Re: Distinct linear orderings on Z

From: aeo6 (aeo6_at_cornell.edu)
Date: 03/28/05 

Date: Mon, 28 Mar 2005 14:26:59 -0500

imaginatorium@despammed.com said:
> aeo6 Tony Orlow wrote:
> > Stephen J. Herschkorn said:
> > > And Tony fails to justify his prioritzation of *his* subjective
> > > intuitions. That is, which intuitions must be maintained and which
> may
> > > be discarded? For example, I have stopped waiting for a coherent
> > > resolution of how "bigulosity" resolves the following paradox:
> Compare
> > > the "sizes" of the following pairs of sets:
> > >
> > > a) {0, 1, 2,...} vs. {(0,1), (1,2), (2,3),...}
> > > b) {1, 2, 3,...} vs. {(0,1), (1,2), (2,3),...}
> > > c) {0, 1, 2,...} vs. {1, 2, 3,...}
> > You are using two separate mappings at once, that differ by an offset
> of 1. It
> > boils down to a mapping function, as you well know, of f(x)=x+1,
> which
> > indicates that **the** mapped set has one less element.
>
> "The"?? Stephen's example looks like this, symbolically:
>
> a) P vs. Q
> b) R vs. Q
> c) P vs. R
>
> You claim (if I can use b(X) for the bigulosity of set X) that:
>
> b(P) = b(R) + 1
>
> Is this right? Please confirm.
Yup.
>
> Well, what is b(Q)? Is it equal to:
>
> b(P) ?
> b(R) ?
> incomparable?
> something else entirely?
Q is a compound set of sets, with two elements per element. Each element of P
or R maps to each pair in Q by two different mapping functions. We map P->Q via
f(x)=(x,x+1) and R-> via f(x)=(x-1,x). Now lets compare the mapping functions.
The mapping from P to Q is one less for each sub-element in Q than the mapping
from R to Q. This is the source of the confusion: doing two mappings at once.
It would appear to me that, since we would like to handle such sets of sets, we
would probably want to assign a set of two bigulosities, since we really have
two mapping functions and two series of numbers. Let's say the bigulosity of Q
is (|N|,|N|-1)
>
> Brian Chandler
>
> 

-- 
Smiles,
Tony

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