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Posted 15 April 2008

The setbuilder notation has a bug: for some assertions P(x), the notation  does not define a set.

#### Example

Let P(x) be the assertion "x is a set".  Then if  were a set, it would be the set of all sets. However, there is no such thing as the set of all sets. This can be proved using the theory of infinite cardinals, but not here.   (See the discussion following Theorem 5 in Suber’s presentation of set theory.)

## Russell’s impossible set

I now give another example of a definition  which does not give a set, and I will prove this time that it does not give a set. It is due to Bertrand Russell.  He took P(x) to be " x is a set and x is not an element of itself." This gives the expression "  ”.

Now let’s prove by contradiction that that expression does not denote a set.

Suppose  is a set.  Call is S.  There are two possibilities:

(i)              Then by definition of S, S is not an element of itself, i.e., . This follows from the first part of the method of comprehension.  So it is not possible that .

(ii)             .  In this case, P(S) is true, so by the second part of the method of comprehension, we know that SS. So it is not possible that .

It follows by this contradiction that S is not a set.