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Re: Why not Hausdorff's ordered pairs
Posted:
Feb 24, 2012 2:46 PM
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On Feb 24, 10:25 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Feb 24, 2:34 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
>
>
> > you added the braces around {a} {b} {c} for a reason, so you could
> > determine the {position,{data}} order.
>
> EXACTLY! what I see it nice about it is how it simplifies sequences
> so for example a sequnce S= 7,63, 2
>
> Instead of implementing this as S= { {{1},{1,7}} , {{2},{2,63}} , {{3},
> {3,2}} }
> which is the customary way using Kuratowski pairs based sequence.
> We can simplify that to S= { {1,{7}} , {2,{63}} , {3,{2}} } which is
> much
> simpler really.
Not in sets it isn't!
Instead of a duplicate redundancy on the numerical index n,
you repeat it n times.
To remove the duplicate index, you have to slice the index into an n
elements set.
{ {3},{3,C} } K Pair
VS
{ {2, 1, 0}, {C} } Z Pair
K Pairs use SET SUBSTRACTION to identifiy the first ordinate/index
Your system uses SET.SIZE>1 member sets to identify the index
Note in K Pair, Naturals can be represented O(n) {{{{{}}}}}
where Von Neuman Numbers are size & time expensive O(n^2) for no
reason.
Breaking your 3rd Ordinate down further..
{ { { {} {{}} }, {{}}, {} }, {C} }
VS K Pairs
{ { {{{}}} }, { {{{}}}, C} }
You are forced to construct arbitrary DISTINCT SETS
in order to count them to represent the Natural by it's SET SIZE =
MEMBER COUNT
{ DIFFERENT, DIFFERENT, DIFFERENT, DIFFERENT } = 4
In pure sets, this is 4 levels of recursion O(n^2) data structure.
You couldn't represent a million digits of Pi on a modern computer in
your system.
Herc
--
http://Matheology.com
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