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THIS QUESTION IS NOT A DUPLICATE OF THIS ONE!!!

So I would like to discuss the following proof of transfinite induction which is taken by the text Introduction to Set Theory wroten by Karell Hrbacek and Thomas Jech: if you like here you can find the original text of the proof.

Theorem

Let be $\pmb P(x)$ a property and let we suppose that if $\alpha$ is an ordinal number such that $\pmb P(\beta)$ is true for all $\beta\in\alpha$ then also $\pmb P(\alpha)$ is true: so if this happens then $\pmb P(\alpha)$ is true for all ordinal.

Proof. So if $\pmb P(x)$ was not true for some $\alpha$ then the set $$ F:=\big\{\beta\in\alpha+1:\neg P(\beta)\big\} $$ was not empty so that by the well ordering of $\alpha+1$ it would have a minimum element $\beta_0$ and this would be such that $\pmb P(\beta)$ is true for all $\beta<\beta_0$ so that by the hypothesis $\pmb P(\beta_0)$ should be true and clearly this is impossible: so we conclude that $\pmb P(x)$ is true for all ordinals.

So if $\beta_0$ is not zero then surely any ordinal $\beta$ less than $\beta_0$ verify $\pmb P(x)$ but in my opinion this is surely true only assuming that $P(0)$ is true: however any author assumes as hypothesis that $0$ verify the property $\pmb P(x)$ so that I ask clarification about; moreover the same problem exist in the linked proof where I think it is necessary to assume that $\pmb P$ is true for the minimum of $C$ where I point out $C$ is a general well ordered set. So could someone explain why the theorem does not assume that $P(0)$ is true, please?

Antonio Maria Di Mauro
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  • I find your question confusing. I don't understand what it is. Do you have a problem with the given proof? In what step concretely? Where did you get this proof? It's wrong (but not by any reason that you seem to mention). Did you copy it over from somewhere and translate it yourself? If so, I suggest you also include the original proof. – Git Gud Mar 18 '22 at 14:40
  • I found this proof in the text Introduction to set theory by Karell Hrbacek and Thomas Jech: if you like I can add as image the original proof but it is exactly equal to the proof I linked. Anyway the problem is that I think it is necessary to assume that $P(0)$ is true because otherwise the proof is invalid: I explain this at the end of the question. So what do you think about? – Antonio Maria Di Mauro Mar 18 '22 at 14:43
  • What edition is this? I'm looking at the third edition and the proof is different in at least three ways: it doesn't have two of the errors in the proof you included, it doesn't use set builder notation. About your question, sorry, I don't get it. Can you be specific in what step of the proof you think it's wrong? – Git Gud Mar 18 '22 at 14:50
  • Third edition theorem 4.1 at the page 114. – Antonio Maria Di Mauro Mar 18 '22 at 14:52
  • Now I add the original text as you asked. – Antonio Maria Di Mauro Mar 18 '22 at 14:55
  • So what can you say now about? – Antonio Maria Di Mauro Mar 18 '22 at 14:55
  • That's exactly what I was looking at. I'll fix your post. Edit: See the differences to the previous version. – Git Gud Mar 18 '22 at 14:56
  • Okay, thanks for your edit. :) – Antonio Maria Di Mauro Mar 18 '22 at 14:56
  • This doesn't resolve my confusion with your question, by the way. I still don't get it, sorry. – Git Gud Mar 18 '22 at 15:00
  • I saw this in the related questions, though. It seems to relate to whatever you're confused about. Does it help? – Git Gud Mar 18 '22 at 15:02
  • So, the question is: if $\beta_0$ was zero then it is not possible to apply the inductive hypotesis because $0$ has not elements and so in this case it cannot exist $\beta\in \beta_0$ such that $P(\beta)$ is true, tha's all. – Antonio Maria Di Mauro Mar 18 '22 at 15:02
  • I saw the question you linked: the accepted answer it seems says exactly that we have to assume that the property $\pmb P$ is true for $0$, right? – Antonio Maria Di Mauro Mar 18 '22 at 15:04
  • So if this was true then is wrong the proof I reported? – Antonio Maria Di Mauro Mar 18 '22 at 15:05
  • No, the linked answer says it is unnecessary. In the case you mentioned ($\beta_0 = 0$), it's even truer in a sense, because it is logically true, regardless of semantics. The statement can be written as $\forall \beta\left(\beta < 0 \to P(\beta)\right)$, which is true. – Git Gud Mar 18 '22 at 15:08
  • So are you saying that if $\beta_0=0$ then $P(\beta_0)$ is true vacuously? – Antonio Maria Di Mauro Mar 18 '22 at 15:11
  • Ah, perhaps I understood: if $P(0)$ was not true then there must exist any $\beta<0$ such that $\pmb P(\beta)$ is not true but this is impossible so that $P(0)$ is true and thus $\beta_0$ cannot be $0$, right? – Antonio Maria Di Mauro Mar 18 '22 at 15:22
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    @AntonioMariaDiMauro. Yes. Exactly................ – DanielWainfleet Mar 18 '22 at 17:07

1 Answers1

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Suppose that $P(x)$ is a property as claimed, i.e., for all ordinals $\alpha$ if for all ordinals $\beta < \alpha$ we have that $P(\beta)$ is true, then it follows $P(\alpha)$ is true.

If we examine this with $\alpha= 0$, then we have the claim "If for all $\beta < 0$, $P(\beta)$ is true, then $P(0)$ is true". This claim is an implication. It's antecedent is "for all $\beta < 0$, $P(\beta)$ is true". As there are no $\beta < 0$, the claim is vacuously true. Hence, we get to conclude the consequent: that $P(0)$ is true.

Thus, this formulation of transfinite induction does imply that $P(0)$ is true.

I have found that students prefer the following equivalent formulation: If $P(x)$ is a property such that

  • $P(0)$ is true.
  • For each ordinal $\alpha$, if $P(\alpha)$, then $P(\alpha + 1)$.
  • If $\beta$ is a (non-zero) limit ordinal and $P(\alpha)$ holds for all $\alpha < \beta$, then $P(\beta)$ hold.

Then $P(x)$ is true of all ordinals.

James
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  • Okay, so I understand that $P(0)$ is vacuously true: that is if $P(0)$ was false then I should can reject the sencente $$\text{if }\pmb P(\beta),\text{is true for all }\beta<0,\text{then }\pmb P(0),\text{is true}$$ however this is impossible because $0$ has not any element, RIGHT? So I argue that the formulation of transfinite induction implies that $P(0)$ is true but I am confuse when you claim $$\text{Thus this formulation of transfinite induction does not imply that},\pmb P(0),\text{is true}$$ so what you want mean? Could you explain, please? – Antonio Maria Di Mauro Mar 18 '22 at 15:42
  • Forgive my confusion. – Antonio Maria Di Mauro Mar 18 '22 at 15:43
  • I think you inserted a "not" that I did not write? – James Mar 18 '22 at 15:47
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    You are right: sorry, but I got little sleep last night. So question upvoted and approved: thanks so much for your assistance!!! – Antonio Maria Di Mauro Mar 18 '22 at 15:50