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Topic: Matheology § 246
Replies: 246   Last Post: Apr 26, 2013 2:37 AM

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 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 246
Posted: Apr 14, 2013 3:57 AM

On 14 Apr., 01:53, Virgil <vir...@ligriv.com> wrote:

> > > The union of the set of all FISONs is |N, but the sequence of all FISONs
> > > does not contain |N, even though |N is its limit.

>
> > A) {1} U {1, 2} U {1, 2, 3} U ...
> > C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
> > 3} U ...

>
> > The sequence C of all unions of FISONs does not contain |N?
>
> C is no more that the sequence of FISONs which makes it a strictly
> increasing ordered sequence, ordered by inclusion, whose LIMIT is |N.

The union of C is |N. Therefore all natural numbers must be in C.
According to the construction of C all that is in C is in one of its
subsets.

Further:
D = 1, 2, 3, ...
and
E = {1}, {2}, {3}, ...
are also strictly increasing ordered sequences. Do they contain all
natural numbers? I.e., can every natural number be found in D and E.
If not, how can the union of E be |N.
If E contains all natural numbers, why can't every natural number be
found in C which is a super-sequence of E?
>
> A strictly increasing infinite sequence like cannot have its limit as
> any one of its terms as there is always a larger term.

I agree. 1, 2, 3, ... is not its limit - since there is no limit.
"Forall n" is not the same as "From 1 to oo".

>
> > Why should
> > it stop before reaching this goal

>
> It doesn't stop, but if it ever reached its limit, being a strictly
> increasing sequence, it would then have to pass that limit.

So it is. Therefore there cannot be limit, at least it cannot be
reached. There is no "all n", because in the table

1
2, 1
3, 2, 1
...

it would exist in the whole table, it would exist in every column, but
it would not exist in any line. That is obviously rubbish that a six-
year-old child would recognize.

>
> Which may well happen in Wolkenmuekenheim, but nowhere else.
>

> > if it can be accomplished by A? But
> > if it stops in fact, then also A cannot reach this goal, because C
> > contains A .

>
> A is a set which is not
> equal to or contained in any term of sequence C, but only in C's limit,
> which is not a member of C or a subset of C.

So it must differ from all sets of the sequence C. What is that
difference?

According to Zermelo's first and obviously most important Axiom 1 we
have
forall A forall C (forall X (X e A <==> X e C) ==> A = C)
All n are in A and in C, no?
>
> > Otherwise there should be a natural number or a FISON
> > that is in A but not in C.

>
> The only way that could be true would be if the limit of a strictly
> increasing sequence could be a member of the sequence,

Not at all! The only way that could be true would be if the limit of C
contained some natural number that is not in any of the sets of C.

Regards, WM

Date Subject Author
4/12/13 mueckenh@rz.fh-augsburg.de
4/12/13 fom
4/12/13 JT
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4/12/13 Virgil
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