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Re: Matheology § 200
Posted:
Jan 27, 2013 8:42 AM
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On 27 Jan., 14:04, William Hughes <wpihug...@gmail.com> wrote:
> On Jan 27, 9:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 27 Jan., 01:55, William Hughes <wpihug...@gmail.com> wrote:
>
> > > On Jan 26, 10:51 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > The set of indexes is 1 + the decadic
> > > > logarithm of L.
>
> > > Nope, 1+ the decadic logarithm of L is the cardinality
> > > of the set of indexes.
>
> > I know.
>
> So do not make silly statements
> about what the set of indexes is.
>
> You are taking the limit of
>
> 1+decadic logarithm of L
>
> You are not taking the limit of sets
> but the limit of cardinalites of sets.
You are wrong. Look it up, in case you have forgotten. For instance
here
http://www.hs-augsburg.de/~mueckenh/GU/GU12c.PPT#403,25,Folie 25
The sets are {2, 1}, {2}, {4, 3, 2}, {4, 3}, {6, 5, 4, 3}, {6, 5,
4}, ...
This is a sequence of sets. If the digits in their order are
interpreted as a number (the usual and most natural way to do), then
the limit is oo. The cardinality of the indexes of this limit in
analysis is aleph_0.
The sequence of cardinalities is 2, 1, 3, 2, 4, 3, ... The limit of
this sequence is aleph_0 too.
> The limit you calculate is not a limit set, nor the
> cardinality of a limit set.
Analysis shows that the cardinality of the digits is 1 + logn. This
does not break down for n = oo.
Regards, WM
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