Explain why transfinite induction does not assume that a property must be true for zero.

inductionordinalsset-theorysolution-verificationtransfinite-induction

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?

Best Answer

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.

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