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Burali-Forti paradox

Paradox in set theory

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Through Set theory, Mathematics and Gottlob Frege

Stated in terms of von Neumann ordinals

We will prove this by contradiction.

  1. Let Ω be a set consisting of all ordinal numbers.
  2. Ω is transitive because for every element x of Ω (which is an ordinal number and can be any ordinal number) and every element y of x (i.e. under the definition of Von Neumann ordinals, for every ordinal number y < x), we have that y is an element of Ω because any ordinal number contains only ordinal numbers, by the definition of this ordinal construction.
  3. Ω is well ordered by the membership relation because all its elements are also well ordered by this relation.
  4. So, by steps 2 and 3, we have that Ω is an ordinal class and also, by step 1, an ordinal number, because all ordinal classes that are sets are also ordinal numbers.
  5. This implies that Ω is an element of Ω.
  6. Under the definition of Von Neumann ordinals, Ω < Ω is the same as Ω being an element of Ω. This latter statement is proven by step 5.
  7. But no ordinal class is less than itself, including Ω because of step 4 (Ω is an ordinal class), i.e. Ω ≮ Ω.

We have deduced two contradictory propositions (Ω < Ω and Ω ≮ Ω) from the sethood of Ω and, therefore, disproved that Ω is a set.

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