# No Largest Ordinal

## 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.