Re: Well Ordering the Reals

MoeBlee said:
> 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.
You mean YOU have absolutely no idea what I'm talking about. Okay enough of the
social plattitudes....
>
> 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.
Oh, somehow I was under the impression that Peano's fifth postulate was
considered an axiom. Maybe that's because it's been mentioned as such here a
million times and on a million web sites. But I guess that's wrong, eh? Show me
the derivation that proves that inductive proof is valid, and in that context,
explain to me why it is only valid for finite iterations, or under what
conditions it is valid for infinite iterations.
>
> Nor are there an actual infinite sequence of implications. Proof by
> induction and definition by recursion are done in a finite number of
> steps.
Can you justify that statement? Why can't we talk about actually infinite
iterations?
>
> As to adding to n, how many billion times do you have to be told, in
> this context, n is not infinite?
Yes, you declare that induction only works for finite iterations, and when
asked why, you say because naturals are finite, and when asked to prove this,
you use induction. Ouroboros strikes again!

>
> 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.
>From the eternal ether of mathematical fact.
>
> 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:
What I said is correct. A finite is less than any infinite.
>
> Df. x is finite iff there exists a bijection between x and a natural
> number.
Are we bijecting single numbers now? I thought bijections were between sets.
>
> Df. x is infinite iff x is not finite.
Why was it wrong when I said the same thing? Hmmm....
>
> 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).
Ha ha ha. What a proof. Define finite based on natural numbers, and then prove
tht natural numbers are finite basedon that definition. Okay, so pi is not
finite? Your circular nonsense is worse than anything I have ever concocted.
>
> 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!
Yay!! (snore)
>
> 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.
>
Thanks.
> MoeBlee
>
>

--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

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