Editing Burali-Forti paradox (section)

==Stated in terms of von Neumann ordinals==

We will prove this by contradiction.

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

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