$, $(B,<\restriction_B)$ is also well-ordered.

  2. Every set of ordinals is a subset of an ordinal.

So without using the axiom of choice we have that every set of ordinals is well-ordered.

Asaf Karagila
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  • so, how to show that $\bigcup_{i\in I}(\alpha_i+1)$ is an ordinal? – Dacian Bonta Dec 03 '15 at 19:31
  • By showing that it satisfies the definition of an ordinal, of course. – Asaf Karagila Dec 03 '15 at 19:37
  • transitive set of transitive sets, that is clear; then use regularity to show well ordering? – Dacian Bonta Dec 03 '15 at 20:15
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    You don't need to use regularity to show it's a well-ordering. If $A$ is a non-empty subset of this union, pick any $\alpha\in A$, then $A\cap\alpha$ is a non-empty subset of $\alpha$ or $\alpha$ is the least element of $A$. – Asaf Karagila Dec 03 '15 at 20:35
  • sorry for my lack of understanding. I understand that the least element of A would have an empty intersection with A, and none of the others. But, why does this element with an empty intersection exist? – Dacian Bonta Dec 04 '15 at 19:59
  • It doesn't, then there is an infinite descending sequence of ordinals. – Asaf Karagila Dec 04 '15 at 20:00
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    well, all $\beta\in A$ less than some $\alpha\in A$ would be also elements of said $\alpha$ and even I know that an ordinal is well ordered. Thus those $\beta$ would have a least among them which would also be a least for the entire $A$. Ergo the original set of ordinals is well ordered. Thank you very much! – Dacian Bonta Dec 04 '15 at 20:59