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Re: Matheology § 203
Posted:
Feb 3, 2013 3:26 AM
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On Feb 3, 8:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 3 Feb., 00:22, William Hughes <wpihug...@gmail.com> wrote:
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> > On Feb 2, 11:58 pm, William Hughes <wpihug...@gmail.com> wrote:
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> > > On Feb 2, 11:42 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
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> > > > On 2 Feb., 23:36, William Hughes <wpihug...@gmail.com> wrote:
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> > > > > On Feb 2, 11:15 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
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> > > > > > On 2 Feb., 20:11, William Hughes <wpihug...@gmail.com> wrote:
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> > > > > > > > > > > Can a potentially infinite list
> > > > > > > > > > > of potentially infinite 0/1
> > > > > > > > > > > sequences have the property that
> > > > > > > > > > > if s is a potentially infinite 0/1
> > > > > > > > > > > sequence, then s is a line of L
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> > > > > <snip>
> > > > > > For every s: There is alsways a list that contains the first n bits of
> > > > > > s.
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> > > > > Is there a single line which contains s
> > > > > Yes or no
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> > > <snip>
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> > > > There is no complete s.
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> > > Then the answer is no
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> > Indeed, since there is no single line, l,
> > such that every initial segment of s is contained
> > in l, we do not even have to talk about complete s.-
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> In fact we can say that in a suitable list "every" initial segment of
> s is contained in some line, since there is no s(n) = (s1, s2, ...,
> sn) missing. But there is no sensible way of saying "all" initial
> segment.
We can say "every line has the property that it
does not contain every initial segment of segment of s"
There is no need to use the concept "all".
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