The Russell Paradox ( { x | p(x) } )

The Russell Paradox...

Russell started out with a simple Axiom, the Axiom of Comprehension. This stated that by defining something we could assume it is a set. For example as in the title, S = { x | p(x) }. That is to say, we define S as the set containing elements x that adhere to the rule p(x). For example, S = { x | x is in R^3 and || x || = 1 } (this of course maps out the sphere of radius 1 in three space =) ).

Russell's Paradox goes as such: Let R = { x | p(x) } where p(x) is the condition for x to be contained in R. In this case chose p(x) to mean that x is not in R.

The question posed is simple: Does R contain itself as an element or not?

Well, suppose R is an element of R.
Then R does not satisfy p, and thus p(R) is false.
Therefore, R is not an element of R. Contradiction!

Well, suppose R is not an element of R.
Then R does satisfy p, and thus p(R) is true.
Therefore, R is an element of R. Contradiction!

PARADOX! Oh nose.

Cheers. =)

-M

 

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