Burali-Forti paradox - Wikipedia

In set theory, a field of mathematics, the Burali-Forti paradox demonstrates that constructing "the set of all ordinal numbers" leads to a contradiction and therefore shows an antinomy in a system that allows its construction. It is named after Cesare Burali-Forti, who, in 1897, published a paper proving a theorem which, unknown to him, contradicted a previously proved result by Cantor. Bertrand Russell subsequently noticed the contradiction, and when he published it in his 1903 book Principles of Mathematics, he stated that it had been suggested to him by Burali-Forti's paper, with the result that it came to be known by Burali-Forti's name.

We will prove this by reductio ad absurdum. Let {\displaystyle \Omega } be a set that contains all ordinal numbers. {\displaystyle \Omega } is transitive because for every element {\displaystyle x} of {\displaystyle \Omega } (which is an ordinal number and can be any ordinal number) and every element {\displaystyle y} of {\displaystyle x} (i.e. under the definition of Von Neumann ordinals, for every ordinal number {\displaystyle y<x}), we have that {\displaystyle y} is an element of {\displaystyle \Omega } because any ordinal number contains only ordinal numbers, by the definition of this ordinal construction. {\displaystyle \Omega } is 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 {\displaystyle \Omega } 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 {\displaystyle \Omega } is an element of {\displaystyle \Omega }. Under the definition of Von Neumann ordinals, {\displaystyle \Omega <\Omega } is the same as {\displaystyle \Omega } being an element of {\displaystyle \Omega }. This latter statement is proven by step 5. But no ordinal class is less than itself, including {\displaystyle \Omega } because of step 4 ({\displaystyle \Omega } is an ordinal class), i.e. {\displaystyle \Omega \nless \Omega }. We have deduced two contradictory propositions ({\displaystyle \Omega <\Omega } and {\displaystyle \Omega \nless \Omega }) from the sethood of {\displaystyle \Omega } and, therefore, disproved that {\displaystyle \Omega } is a set.