Re: Am I a crank?

MoeBlee wrote:
Tony Orlow wrote:
Hi Aatu -
You say here that induction follows from our mathematical picture of the
naturals,

He's talking about a particular axiom of induction in PA. I agree that
it essential to our understanding of the naturals. But you should
understand that induction is even more general. Inductive sets are a
certain kind of set. Roughly, an inductive set is a set that is the
intersection of a class of sets each of which is closed under a given
relation. So the naturals with the relation of successor is just an
example of an inductive set since the set of naturals is the
intersection of the class of sets that are closed under the sucessor
relation (the class is not a set in Z set theories, so the formulation
in Z set theories must find a workaround that). (Also, if you read
anything about this, you'd understand that the intersection approach
has its counterpart as a certain kind of union of sets. See Enderton's
'A Mathematical Introduction To Logic', in which his explanation of
this excels.)

Yes, that's what I'm saying. Thanks again for the reference.


but isn't our mathematical picture of the naturals based on
the inductively defined set given by the Peano axioms?

I don't speak for Koskensilta, but this may work in many directions -
first order PA may be thought of as a formalization of our mathematical
understanding, so that it is not required to reverse this as you are
doing. But also, first order PA is embedded in set theory so that all
of the axioms of first order PA are theorems but not axioms of set
theory.

Okay.


It seems to me
that inductive proof is one with a loop of logical implication, such
that one fact implies another, which in turn implies another, ad
infinitum.

That's why it's not a loop. A loop comes back around to the beginning.
Not so with induction on the naturals.

x=0;
while(finite(x))
{ add_to_list(x);
x++;
}

That's a loop generating the naturals.


The existence of a natural implying the existence of a next
natural is but an example of this kind of logical construction. Does
that seem like a wrong perspective to you?

In what context? Generally, I don't see existence of naturals as
arrived upon inductively. In first order PA, the natural numbers are
not mentioned in the theory itself. The natural numbers and the system
of them are a model of the theory, and each natural number is a member
of the universe of that model, but it is not the theory itself that
proves the existence of each natural number. Then, in set theory, any
given natural number can be proven to exist without recourse to
induction. We use induction to prove that certain PROPERTIES hold of
every natural number, but I think looking at existence itself as proven
that way is odd at best.

I don't see it as prof of existence as much as definition out of thin air, which is fine for number systems.


Secondly, I would like your opinion on inductive proof in the infinite
case.

How many times have I already posted to you that there IS transfinite
induction?

Many, but do you think that's what I'm talking about?


I am aware that this concept is not compatible with transfinite
cardinalities or limit ordinals,

No, that's incorrect. I've been telling you that for months now. There
is transfinite induction.

Yes, but is that what I'm talking about? If it were allowable to prove that 2x>x for all x>0, and omega or aleph_0 were considered greater than 0, then 2*aleph_0>aleph_0 would be a fact. However, it's not in the standard construction.


but independent of that, does the
following make any sense to you? I see any set size as a count,
something in the same sequence as the naturals, albeit possibly
infinitely far beyond them. It's a quantity. When we prove that, for
instance, x^2>2x for all x>2, it seems to me that any infinite quantity
is a number greater than 2, and that this inequality would hold for any
infinite x as well.

Then get a set of axioms that prove that.

Yessir!


If we put the infinite numbers on the same line as
the finites in this way, and extend inductive proof to also hold for the
infinite case, then not only can we say that x < 2x < x^2 < 2^x < x^x
for finite x>2, but also for all infinite x.

We do have transfinite induction in Z set theories. It just does not
prove what you want it to prove here. So get some axioms that prove
what you want them to prove.

MoeBlee


ToeKnee
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