# Proof of the Principle of Transfinite Induction for On

Principle of Transfinite Induction for On:

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

Proof:

(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?

• Don't shove all the set theory tags you can find, and then some. Use tagging appropriately and responsibly. Dec 30, 2014 at 22:41
• 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. Dec 31, 2014 at 11:54

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 $C$.
• 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. Dec 31, 2014 at 12:00
• I dont see how the intersection of $C$ and $\alpha\in C$ could ever be the empty set Dec 31, 2014 at 12:01
• Yes, if $\alpha<\alpha_0$, namely $\alpha\in\alpha_0$ then $\alpha\notin C$, since $\alpha_0$ is the minimal element of $C$. Dec 31, 2014 at 12:03
• Sorry for being pedantic, but is $C\cap\alpha_0=\varnothing$ if $\alpha_0$ is the minimal element of $C$? Dec 31, 2014 at 12:08
• 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! Dec 31, 2014 at 12:20