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From: Virgil on 12 Jun 2010 16:31
In article
<5966b867-ca12-411f-ba97-43df2ab3a253(a)b35g2000yqi.googlegroups.com>,
WM <mueckenh(a)rz.fh-augsburg.de> wrote:

> On 12 Jun., 01:35, Owen Jacobson <angrybald...(a)gmail.com> wrote:
>
> > >>>>> 0.1
> > >>>>> 0.11
> > >>>>> 0.111
> > >>>>> ...
>
> >
> > >> For all elements n of N, let F_n be a function from N to {0, 1} as
> > >> follows:
> >
> > >> � F_n(x) = 1 if x <= n
> > >> � F_n(x) = 0 if x > n
> >
> > >> Let G be a function from N to {0, 1} as follows:
> >
> > >> � G(x) = 1
> >
>
> > >> Which of the following statements accurately characterizes your
> > >> argument?
> >
> > >> 1. If N exists, then there exists some natural number z in N such that
> > >> L(z) = F_z = G.
> >
> > > Yes. But z does not and cannot exist. Therefore N cannot exists as a
> > > completed entity.
> >
> > Can you frame an existence argument for a number z in N with the
> > properties above, either by deriving it directly,
>
> Proof by complete induction. If the existence of all nodes 1 in
> different lines of the list is claimed, then at least two lines must
> be shown which, with respect to their contents of nodes 1, cannot be
> replaced by one line. This is obviously impossible.
>
> On the contrary, it can be proved by complete induction, hence for
> every line of the list, that all nodes which are in that line and in
> all previous lines, also are in the next line. (Complete induction
> holds for all natural numbers, though not for the set of them. But
> that ist not required here.)
>
> > or by showing that it
> > must exist, in terms of ZFC?
>
> ZFC is rubbish. I do not use it.
> >
> > Try to avoid analogies:
>
> Try to use complete induction.
>
> > you don't actually need binary trees or decimal
> > expansions to frame your argument, and your argument would be much
> > clearer without them.
>
> At present we are not in the binary tree. We have only a very simple
> list. See above.
>
>
> > Framing your argument in terms of functions would
> > help me, obviously, but framing it in terms of sets and subsets would
> > work as well.
>
> Your approach stands above. I did not delete it. As far as I can see
> we discuss it.
>
> > It's not at all obvious from the arguments you've
> > presented that z must exist.
>
> Why do you resist complete induction? Please understand: Complete
> induction works for every line of the list above. Therefore, if all
> 1's are ther, then they all must be in one and the same line together.
>
> > It's very obvious that you're convinced it
> > must exist, but your argument is so buried in your "tree" analogy and
> > in invented terms like "cover" that it's very hard to follow.
>
> Please forget the tree. Consider this list
>
> 0.1
> 0.11
> 0.111
> ...
>
> and consider this line
> 0.111...
>
> If the latter exists, then it must exist also in the list. That has
> been shown by induction And it can be shown by super task. (Some
> people deny super-task, but then they should also deny finished
> infinity.)
>
> Proof by construction of an infinite set. Construct the above
> list, but remove always line number n after having constructed the
> next line number n + 1. Then the list shrinks to a single line but
> this single line contains the same as the list because never anything
> has been removed that had not been added before to an existing line.

Trivially false! While it may require that there be some SET of lines
not replaceable with a single line, there is no requirement that such a
set be finite. And any infinite set of such lines suffices.

> >
> > > But there are only countably many finite definitons.
> >
> > That was never in dispute. Look up "computable numbers," for example.
>
> But it is frequently disputed that all real numbers form a countable
> set. Therefore besides finite definition, there must be infinite
> sequences "standing alone". But there is no infinite sequence without
> finite definition.

Any function having N as domain is an infinite sequence, but there is
nothing that requires such a function to be explicitly defined or known
when considering the properties of such functions.
From: Virgil on 12 Jun 2010 16:43
In article
<4e3b27c6-93e2-4aa7-b177-6664ae742a8a(a)x21g2000yqa.googlegroups.com>,
WM <mueckenh(a)rz.fh-augsburg.de> wrote:

> On 12 Jun., 15:53, Gus Gassmann <horand.gassm...(a)googlemail.com>

> > However the general gist of these "proofs" is this:
> >
> > 1. For every integer n, card (T_n) < b_n for some structure T_n and
> > real number b_n
> > 2. T_n converges to some T in some intuitive sense that might even be
> > made rigorous.
> >
> > The fallacy (or circular argument, to be precise) is then to infer
> > from this that
> >
> > card(T) <= lim (b_n).
>
> For some reciprocal values of integer sums, i.e. convergent series, we
> conclude:
> If for all n in N: SUM(k= 1...n) a_k < SUM(k= 1...n) b_k, then
> SUM(k= 1...oo) a_k =< SUM(k= 1...oo) b_k.
>
> Why is that assumed? Nobody really knows what happens in the limit.
> Nobody ever could try. Why is this definition assumed to be true?

