Re: Well Ordering the Reals

Tony Orlow wrote:

> If you want to know where I think the root problems lie, I can reiterate
> something I think modern mathematicians need to consider. One of the very root
> problems in set theory as I see it is a misunderstanding of the nature of
> inductive proof. In a traditional proof, one starts with some set of premises
> assumed true, and given logical rules of deduction, creates a chain or network
> of implications that finally imply the result. In a recursive proof such we
> have in induction, we have an implication that always implies another
> implication, and so we have an infinite chain of implications, thus covering
> the infinite linear set, generally considered the naturals. When induction is
> used to prove that every natural is finite it's a mistake. While it is true
> that adding 1 to a finite number yields a finite number, that is a finite
> operation. If you add 1 n times, then you have added n to the value of the
> largest element. If n is infinite, then you have added an infinite value. In
> order for induction to prove anything in the infinite case, the property being
> proven must be in the form of an equality, and any inequality must be shown not
> to converge to zero at infinity. When one says "x is finite", what one means is
> "x<oo", where oo is any infinite value. The limit of x, as x apporaches oo, is
> oo, in which case x<oo is false. Similarly, induction can be used to prove that
> 1/x>0, but since the limit of 1/x at oo is 0, this proof does not hold for the
> infinite case. Once this issue is noted, some other problems can be addressed.

Since I think you do have innate intelligence, it's unfortunate that
what needs to be said about this, as with virtually every one of your
posts, is not mathematical, but rather directly personal: You have no
idea what you're talking about.

Induction and recursion are not root problems of set theory, since they
are theorems of set theory, not axioms of it. If you object to
induction and recursion in set theory, then the roots are the axioms,
not the theorems that are derived from the axioms. You don't know root
from trunk from branch from stem in this matter. Please read a book.

Nor are there an actual infinite sequence of implications. Proof by
induction and definition by recursion are done in a finite number of
steps.

As to adding to n, how many billion times do you have to be told, in
this context, n is not infinite?

And this buisness about a property must be an equality, convergence to
infinity, et. al, is just straight from where you pull the rest of
your pseudo-mathematics.

Then you say, "When one says "x is finite", what one means is x<oo",
where oo is any infinite value." No, that is not what is meant. You've
been told a billion times the definitions of finite and infinite
already. One more time:

Df. x is finite iff there exists a bijection between x and a natural
number.

Df. x is infinite iff x is not finite.

Df. x is Dedekind infinite iff there exists a bijection between x and a
proper subset of x.

Df. x is Dedekind finite iff x is not Dedekind infinite. (The
pigeonhole principle.)

Th. If x is Dedekind infinite, then x is infinite.

Th. If the axiom of choice and x is infinite, then x is Dedekind
infinite.

The proof that every natural number is finite is trivial: A natural
number n is finite since the identity function on n is a bijection
between n and a natural number (viz. itself).

The proof that no natural number is Dedekind infinite, which is
basically the proof of the pigeonhole principle for natural numbers, is
a little more tricky. But you can take a look at any set theory book
for it. Enderton's book even gives you a picture!

The proof of the induction principle, for any inductive set, not just
the set of natural numbers, is a trivial consequence of the definition
of an inductive set, for each form of induction (whether from the
successor operation or from the formula building operations, etc.).
And that the set of natural numbers is an inductive set is a trivial
consequence of the definition of the set of natural numbers. The proof
of the definition by recursion theorem, for appropriate contexts, not
just the natural numbers, is complicated, but found in any book on set
theory. The proof of the definition by transfinite recursion theorem is
involved, but found in any book on set theory.

MoeBlee

.

Relevant Pages

  • Re: Well Ordering the Reals
    ... One of the very root ... >> have in induction, we have an implication that always implies another ... >> the infinite linear set, generally considered the naturals. ...
    (sci.math)
  • Re: Logarithm of transfinite numbers
    ... naturals, the nth naturals will ALWAYS be equal to n. ... We claim that induction on peano sets only works according to the ... There is nothing in the Peano axioms which allows TO's "infinite case" ...
    (sci.math)
  • Re: Orlow cardinality question
    ... > using induction, but I can't prove things about BEING infinite. ... Not the set size of the reals. ... For each finite natural, n in N, let n* represent the set of naturals up ...
    (sci.math)
  • Re: Well Ordering the Reals
    ... >>> offer about two infinite series apply only to series both of which converge. ... > No. Standard induction is defined to apply to *every element of* the entire set ... >> have an inductive proof of some kind of inequality based on a difference which ... >> course the inductive proof that the naturals are all finite is based on an ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> root. ... > They have turned it into a discussion now of infinite binary trees, ... > there are infinitely long paths, but all nodes are a finite distance ... > that there are infinitely many naturals, but every one has a distinct ...
    (sci.math)
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