[FOM] FOM Cantor's argument

Neil Tennant neilt at mercutio.cohums.ohio-state.edu
Sun Feb 9 18:53:19 EST 2003

On Sun, 9 Feb 2003, steve newberry wrote:

> There are two caveats w.r.t. Cantor's argument.
> (1) It may or may not be constructive, but it surely is NOT predicative.

There is an easy argument for Cantor's Theorem that is both constructive
and predicative. See

> (2) As ordinarily presented, it is a "proof by contradication". That is, we 
> seek to
> prove B. We assume ~B. We show that  ~B  leads to a contradiction. We
> interpret the contradiction as a FALSE proposition, hence to be rejected, 
> and then
> by Modus Tollens, [.:B => C. & ~C. => ~B.] obtain ~~B, or classically, B. 
> Q.E.D.

This is not the strategy at all. The assumption for reductio is not a
negation. Rather, it is the assumption that there is a function mapping
every member of the given set onto all its subsets.
> So where's the problem? Certainly the negation of a contradiction is a 
> tautological
> truth, a u-valid proposition. That is the premiss. But what of the 
> conclusion? If the
> conclusion is also a tautology, then no problem. But in this case, the 
> conclusion is
> most emphatically NOT a tautology. It is quintessentially **contingent**. 
> It is a
> matter of FACTUAL truth or falsehood. We know already from the 
> Loewenheim-Skolem
> Theorem that models exist in which the Cantor Theorem is false. Hence it is 
> necessarily
> contingent.

But in no model of set theory is Cantor's Theorem false. So what you say
here is irrelevant for establishing the contingency, necessary or
otherwise, of Cantor's Theorem. If one takes the axiom of separation as a
necessary truth about sets, alongside whatever even more obvious
set-theoretic principles are used in the proof, then Cantor's Theorem is
necessarily true. Note that the proof of Cantor's Theorem need not make
any appeal to the axiom of power sets.

Neil Tennant

More information about the FOM mailing list