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Principle of transfinite induction for ordinals:

Let $A$ be a subset of the ordinal $\alpha$ such that

  1. $0 \in A$
  2. for all $\beta \in \alpha$ if $\beta \subseteq A$ then $\beta \in A$

Then $A=\alpha$

Isn't the first condition $0\in A$ implicit in the second condition?

"For all $\beta \in \alpha$ if $\beta \subseteq A$ then $\beta \in A$" means that, for all the elements $\beta$ in $\alpha$ (therefore $0$ included) this holds: $(\delta \in \beta$ ⇒ $\delta \in A) ⇒ \beta \in A$ the antecedent of this is vacuously true for $\beta=0$ and therefore if this property has to hold that means that $0\in A$ which is just the first condition. Isn't this correct? Is the condition $0 \in A$ there just to help avoid mistakes or it's actually needed?

Asaf Karagila
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Andrea Burgio
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1 Answers1

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The first condition is not implicit in the second: If $\alpha=0$, then clearly $0\notin A$ because $0\notin\alpha=0$, but the second condition is vacuously fulfilled. Therefore $\alpha=0$ is a counterexample to the claim that the second condition implies the first.

However it is the only counterexample, and for $\alpha=0$, $A=\alpha$ trivially holds. Thus the theorem obtained by removing the first condition also is true.

Therefore the first condition, while not implicit in the second, is still unnecessary.

celtschk
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  • But I've never said that the second condition is implicit in the first one I asked if the first condition is implicit in the second, how is it not? What is wrong with what i wrote? – Andrea Burgio Aug 24 '20 at 10:46
  • @AndreaBurgio: I mistyped; fixed. The how is exactly as I wrote. What's wrong in what you wrote is exactly the assumption that $0\in\alpha$ always holds. – celtschk Aug 24 '20 at 10:47
  • You mean that in the case of $\alpha = ∅$ the first condition is false and the second condition is true right? And therefore the implication "(for all $β∈α$ if $β⊆A$ then $β∈A$) $⇒ 0∈A$" is false since the antecedent is true and the consequent is false right? – Andrea Burgio Aug 24 '20 at 10:57
  • @AndreaBurgio: Yes, exactly. – celtschk Aug 24 '20 at 11:01
  • @AndreaBurgio: It's not quite that. If $\alpha=0$, then $A$ must be empty, so the whole thing is vacuously true anyway. – Asaf Karagila Aug 24 '20 at 11:03
  • Thank you! and just to make sure, the theorem still holds because the antecedent of the whole theorem "($A ⊆ \alpha$ $∧$ (for all $β∈α$ if $β⊆A$ then $β∈A$) $∧ 0∈A$))⇒..." is false since $0\in A$ makes the conjunction false, right? – Andrea Burgio Aug 24 '20 at 11:06
  • @AsafKaragila: “The whole thing” here is quite ambiguous. The statement Andrea Burgio was talking about in the previous (now, pre-previous) comment was (condition 2 $\implies$ condition 1), which for $\alpha=0$ is not vacuously true, but false because the antecedent is vacuously true (the empty set has no elements) and the consequent is false ($A$ does not contain any elements, in particular not $0$). Of course the full statement in the question still is true, and also remains true if the first condition is removed. – celtschk Aug 24 '20 at 11:07
  • @AsafKaragila yeah also that, thank you – Andrea Burgio Aug 24 '20 at 11:10
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    @AndreaBurgio: Yes, for $\alpha=0$, $A$ is empty, and therefore $0\in A$ is false. – celtschk Aug 24 '20 at 11:11
  • alright thank you! – Andrea Burgio Aug 24 '20 at 11:14