John Edser wrote:

>>>>>JE:-
>>>>>I have commented on the quotes, in detail.
>>>>>For example, "set C of all elements that
>>>>>are in both A and B" means that all the
>>>>>set elements in the intersection C are strictly
>>>>>_equivalent_ elements and not just elements
>>>>
>>>>>from one set, i.e not "the same" elements.
> 
> 
>>>>BOH:-
>>>>This is opaque to me.  Are you saying that the elements in the
>>>>intersection C are not also in sets A and B?
> 
> 
>>>>JE:-
>>>>The concept of equivalence is simple.
>>>>Two things may be defined equivalent.
>>>>This does not make them the one, same,
>>>>thing, but it does mean either of them
>>>>can represent the other. All I am saying
>>>>is that the elements in intersection C are equivalent
>>>>set elements to the set elements in set A and B.
> 
> 
>>>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, 

i.e. In which case it can't be in the intersection of all of the sets. 
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."


>>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.
> 
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:-
>>Do you agree that all fitness set elements are equivalent
>>but are _not_ just the one, same, fitness element?
> 
> 
> BOH:-
> I want to nail down the concept of intersection first, before I get onto 
> anything else - and we've been on that since June.
> 
> JE:-
> It would have been easier for you if you 
> had referred to Venn Diagrams within a junior
> High School text.
> 
> Please answer the question:
> Do you agree that all fitness set elements are equivalent
> but are _not_ just the one, same, fitness element?
> 
Please read what I wrote again.  I haven't changed my opinion over the 
weekend.

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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