John Edser wrote:
>>>>BOH:-
>>>>Now can you answer my question - are you saying that the
elements in the
>>>>intersection C are not also in sets A and B?
>>>>JE:-
>>>>All the elements in the intersection C are also in sets
>>>>A and B.
>>>
> 
>>>BOH:-
>>>Thank you.  After many months, you finally say this.
>>
> 
>>>JE:-
>>>Rubbish.
>>>I have never disagreed that "All the elements in the
>>>intersection C are also in sets A and B" as a "how"
>>>proposition. Please provide a quote of myself where I
>>>have stated, either implicitly or explicitly that "All the
>>>elements in the  intersection C are NOT also in sets
>>>A and B"!
>>
> 
>>BOH:-
>> From the 3rd of September this year:
>>
>>Total set intersection does NOT mean the
>>same 4 offspring are being reproduced
>>by more than one parent, it means that 4 offspring were
>>reproduced by one parent and another, different, 4
>>offspring were reproduced by another, non related parent
>>within the intersection. Neither of these
>>2 parents were sexual partners. When you intersect
>>6 with 4 the intersection contains 4 offspring
>>from TWO SETS not just one set, producing a total
>>8, with 2 offspring not intersected. The total remains
>>10 offspring (6+4 = 10). You have to count the 4
>>in the intersection _twice_ not just once because 2
>>sets are being intersected.
>>You seem not to understand that the intersection
>>contains a _mixture_ of ALL the reproductions of
>>every parent in one population. You seem to think
>>that a mixture of parental fitness elements must
>>be an invalid category; it isn't. It is simply
>>the category of fitness comparison.
>>
>>The second paragraph clearly states that the intersection contains the
>>elements of all parents - i.e. each individual in it must be the
>>offspring of every parent.  Some orgy!
> 
> 
> 
>>JE:-
>>No sex was supposed, i.e. all reproduction
>>was asexual. Each individual in the
>>intersection was NOT the offspring of every
>>parent,
> 
> 
> BOH:-
> i.e. In which case it can't be in the intersection of all of the sets.
> 
> JE:-
> Yes it can because every unit of fitness is
> _equivalent_ to every other, irrespective
> of the parent that produced it. 

But in the absence of sex, each unit of fitness can only have one 
parent, surely?

Only one
> rule exists that enables set intersection
> or set union: the sets must contain equivalent
> elements.
> 
> 
> BOH:-
> So, if we define A and B as sets of offspring of one parent, and C is
> the intersection (under your definition), then the following statement
> of false: "All the elements in the intersection C are also in
sets A and B."
> 
> JE:-
> Incorrect.
> All the fitness elements in C are also in sets A and
> B because they are _equivalent_ fitness elements. 

By "equivalent", do you mean "they have the same properties
as elements 
in both A and B"?


> 
> 
> 
> 
> 
>>>BOH:-
>>>Just to check,
>>>does this means that you do agree with the following definitions of
>>>intersection (copied from a few days ago):
>>> "Given two sets A and B, the intersection of A and B,
written A INT B,
>>>is the set C of all elements that are in both A and B."
>>> From the second site:
>>>"A INT B: A intersection B is the set of all elements that
are in both
>>>sets A and B."
>>> From the third site:
>>>"Intersection - Denotes the set of elements that are
members of all the
>>>sets under consideration."
>>>(I've written INT for the intersection symbol, as it doesn't appear in
>>>ASCII).
>>
> 
>>>JE:-
>>>Yes, as "how" propositions. The "why" proposition
>>>has not been dealt with, in the above.
>>
> 
> BOH:-
> But,as you say, the "why" is definitional.  So if, for example, I have
> sets of offspring from parents, then the intersection between two of
> these sets contains the elements that are offspring of both parents.
> Why this should be so is a matter for biology, but the mathematics is clear.
> 
> JE:-
> The "intersection between two of
> these sets" does NOT contain "the elements
> that are offspring of both parents"!
> 
But is each set not the sets of elements from one parent?



Bob

-- 
Bob O'Hara

Rolf Nevanlinna Institute
P.O. Box 4 (Yliopistonkatu 5)
FIN-00014 University of Helsinki
Finland
Telephone: +358-9-191 23743
Mobile: +358 50 599 0540
Fax:  +358-9-191 22 779
WWW:  http://www.RNI.Helsinki.FI/~boh/
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