This is Asaf's answer from this link: How do we know an $ \aleph_1 $ exists at all?

I don't understand this sentence that is;

From each equivalence class choose the representative which is an ordinal (which does not require any form of choice, as the equivalence classes can be described without the axiom of choice, as well as being an ordinal). The set of representatives is a set of ordinals, we take its union.

Here's what I think this means. Please tell me I'm following this argument correctly.

Let $X$ be the class of all the well orderings of $\omega$

Let $[G]$={$F \in X$|$F$ is isomorphic with $G$} for every $G\in X$.

Then we 'choose' representatives from each $[G]$ and take a union.

I see this is definitely a choice since there might be infinitely many [G]'s. Why is this not a choice??

  • $\begingroup$ I think I will revise my answer on that thread to make this construction clearer. $\endgroup$ – Asaf Karagila Jul 22 '12 at 12:53
  • $\begingroup$ @Asaf Thank you. Excuse me keep asking different questions, but it's hard for me to understand the part deriving a contradiction too.(that is, i know that $\alpha$ is a supremum but why ordinals less than $\alpha$ is isomorphic to some order relation that well orders $\omega$? And why that $\alpha + 1$ is countable is a contradiction?) $\endgroup$ – Katlus Jul 22 '12 at 13:09
  • $\begingroup$ @Asaf Or did you mean this? Let $X$={$\beta \in OR$|$\alpha$ is equipotent with $\omega$}. Let $\gamma$ be the least element of $X$. Is this $\gamma$ same as $\alpha$ in your argument? $\endgroup$ – Katlus Jul 22 '12 at 13:24
  • $\begingroup$ I dunno why edit button is unavailable to me now.. I meant 'union' not 'least' above. $\endgroup$ – Katlus Jul 22 '12 at 13:26

Since your choice comes from the collection of ordinals, you can specify that you want the least such ordinal that satisfies the condition. This ordinal exists and is unique, so there is no choice involved. (Of course as Asaf's answer says it turns out that there is only one ordinal that satisfies the property, but you can also make this selection without knowing that fact)

  • $\begingroup$ Now i got it. Thank you. Would you please answer my comment too please? $\endgroup$ – Katlus Jul 22 '12 at 13:12
  • $\begingroup$ @Katlus To which comment are you referring? $\endgroup$ – Francis Adams Jul 22 '12 at 16:15

There is a theorem which states that every well-ordered set is isomorphic to a unique ordinal.

Since the equivalence relation is that of order isomorphism all the elements in a given isomorphism class are isomorphic to a unique ordinal. We thus have eliminated the need to choose, since there is only one possible choice.

The full point I the sentence was that the statements saying that two ordered sets are isomorphic, and that a certain set is an ordinal do not require the axiom of choice. Thus this makes sense to do these constructions in ZF.

  • $\begingroup$ Perhaps a typo? Did you mean: The full point in the sentence, or The full point of the sentence...? $\endgroup$ – Martin Sleziak Dec 16 '13 at 9:38
  • $\begingroup$ Martin, thanks. I'm trying to recall where I was at that time and why my post had those typos. Most likely on my way to lunch and I typed that from my phone... Still won't explain why the autocorrect didn't kick in, though. $\endgroup$ – Asaf Karagila Dec 16 '13 at 11:54

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