In mathematics, such a thing would not be assumed, it would either be
proved or junked.
> >
> > > Do you believe that there is an abyss "after all"? Do you think along
> > > the lines of the medieval sophists who expected to fall from the earth
> > > after passing its edge?

WM seems to.
> >
> > > Well, your reasoning sounds like that. Don't worry. There is no abyss
> > > and there is no dragon and there is no discontinuity .
> >
> > As a mathematician, I demand proof of this "fact",
>
> But as a "matheologician"

Such ad homs are fallacious.



>
> You can be happy, that such "mathematics" fortunately has no influence
> on real mathematics.

It may have no effect in physics, but physics, fortunately, only
involves a fairly small part of real mathematics, and has little to no
influence over the greater part of real mathematics.
>
> Regards, WM
From: WM on 12 Jun 2010 16:58
On 12 Jun., 20:17, Owen Jacobson <angrybald...(a)gmail.com> wrote:


F_n(x) = 1 if x <= n
F_n(x) = 0 if x > n



n F_n(x)

1 1000...
2 11000...
3 111000...
.... ...

>
> > Proof by complete induction. If the existence of all nodes 1 in
> > different lines of the list is claimed,
>
> This is exactly the sort of analogy-based presentation I'd like to
> avoid. I *think* what you mean is something similar to
>
> 1. ∀x ∃n F_n(x) = 1
> where x, n ∈ N

No. This would yield only for every finite x a finite n.
The claim of set theorists is that there is an infinite number of 1's
in the whole list, but that there is no F_n with infinitely many 1's.
My claim is that *if this was true* then there must be a function,
call it F_oo, with F_oo(x) = 1 for all x.
Hence
E oo A x F_oo(x) = 1.
Of course this F_oo cannot be fixed. But I can prove that all !'s
remaain in the list, when all F_n with finite n are removed.
>
>
> > Try to use complete induction.
>
> Define "complete induction". In particular, how is it different from
> the following:
>
> 4. φ(0) ∧ (∀n φ(n) → φ(S(n))) → ∀x φ(x)
> where 0, n, x ∈ N and S(n) and N are defined in the usual way

In exactly that way I prove that for all n F_n(x) can be removed from
the list and nevertheless all 1's remain the list.
>
> Note that this induction axiom schema does not allow you to prove, from
> φ(0), that φ(N).

I don't need that.
I prove that A n φ(x) where φ(x) : F_n can be removed from the list
without loss of any 1.
Proof:
F_1 can be removed, because its 1 appears in F_2.
F_n can be removed, because its 1's appear in F_(n + 1).
Therefore every finite function can be removed.

But there would not remain anything. Hence the original claim of
finished infinitely many 1's must be wrong.

Regards, WM
From: Virgil on 12 Jun 2010 17:50
In article
<adf3fd69-402e-4ad2-8cc8-aa543e1b2b70(a)y4g2000yqy.googlegroups.com>,
WM <mueckenh(a)rz.fh-augsburg.de> wrote:


> F_1 can be removed, because its 1 appears in F_2.
> F_n can be removed, because its 1's appear in F_(n + 1).
> Therefore every finite function can be removed.

You have only proved that an F_n can be removes PROVIDED some F_m, with
m > n remains unremoved, which is false when ALL F_k are removed.

WM fakes it again.
From: Owen Jacobson on 13 Jun 2010 01:15

On 2010-06-12 16:58:50 -0400, WM said:

> On 12 Jun., 20:17, Owen Jacobson <angrybald...(a)gmail.com> wrote:

For every natural number n,
F_n : N → {0, 1}
> F_n(x) = 1 if x <= n
> F_n(x) = 0 if x > n
>
>
> n F_n(x)
>
> 1 100000...
> 2 110000...
> 3 111000...
> ... ...

