No Largest Ordinal

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Theorem

Let $a$ be a set of ordinals.

Then:

$\forall x \in a: x \prec \left({\bigcup a}\right)^+$


Proof

For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably.

We are justified in doing this because of Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.

\(\displaystyle x\) \(\in\) \(\displaystyle a\) By Hypothesis
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(\subseteq\) \(\displaystyle \bigcup A\)
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(\preceq\) \(\displaystyle \bigcup A\)
\(\displaystyle \) \(\prec\) \(\displaystyle \left({\bigcup A}\right)^+\) Ordinal is Less than Successor
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(\prec\) \(\displaystyle \left({\bigcup A}\right)^+\)

$\blacksquare$


Remark


This theorem allows us to create an ordinal strictly greater than any ordinal in the set. Thus, this is another means of proving the Burali-Forti Paradox. If the ordinals are a set, then we may construct an ordinal greater than the set of all ordinals, a contradiction.


Sources