[FOM] Might there be no inaccessible cardinals?

Rupert McCallum rupertmccallum at yahoo.com
Fri Nov 2 17:54:12 EDT 2007

--- Vaughan Pratt <pratt at cs.stanford.edu> wrote:

> I ran across an interesting paper by Jesus Mosterin titled "How set 
> theory impinges on logic" at
> The following sentence caught my eye.
> "In ZFC we can neither prove nor disprove the existence of
> inaccessible 
> cardinals."
> Can anyone (Mosterin perhaps?--I don't have his email address)
> enlighten 
> me as to the meaning of "cannot" ("can neither") here?
> In the case of "prove" there is no question: "cannot" means it is 
> impossible, since there are models of ZFC too small to include an 
> inaccessible cardinal.
> At first I assumed that he meant the same for "disprove."  But I 
> couldn't immediately come up with an equally convincing argument, nor
> was Google of much help.
> Has this been shown?  Or did Mosterin merely mean that we can't 
> *currently* disprove the existence of inaccessible cardinals?
> Vaughan Pratt

If ZFC+Con(ZFC) is consistent, then it is not possible to prove in
ZFC+Con(ZFC) that ZFC+"there is an inaccessible cardinal" is
consistent. This follows from the second incompleteness theorem. A
belief that the existence of an inaccessible cardinal is consistent has
to be an article of faith.

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