# Transfinite Induction/Principle 1

## Theorem

Let $\On$ denote the class of all ordinals.

Let $A$ denote a class.

Suppose that:

For all elements $x$ of $\On$, if $x$ is a subset of $A$, then $x$ is an element of $A$.

Then $\On \subseteq A$.

## Proof

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Aiming for a contradiction, suppose that $\neg \On \subseteq A$.

Then:

$\paren {\On \setminus A} \ne \O$

From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.

By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a strictly minimal element $y$ under $\in$.

By Element of Ordinal is Ordinal, $y$ must be a subset of $\On$, the class of all ordinals.

However, from the fact that $y$ is a strictly minimal element under $\in$ of $\On \setminus A$:

$\paren {\On \setminus A} \cap y = \O$

So by its subsethood of $\On$:

$\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$

Therefore $y \subseteq A$.

However, by the hypothesis, $y$ must also be an element of $A$.

This contradicts the fact that $y$ is an element of $\On \setminus A$.

Therefore $\On \subseteq A$.

$\blacksquare$

#### Law of the Excluded Middle

This proof depends on the Law of the Excluded Middle, by way of Reductio ad Absurdum.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this proof from an intuitionistic perspective.