Re: Set Theory: Should you believe?

george wrote:

<snip - things directed at someone else's remarks>



Here's a system which captures, I think, Norm's intutition. Consider
second-order PA without the successor axiom (that every number has a
successor) - call this F.

There is no such thing.
PA has well-defined axioms and NOT ONE of them
says "every number has a successor".

There are many different ways of writing down PA. Look for instance at
Richard Heck, "Cardinality, Counting, and Equinumerosity," where he
writes,

"Stated in a form that is useful for comparison with Frege arithmetic,
the Peano axioms are:

.... 6. (x)(Nx ==> (there exists y) xPy)."

Or look at John Burgess, "Fixing Frege" (p. 25), where he writes, "The
so-called Peano postulates ... consist of ...", and he then lists the
successor axiom as axiom no. 1.

The fact that every
number has a successor is, in PA, a consequence of a
LINGUISTIC PARADIGM that insists that if t is a term in the
language and s is a unary function in the signature of the
language, then s(t) MUST be a well-formed term, for ALL
well-formed terms t. I.e., if t is in the language then s(t)
must be as well. This IS NOT an axiom OF PA (it is part
of the definition of a first-order language, and goes up to 2nd
"automatically"), so there cannot be a version of PA with this
"removed".

Instead of assuming a functional term s(t), you assume a relationship
Sx,y. Instead of stating axioms using s(t), one states axioms using
Sx,y. The Successor Axiom is then needed as an axiom to get the usual
force of the Peano Axioms. It can then be removed.

The ONLY way to remove s(.) from the list of axioms
of PA is to remove all the axioms that USE s(.).
Since s(x) is
also expressible as x+1, I don't see how this is EVER going to
be coherent.



F does not assume there is a maximum number
(i.e. a number which does not have a successor) - it is simply agnostic
about the matter, neither assuming nor not assuming it.

I repeat, if you have successor IN the language AT ALL, this is
NOT *linguistically* POSSIBLE.

It is certainly possible if you use a successor relationship...

.

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