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Re: Set Size to represent Natural Numbers { {}{}{}{}{} }
Posted:
Feb 25, 2012 5:10 PM
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On Feb 26, 7:28 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Feb 26, 7:15 am, Zuhair <zaljo...@gmail.com> wrote:
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> > On Feb 25, 11:09 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
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> > > On Feb 26, 6:03 am, Zuhair <zaljo...@gmail.com> wrote:
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> > > > On Feb 25, 10:27 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
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> > > > > On Feb 26, 3:52 am, Zuhair <zaljo...@gmail.com> wrote:
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> > > > > > On Feb 24, 9:15 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
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> > > > > > > If we take {} as being an elementary SCORE / SCRIBE symbol,
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> > > > > > > then we can do maths caveman style, scribbling on walls
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> > > > > > > //// + // = //////
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> > > > > > > 1 = { {} }
> > > > > > > 2 = { {},{} }
> > > > > > > 3 = { {},{},{} }
> > > > > > > 4 = { {},{},{},{} }
> > > > > > > ..
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> > > > > > > So the Natural Numbers are merely a count of all possible set sizes!
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> > > > > > > Herc
> > > > > > > --http://Matheology.com
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> > > > > > hmmm..., Ok lets violate Extensionality in order to allow distinct
> > > > > > empty objects, the problem is that we would have many of those sets
> > > > > > per size, so for example we would have
> > > > > > infinitely many { {} } sets, so how do we pick The one representing
> > > > > > number 1?
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> > > > > A repeatable element or 2 might be useful.
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> > > > > A = B IIFF A(x): x e A <-> x e B
> > > > > VS
> > > > > A = B IFF A(x) (x =/= {}) -> (x e A <-> x e B)
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> > > > > NOW
> > > > > { {}.{},{}, a, b, c } =/= { {}, b, a, c }
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> > > > > BUT
> > > > > { a, b, c } = { b, a, c }
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> > > > > Herc
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> > > > Yes I already said that, "if we violate Extensionality" then we can
> > > > have multiple distinct empty objects, but this is not the problem, the
> > > > problem is that we would have multiple objects that are empty, so we
> > > > will have multiple {{}} objects, now which one of those is 1???
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> > > then we can do maths caveman style, scribbling on walls
> > > //// + // = //////
> > > 1 = { {} }
> > > 2 = { {},{} }
> > > 3 = { {},{},{} }
> > > 4 = { {},{},{},{} }
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> > > Since a simple replacement for Extensionality allows
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> > > 1 =/= 2 the above method should suffice.
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> > > Herc
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> > No you can't, I already asked you the question many times, how do you
> > pick the {{}} for which the symbol 1 stands? you have infinitely many
> > {{}} sets, so which one of them is 1. You didn't answer till now this
> > simple question? I gave you a suggestion which you simply ignored just
> > to repeat your assertion again and again. You wrote 1= {{}} My
> > question again
> > which {{}} of the infinitely many {{}} sets you mean? The same
> > question applies to the rest of 2,3,4,.... My suggestion is that 1 is
> > the set of all {{}} sets, 2 is the set of all { {},{} } sets... etc...
> > This would be understandable because those sets would be unique
> > because according to your suggested weak extensionality there will be
> > unique sets of those, i.e.
> > there will be unique sets of all {{}} sets, also a unique set of all
> > { {},{} } sets, etc...
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> > In symbols again: my alternative suggestion for your approach is:
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> > 1={ {{}} , {{}} , .....} = {{x}| ~Exsit y. y e x}
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> > 2= { { {},{} } , { {},{} } , .....} = { {x,y}| ~Exist z. z e x &
> > ~Exist u. u e y}
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> > etc...
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> > Zuhair
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> Some further axioms on "="
> would be required to ensure
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> {{}} = {{}}
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> and
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> {{},{}} = {{},{}}
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> Where is the problem with
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> {{}{}{}} + {{}{}} = {{}{}{}{}{}}
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> 'there are oo-many {{}} sets' makes no sense to me.
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> Herc
A simpler system might be
{...{}...} = {{}{}{} ..{}}
Since each is a form of redundancy.
Herc
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