FOM: con ZFC

[Randall Holmes]

Kanovei said:

Con ZFC is believed to be true 
not as a formal statement of any sort 
but rather as a prediction that any 
contradiction in ZFC will never be found in 
the practical activity of mathematicians. 
For instance if you succeed to prove 
some A assuming Con ZFC this will 
mathemtically mean that you have proved 
Con ZFC --> A, but not that you have proved A.

Holmes replies:

This is the attitude of some.  On the other hand, some probably think
that it is clear that there is an inaccessible cardinal (by an
extension of the same intuition that makes causes most of us to think
ZFC is OK in the first place) and so Con(ZFC) is simply true.

I'm of several minds about ZFC being "true" (in the sense that there
is any model of second-order ZFC) but I'm quite sure that it is
consistent; I believe Con(ZFC) as a statement of arithmetic to be true
as a formal statement, and I think that there are many who are of this
opinion.

I think that there are a wide variety of "informal" beliefs held
by mathematicians. 

And God posted an angel with a flaming sword at | Sincerely, M. Randall Holmes
the gates of Cantor's paradise, that the       | Boise State U. (disavows all) 
slow-witted and the deliberately obtuse might | holmes at math.idbsu.edu
not glimpse the wonders therein. | http://math.idbsu.edu/~holmes