Principle of Transfinite Induction for On:

If $C\neq\varnothing, C\subseteq$ On then $\exists\alpha\in C\forall\beta\in C[\alpha\in\beta\vee\alpha=\beta]$


(1) As $C\neq\varnothing$, let $\alpha_0\in C$ .

(2) If for no $\beta\in C$ do we have $\beta\in\alpha_0$ then $\alpha_0$ was the $\in$-minimal element of $C$.

(3) Otherwise we have that $C\cap\alpha_0\neq\varnothing$. As $\alpha_0\in$ On, by definition $\in$ wellorders $\alpha_0$.

(4) Hence $C\cap\alpha_0$ is non-empty and so has an $\in$ minimal element $\alpha_1$; and then $\alpha_1$ is the minimal element of $C$.

I understand the general argument, but I am confused about $C\cap\alpha_0\neq\varnothing$ on line (3).

If $\alpha_0$ is the $\in$-minimal element of $C$ does that mean $C\cap\alpha_0=\varnothing$? If so, why?

  • $\begingroup$ Don't shove all the set theory tags you can find, and then some. Use tagging appropriately and responsibly. $\endgroup$
    – Asaf Karagila
    Dec 30, 2014 at 22:41
  • $\begingroup$ Ok sorry, I dont really know what one it comes under and I want to make sure I include the correct one. But I'll be more careful next time. $\endgroup$ Dec 31, 2014 at 11:54

1 Answer 1


Note that (3) begins with "Otherwise", meaning if $\alpha_0$ wasn't the minimal element of $C$, then $C\cap\alpha_0\neq\varnothing$ and in which case $\alpha_1=\min\alpha_0\cap C$ is also the minimal ordinal in $C$.

  • $\begingroup$ Thanks for your answer, my confusion lies in that if $\alpha_0$ is the $\in$-minimal element of $C$ does that mean $C\cap\alpha_0=\varnothing$? I cant understand this part. $\endgroup$ Dec 31, 2014 at 12:00
  • $\begingroup$ I dont see how the intersection of $C$ and $\alpha\in C$ could ever be the empty set $\endgroup$ Dec 31, 2014 at 12:01
  • $\begingroup$ Yes, if $\alpha<\alpha_0$, namely $\alpha\in\alpha_0$ then $\alpha\notin C$, since $\alpha_0$ is the minimal element of $C$. $\endgroup$
    – Asaf Karagila
    Dec 31, 2014 at 12:03
  • $\begingroup$ Sorry for being pedantic, but is $C\cap\alpha_0=\varnothing$ if $\alpha_0$ is the minimal element of $C$? $\endgroup$ Dec 31, 2014 at 12:08
  • 1
    $\begingroup$ No. If $\alpha<\alpha_0$ then $\alpha\notin C$. But remember that $\alpha<\alpha_0$ is just a fancy way of writing $\alpha\in\alpha_0$. Therefore for every element of $\alpha_0$, that element is not an element of $C$. This is just a long way of saying that $\alpha_0\cap C$ is empty! $\endgroup$
    – Asaf Karagila
    Dec 31, 2014 at 12:20

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