Re: Set Theory: Should you believe?


abo wrote:
There are many different ways of writing down PA.

Not really.
As Aatu Koskensilta recently said, the literature is full of
"inequivalent versions" of Godel's 1st Incompleteness Theorem,
but that does NOT mean that there even CAN exist inequivalent
versions of ONE theorem. The reason WHY Peano Arithmetic
is PROPERLY named PEANO Arithmetic is because PEANO
wrote it down THAT way. OTHER ways of writing "it", by OTHER
people, ARE NOT Peano Arithmetic, ESPECIALLY if they differ
in some way that turns out to be IMPORTANT.

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)."

But the whole point is, THAT is just UTTER BULL***.
It is a DEFINING feature of the ACTUAL Peano Arithmetic
that there simply IS NO SUCH PREDICATE as Nx!
Indeed, in the WHOLE of 1st-order logic (NOT JUST
something as limited as PA, but going all the way up to
the FULL expressive power of ANY recursive axiom-set
WHATSOEVER), it is NOT POSSIBLE to define Nx!
The only way to define Nx in PA is the same way you would
define Set(x) in ZFC: EVERYthing in the domain is a number,
and PA CANNOT TELL which of them have finitely many
predecessors (i.e. are natural) and which have infinitely
many. If you have a theory that CAN tell this then that
theory simply IS NOT PA.


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.

You need to quote that.
If it is 2nd-order then it is irrelevant.


Instead of assuming a functional term s(t), you assume a relationship Sx,y.

One can, in general, eliminate 1st-order functors that way, but that
REQUIRES EQUALITY support. In PA, you DO NOT HAVE this.
PA is stated in FOL withOUT equality because you can DEFINE
equality in PA. In PA, x=y <=df=> ~(x<y v y<x).

Instead of stating axioms using s(t), one states axioms using
Sx,y.

You CAN'T do that unLESS you can state that the y making Sxy true
is unique for x. More to the point, eliminating all functors will
eliminate all your named terms (including 0 if you really want
to go all the way). This is NOT STILL Peano Arithmetic.
It is, again, a DEFINING feature of PA that it DOES have names
(numerals) for all the naturals!

The Successor Axiom is then needed as an axiom to get the usual
force of the Peano Axioms.

Of course, but just because something has "the force of the Peano
axioms" doesn't mean it is Peano Arithmetic. It may have the force,
but there are other things it may not have -- LIKE NUMERALS --
things that are indispensable to the ACTUAL Peano Arithmetic.

It can then be removed.

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...

You mean a successor relation.
But "successor" ISN'T a relation, at least not in PA.
It's a function. A LINGUISTIC function, NOT a defined one
and NOT a set-theoretical one. In FOL we sometimes call it
a functor, though that gets ambiguous if there is a category
theorist in the room.

.

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