That table is a handy illustration, but it is not the definition of the
functions F_n; the equations above are.

And, for completeness:

G : N → {0, 1}
G(x) = 1

L : N → {F_x | x in N}
L(x) = F_x

>>> Proof by complete induction. If the existence of all nodes 1 in
>>> different lines of the list is claimed,
>>
>> This is exactly the sort of analogy-based presentation I'd like to
>> avoid. I *think* what you mean is something similar to
>>
>> 1. ∀x ∃n F_n(x) = 1
>> where x, n ∈ N
>
> No. This would yield only for every finite x a finite n.
> The claim of set theorists is that there is an infinite number of 1's
> in the whole list, but that there is no F_n with infinitely many 1's.
> My claim is that *if this was true* then there must be a function,
> call it F_oo, with F_oo(x) = 1 for all x.

Yes, I'm aware that that's your claim, but I still see no proof. The
function F_oo(x) = G(x) = 1 exists, but nothing about the set of
functions labelled F_n that implies that it exists, or that there is
some (natural) number x such that L(x) = F_oo = G.

> Hence
> E oo A x F_oo(x) = 1.
> Of course this F_oo cannot be fixed.

It certainly can be fixed, but it's not possible for your 'oo' to be a
natural number. However:

For all _ordinal_ numbers n, there exists a function F_n(x) from the
natural numbers to the set {0, 1} such that:

F_n(x) = 1 if x <= n
F_n(x) = 0 if x > n

For all natural numbers x, x < ω, therefore F_ω(x) = G = 1 (where ω is
the first limit ordinal greater than zero).

However, ω is provably not a natural number.

>>> Try to use complete induction.
>>
>> Define "complete induction". In particular, how is it different from
>> the following:
>>
>> where 0, n, x ∈ N and S(n) and N are defined in the usual way
>
> In exactly that way I prove that for all n F_n(x) can be removed from
> the list and nevertheless all 1's remain the list.

A vague description of a mechanical process on a physical object is not
a mathematically-useful definition. Can you define "complete induction"
or at least distinguish it properly from "induction" as given in my
last post (or in any presentation of Peano arithmetic)? You appear to
be relying on "complete induction" being able to prove statements that
induction, as normally defined, cannot prove, so there must be some
distinction between the two.

>> Note that this induction axiom schema does not allow you to prove, from
>> φ(0), that φ(N).
>
> I don't need that.
> I prove that A n φ(x) where φ(x) : F_n can be removed from the list
> without loss of any 1.

This is incoherent. "A n φ(x)" has an unbound variable (x) and does not
use the only bound variable (n) at all. Your description of the
predicate φ has the same problem - 'n' simply appears out of nowhere in
it. It's possible to figure out what you meant, but as a professor you
should really be doing a better job with your notation.

> Proof:
> F_1 can be removed, because its 1 appears in F_2.
> F_n can be removed, because its 1's appear in F_(n + 1).
> Therefore every finite function can be removed.

What do "removed" and "appears" mean in the context of the three groups
of functions (L, G, and F_n) mentioned? These are not standard terms
(actually, they appear to be more analogies, which I'd still like to
avoid); you should give them clear, formal definitions as you would any
other new term. Remember, for now we have no lists, no trees, and no
lines -- we only need functions on sets.

Here's my stab at your "appears":

∀n ∀m (n <= m) → (∀x (F_n(x) = 1) → (F_m(x) = 1))
where n, n, x ∈ N

Which, again, implies nothing about the relationship between the
functions F_n (where n is a natural number) and the function G.

Or perhaps you meant something like:

Given a function l from N to the set of functions from N to {0, 1}, we
define A_l:

A_l(x) = 1 if ∃y l(y)(x) = 1
A_l(x) = 0 if ∀y l(y)(x) = 0

Then we can say, for example, that for every natural number x, A_L(x) =
1. This might give you some traction on defining "remove" in as (*deep
breath*) a function which takes a functions from N to the set of
functions from N to {0, 1} and gives another function from N to the set
of functions from N to {0, 1}... I think I got that right[0].

I'm also still interested in an answer to this question from my last post:

> ZFC is rubbish. I do not use it.

Without a set theory, there is no set N. What set theory *are* you
using? What are its axioms and primitives? Where can I read more about
it by other mathematicians?

-o

[0] Normally, I'd give all of these families of functions more
convenient names, but we're not at that point yet.

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