Date: Mar 6, 2013 4:39 PM Author: Virgil Subject: Re: Matheology � 222 Back to the roots In article

<ed7bc401-706f-4a81-8c33-14cb22787151@r9g2000vbh.googlegroups.com>,

WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 5 Mrz., 22:19, Virgil <vir...@ligriv.com> wrote:

>

> > > I prove for every n that line n is not necessary. You should be able

> > > to understand this proof.

> >

> > It is true that no one of 1 or 2 or 3 is neccessary to make the sum of

> > them positive, so by WM's argument, none of them are necessary, and 0 is

> > positive.

>

> Spare your clumsy analogies.

> Here we are asking what lines of the list

> 1

> 1, 2

> 1, 2, 3

> ...

> are required to contain all natural numbers. The first three lines are

> definitively not required. And every mathematician can show that no

> line is required,

While no particular line is required, WM is falsely implying hat no

lines are required at all, whereas infinitely many lines are required.

> hence the concept of a list containing "all natural

> numbers" is nonsense.

Only in Wolkenmuekenheim. Elsewhere, there is no problem with the

existence of an ordered set having a first or smallest element and with

each member having successor member larger than itself.

>

> >

> > But I do not claimed that any line is neccessary.

>

> In fact it is easy to see that no line is necessary in an actually

> infinite list.

So WM claims that no lines are necessary to get all lines?

> >

> > My claim is that any infinite set of lines is sufficient, and that no

> > finite set of lines is sufficient.

>

> Why do you believe that an infinite set was sufficient, when for every

> line we can prove that it does not change the sufficiency of a set?

Given any infinite set of lines, every FIS of d will be contained in at

least one of them (and actually in infinitely many of them).

Given any less than infinite set of lines, there will be many FIS's of d

contained in any of them.

Conclusion: No finite set of lines is sufficient and every infinite set

is sufficient.

> You helplessly claim infinitely many lines, because you hope that

> nobody can show that this claim is nonsensical.

NO! Only your objection to it is nonsensical.

Given any infinite set of lines, every FIS of d will be contained in at

least one of them (and actually in infinitely many of them).

Given any less than infinite set of lines, there will be many FIS's of d

not contained in any of them.

Conclusion: No finite set of lines is sufficient and every infinite set

is sufficient.

> But that is easy. Even

> infinitely many finite lines are not sufficient, because the set

> cannot become actually infinite by adding finite initial sequences.

It can anywhere but in Wolkenmuekenheim.

> In

> every step the set remains finite. This yields a typical potential

> infinity, but forbids an actually, i.e., completed infinity.

But this restriction of adding only one thing at a time does not hold

outsde WMytheology, outsde WMytheology, one can add infinitely many

lines in one step.

>

> You could claim, with the same right, that an infinite set of white

> sheep contains a black sheep.

Only if I were to fall into WMytheology would that hold true.

>

> >

> > So if WM wishes to prove me wrong, he must disprove either

> > (1) that any infinite set of lines is sufficient

> > or

> > (2) that no finite set of lines is sufficient.

> >

> > Neither of which he has done or can do.

> >

>

> How should I? I can prove that an infinite set of white sheep does not

> contain a black sheep. In the same way I can prove that the infinite

> set of FIS 1, 2, 3, ..., n does not contain an infinite FIS and,

> therefore, the sequence (which is the same as the union) does never

> complete an infinite FIS. But you will simply believe the contrary.

Since none of those proofs are valid without the assumption which form

the walls keeping us out of your Wolkenmuekenheim, none of those proofs

hold up outside those walls.

>

> You believe:

> The sequence of FIS does never reach the limit |N.

> The union of FIS does reach the limit |N.

Is there any element of |N which is NOT in some FIS of |N?

Only if there is some such element can the union of all FISs NOT contain

!N, a least outside Wolkenmuekenheim.

> Because you are unable to recognize that in my example the union is

> the sequence.

Since "your example" does not work outside Wolkenmuekenheim, it is of no

interest to anyone not imprisoned there.

>

> >

> >

> > > Can you understand the above proof?

> >

> > I do not see anything by WM that qualifies as a proof of anythng

> > relevant to my claim that any infinite set of lines is sufficient, and

> > that no finite set of lines is sufficient.

>

> If the set L_n = 1, 2, 3, ..., n is not sufficient, then the set L_n+1

> = 1, 2, 3, ..., n+1 is not sufficient. This proves that the infinite

> set of of naturally indexed lines is not sufficient.

Not outside Wolkenmuekenheim it doesn't!

Outside of Wolkenmuekenheim, it only proves that no FIS of |N equal |N.

And where is WM's proof that some mapping from the set of all binary

sequences to the set of all paths of a CIBT is a linear mapping?

WM several times claimed it but cannot seem to prove it.

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