Transfinite Induction/Principle 1/Proof 2

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Theorem

Let $\operatorname{On}$ denote the class of all ordinals.

Let $A$ denote a class.


Suppose that:

For each element $x$ of $\operatorname{On}$, if $\forall y \in \operatorname{On}: \left({ y < x \implies y \in A}\right)$ then $x$ is an element of $A$.


Then $\operatorname{On} \subseteq A$.


Proof

Suppose for the sake of contradiction that $\neg \operatorname{On} \subseteq A$.

Then:

$\left({\operatorname{On} \setminus A}\right) \ne \varnothing$

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

By Ordinal Class is Strongly Well-Ordered by Subset, $\operatorname{On} \setminus A$ must have a smallest element $y$.

Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$.

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


Therefore $\operatorname{On} \subseteq A$.

$\blacksquare$

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.