The original form of Russell’s (Zermelo’s in fact) paradox is not this. The original form is {x|x not member of x}.
That leads to both
x is a member of x
and
x is not a member of x
And that is the original form of the paradox.
No. See for example This discussion. The form you give where it is described as a simple predicate recursion was not the original form of the paradox.
The original form of Russell’s (Zermelo’s in fact) paradox is not this. The original form is {x|x not member of x}.
That leads to both
x is a member of x
and
x is not a member of x
And that is the original form of the paradox.
No. See for example This discussion. The form you give where it is described as a simple predicate recursion was not the original form of the paradox.