# No Largest Ordinal

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## Theorem

Then:

- $\forall x \in a: x \prec \paren {\bigcup a}^+$

## 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.

\(\ds x\) | \(\in\) | \(\ds a\) | by hypothesis | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(\subseteq\) | \(\ds \bigcup A\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(\preceq\) | \(\ds \bigcup A\) | |||||||||||

\(\ds \) | \(\prec\) | \(\ds \paren {\bigcup A}^+\) | Ordinal is Less than Successor | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(\prec\) | \(\ds \paren {\bigcup A}^+\) |

$\blacksquare$

## Remark

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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

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 7.26$