9:44 AM
@MartinSleziak Hello Martin! I was pleasantly surprised to receive a bounty from you on this answer. Thank you! Do let me know if you have any suggestions for me to improve any of my posts! =)

@user21820 Thanks for your message. I wanted to leave a few comments about various proofs of Zorn's lemma in the Set Theory chatroom, but I did not really get around to it. (I should spend some time thinking about this before doing so.)
Originally I learned the proof using transfinite induction - which seemed very natural to me. (I was already familiar with ordinals. Maybe it shows that I learned things in a non-standard order.)
I think it is good that the proof without ordinals is available on the site - that was my reason for looking at this post.

@MartinSleziak Ah I see. In fact, I also was taught the 'standard'(?) proof of Zorn's lemma via transfinite induction along the ordinals. I was never taught that we can get by with much less than ZFC, because my teacher basically jumped straight up to the von Neumann ordinals. I later found the proof I presented in that post completely on my own...

Many authors (teachers) want to use Zorn's lemma in texts (courses) where they cannot assume about students that they know ordinals. So there certainly are courses/texts which include such proof.
Just searching a bit for the terminology used in the Halmos' text or in the Weston's paper shows some places which use thos proofs.

9:59 AM
@MartinSleziak By the way, it's not stated in my post there, but that exact same technique leads to an almost identical proof of the well-ordering theorem:
In mathematical practice, you do not need to know about ordinals, and actually all you need to know is the well-ordering theorem. It turns out that there is in fact a short proof of that theorem along the same lines as this short proof of Zorn's lemma: Take any set S. Let F be a choice-function that maps each strict subset A of S to a member of S not in A. We say that (T,<) is a tower iff T is a subset of S and < is a well-ordering of T and ∀x∈T ( x = F(T[<x]) ). Any two towers agree (i.e. one order-embeds into the other). Union all the towers. — user21820 Sep 7 '20 at 15:52

That probably depends a bit on what you mean by "mathematical practice".

@MartinSleziak Haha, do you have any example of a theorem outside of set theory and logic that requires the use of von Neumann ordinals?

You often need to read proofs/papers by other people. If ordinals are used as a standard notation or proof technique, you'll need some basic understanding of ordinals to be able to understand those proofs.

Ah that's what you mean.

How commonly are ordinals used in various proofs probably depends a bit on the area of mathematics.

10:05 AM
Yea then of course, but that's a chicken-egg issue. If people learnt the well-ordering theorem and Zorn's lemma and didn't learn about ordinals, we would have seen a different distribution of papers. =)
By the way, those proofs may seem funny at first; how can not actually performing any iteration allow us to construct a complete iteration along a well-ordering that we didn't build? The answer is that these proofs rely crucially on the impredicative comprehension supported by ZC, where we can pick out from all well-orderings exactly those which agree with the recursive relation specified by the choice function.

Thanks for your comments! I should probably get to preparing something to eat for lunch.

10:58 AM
@MartinSleziak You're welcome, and have a good lunch!

5 hours later…
3:54 PM
in CURED, Feb 24 at 0:32, by amWhy
@XanderHenderson Well below, for me too, @107. I know, that's why I commented above. It was at 166 when I started today. I'm hoping that additional gold stars after reaching and multiple of 1000 has brought us more reviewers. We should ask @MartinSleziak if there is any query we can run to determine the number of close vote reviewers now, vs. three months ago.
in CURED, Feb 24 at 7:45, by Martin Sleziak
@amWhy I am not sure whether this is really possible. You (or somebody else) can check the related posts on Meta Stack Exchange: How to query the Data Explorer for reviewer stats, Is it possible to estimate the active reviewer population using the SEDE data?
Although we cannot get (as far as I know) the number of reviewers, we can get the number of users who voted to close.
The query has some limits (only the questions which were actually closed are counted, the date of closure is considered - rather than the date when the close votes was cast), but it still could serve as an approximation: Number of close voters per month.
Maybe somebody who is more fluent in SQL than me could have a look whether I haven't overlooked some problem with that query.
I have previously posted also some other queries concerning closures, but they were about other things, not about close voters: chat.stackexchange.com/rooms/19138/conversation/… chat.stackexchange.com/rooms/19138/conversation/… chat.stackexchange.com/rooms/19138/conversation/…
Query showing Top close/reopen voters.