Date: May 2, 2013 6:21 PM
Subject: Re: Matheology � 258
WM <firstname.lastname@example.org> wrote:
> On 2 Mai, 01:39, Zeit Geist <tucsond...@me.com> wrote:
> > On Wednesday, May 1, 2013 1:26:12 PM UTC-7, WM wrote:
> > > > Each line contains only naturals.
> > > Yes. And it is as obvious that the union over all lines has been in
> > > the list before you took it again. Therefore every set that your
> > > approach may yield has been in a line of the list before. If your
> > > union is |N then |N has been in a line before. Contradiction.
> > No, you dumbass, each, when taken singlely, natural number
> > is in at least one line
> Each but not all.
Not all in any one line but all in lines.
> > But that does not mean that all the numbers, the union of N,
> > is a subset of any line.
> All numbers that are in the list, are in one and the same line
NOPE! There are more than any finite number of such numbers in the list
but not more than any finite number of numbers in any one line, so that
WM's claim is false, as is to be expected of his claims.
> That means that the complete list is missing some natural number.
> But all naturals are there. And by construction of the list, all that
> is in the list is in one single line.
> This is true for every finite
> step n. When does it become invalid?
When the list has been completed, and ther are no more lines to be added
> > For instance our union could have been formed by
> > taking all of the even numbered lines and then unioning them.
> There is no last line. If there are all naturals, they must be in
> lines before the last one.
All lines are, by your own admission above, before any last line.
> Is it one line, that contains all naturals?
No, it is infinitely many of them. No finite subset of them suffices
but, curiously enough, ANY infinite subset of lines suffices.
> Or are there more than one line required?
> These questions are nonsense
WM is good at producing nonsense.
But he is so good at it that he rarely manages to produce anything else.