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$1$- We can prove by induction that If $\preceq$ is a total ordering on $A$, every non-empty $\color{blue}{finite}$ subset $S$ of $A$ has a least element and a greatest element.

$2$- But how to prove that If $\preceq$ is a total ordering on $A$, every non-empty $\color{red}{countable}$ subset $S$ of $A$ has a least element and a greatest element.

Here we can't use induction anymore.

Edit:

I've seen this question "Suppose $A$ has the property that every non-empty countable subset has a least element. Show that $A$ is well ordered. (Hint: show first that A is totally ordered.)". So I'm wondering could we prove the second paragraph?

user1035648
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    You can't prove it because it's not true. The set of all natural numbers is a counterexample. – Ethan Bolker Oct 12 '23 at 14:37
  • The property "every non-empty countable subset has a least and greatest element" is equivalent to being finite. This follows from this famous fact about sequences having monotone subsequences. I'm interested to know why you thought it would be true! Is there something else you're trying to prove? There are some generalisations you could make. "Every non-empty subset has a least element" is equivalent to "every decreasing $\omega$-indexed sequence is eventually constant". You can generalise by talking about $\omega_1$, but that is a bit trivial. – Izaak van Dongen Oct 12 '23 at 14:51
  • @EthanBolker, you're right. Let's just consider the least element. In this condition, can we prove? – user1035648 Oct 12 '23 at 19:17
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    @user1035648 Still no. Consider the set of all integers, which is totally ordered. What you need is well ordering. That's what makes induction work. https://en.wikipedia.org/wiki/Well-ordering_principle – Ethan Bolker Oct 12 '23 at 19:20
  • @IzaakvanDongen, I've edited the question and added what drove me to this point. – user1035648 Oct 12 '23 at 19:32
  • @EthanBolker, yes, that's right. Thank you. – user1035648 Oct 12 '23 at 19:33
  • Your edit asks a different question. If that's what you want to know, delete this one and ask that. Show your work and where you are stuck. – Ethan Bolker Oct 12 '23 at 19:38
  • @EthanBolker No, I know the answer to that question. I just wanted to know why they didn't want us to prove in the "non-empty countable" case. – user1035648 Oct 12 '23 at 19:41
  • Well I answered this question, it's "no". I can post that answer if you like, so you can accept it, or you can post the answer yourself and accept, or you can delete the question. Just don't leave it on the unanswered queue. – Ethan Bolker Oct 12 '23 at 19:46
  • @EthanBolker Please answer it. – user1035648 Oct 12 '23 at 20:45

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Consider the set of all integers, which is totally ordered, countable, with neither a maximum nor a minimum element.

What you need is well ordering, not total ordering. That's what makes induction work. See the wikipedia page.

Ethan Bolker
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