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Topic: WMytheology 000
Replies: 10   Last Post: May 4, 2013 5:50 PM

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Posts: 8,833
Registered: 1/6/11
Re: WMytheology 000
Posted: May 2, 2013 5:07 PM
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In article
WM <> wrote:

> On 2 Mai, 10:33, Zeit Geist <> wrote:

> > Bull. Every n is in at least one line so every n is in the union.
> > Even though every line lacks infinitely many numbers

> We have three ways to produce the "last" line.

In a list without a last line, there is no last line, but it is
possible, by forming the set union of all lines as sets, to create a
line not in the list but having as a member every member of every member
of that list.

> I showed you.

No, you did not!

> We know that there is no last line, so there is no last union

If we have the list, we then have all its lines as sets so can form the
union of all those sets to form a union set, which has as a member every
member of every member of that list of lines.

, so
> there are infinitely many unions in the list. But you claim that
> another union will create something that has not been existing before.

Unless you have previously created a union of ALL lines, or of at least
infinitely many of them, such a union will contain more than any of
> > For every natural there is some line which some that specific
> > natural is in, but there in no line the each and every natural is in.

> Can you name (as individuals, not as bigmouthed "all naturals" or "all
> primes") any set of naturals that is not in one single line?

Name me a line by its position in your list and I will name you some
naturals not in it.
> >
> > > Therefore the assumed existence of |N is false.
> >
> > In your limited ontology.

> No. It is false if we calculate the limit of an infinite sequence from
> its finite terms - as it is necessary and usual in mathematics.

In standard mathematics, given any sequence of sets with each set being
a proper subset of its immediate successor, the limit of such a sequence
will exist as the union of all its members, but not be a member of that

If that does not old true in Wolkenmuekenheim, then the mahematicas of
Wolkenmuekenheim is both corrupted and invalid.
> Do you explicitkly deny that limits have to be calculated from the
> finite terms of the sequence?

Only if one's methodology takes into account all infinitely many finite

The limit of any sequence of sets with each set being a proper subset of
its immediate successor exists and is its union set.

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