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Re: Why not Hausdorff's ordered pairs
Posted:
Feb 22, 2012 2:40 PM
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On Feb 23, 4:12 am, Zuhair <zaljo...@gmail.com> wrote:
> On Feb 21, 3:48 pm, Shmuel (Seymour J.) Metz
>
> <spamt...@library.lspace.org.invalid> wrote:
> > In <slrnjk71dv.put.cmen...@philebus.tamu.edu>, on 02/21/2012
> > at 11:53 AM, Chris Menzel <cmen...@remove-this.tamu.edu> said:
>
> > While I find (a,b)={{a},{a,b}} to be more natural for ordered pairs,
> > it does not generalize well to m-tuples.
>
> Actually Hausdorff's generalize much much better than Kuratowski's.
> Compare between
> (a,b,c)={{1,a},{{2,b},{3,c}}} and (a,b,c)={{a},{a,{{b},{b,c}}}}
> which is very Ugly. The nice thing is that Haudorff's admits simple
> tagging of projections according to their order, and this is the
> secret of its naturally looking strength.
>
> Zuhair
Who on Earth has ever practically used the syntax
{{1,a},{{2,b},{3,c}}}
It's a miss match of notation.
Just use (a, b, c)
It doesn't even work!
{ {3,2} , {2,1} , {1,3} }
= { {2,3} , {1,2} , {3,1} }
So much for your proofs!
{{1,a},{2,b}}={{1,c},{2,d}} entails [{1,a}={1,c} & {2,b}={2,d}]
thus
a=c & b=d
or [{1,a}={2,d} & {2,b}={1,c}] thus a=2,d=1,b=1,c=2 thus a=c & b=d.
QED
How are you defining NUMBERS without ORDERED PAIRS?
You can't even write down EQUALS( S(a), a+1 )
It's like saying - HEY LETS USE NEUTRINOS TO COUNT CALORIES INSTEAD!
Herc
--
http://Matheology.com
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