[FOM] NBG, Subsets and Cantor's Theorem

Studtmann, Paul pastudtmann at davidson.edu
Tue Apr 3 13:06:38 EDT 2007


I have a question about the way in which one proves Cantor¹s Theorem that
there is no one to one function from a set onto its power-set within NBG.  I
should say that I am taking as my example of NBG the presentation that
Mendelson gives in his Introduction to Mathematical Logic, third edition.
My question is simply: is the axiom of subsets needed in order to prove
Cantor¹s Theorem within NBG?
 
I will state briefly why I came to the thought that it might be, though I
would not be at all surprised to learn that my mathematical reasoning has
gone awry somewhere ­ it normally does.  I shall also try to keep the
reasoning as informal as possible.
 
Any variable or constant that is in lower case refers to a set.
 
The axiom of subsets states that the intersection, I, of any class, X, and
any set, y, is a set.
 
Suppose there is a 1-1 function, f, from the set, a, onto the power-set of
a, P(a). Then, there exists a class, C, such that for all sets, y, y E C if
and only if y E a and it is not the case that y E f(y), where 'E' is the
membership relation. (By the comprehension schema)
Now, on the assumption that C is in P(a), a contradiction eventually follows
in the familiar way. But how does one prove that C is in P(a)?  Well, as far
as I can tell, what one needs to do is prove that C is a subset of a.  This
requires proving that (i) every member of C is in a, which follows easily
from the defining condition of C; and (ii) C is a set.  Why does one have to
prove that C is a set?  Because if C is not a set, then it could be the case
that everything that is in it is in the set, a, even though C is not in the
power-set of a ­ the power-set axiom in NBG says that every sub-set, x, of
another set, y, is in the power-set of y, not that every sub-class, X, of
another set, y, is in the power-set of y.  Now, with the axiom of subsets,
the fact that C is a set follows easily from the facts that (i) the
intersection of C and a is C, and (ii) a is a set. But without the axiom of
subsets I do not see how one would derive the fact that C is a set.  And so
that has led me to my present question.
 
 
I would not be surprised to find out that I am making some kind of mistake
here ­ my grasp on mathematical fact is flimsy at best.  But I would very
much like to know the answer to the question.  So if anyone could either
tell me the answer or cite some relevant source, I would be very much
obliged.


Paul Studtmann




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