In article <dc0f1cc8-0ece-48ee-97ca-c395fb109ffa@n6g2000vbf.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 13 Feb., 23:22, William Hughes <wpihug...@gmail.com> wrote: > > On Feb 13, 9:03 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > <snip> > > > > > You cannot discern that two potentially infinity sequences are equal. > > > When will you understand that such a result requires completeness? > > > > Nope > > > > Two potentially infinite sequences x and y are > > equal iff for every natural number n, the > > nth FIS of x is equal to the nth FIS of y > > And just this criterion is satisfied for the system > > 1 > 12 > 123 > ... > > For every n all FISs of d are identical with all FISs of line n.
But that is not at all the same thing.
Note that even the actually infinite set of FISs of what you call a merely potentially finite sequence does not contain that sequence as a member.
What WM is claiming that given the infinite sequence of finite sequences l1=1, l2 = 12, l3 = 123, ... and the infinite sequence d = 1,2,3,...
that l1 and l1,l2, and l1, l2, l3 and so on are FISs of d.
But, at least outside of Wolkenmuekenheim, it is not so.
> > And the three points stand for every finite number, but not for all.
If not for all, some must be missing, so which are missing? > > > > Consider the list of potentially infinite sequence > > L1= > > 1000... > > 11000... > > 111000... > > ... > > > > L2= > > 111... > > 11000... > > 111000... > > ... > > > > The diagonals are both > > d=111... > > And again you confuse every with all.
Can WM distinguish between not every and not all? > > > > It makes perfect sense to say that there > > is no line in L1 that is equal > > to d > > Perfect sense?
Far more perfect than WM every makes.
> Do you claim that the list > 1 > 12 > 123 > ... > does not contain every FIS of d?
For the d above, namely d = 111..., your list certainly does not any FIS of that d of more than one digit in length.
> Do you claim that there are two or more FISs of d that require more > than one line for their accomodation?
All of them cannot be accomodated at all > > We can use induction to show that! > > Ponder about this question and then try to make perfect sense.
