John Edser wrote:

>>>BOH: 
>>>So, if no sets are intersected, then there are no intersections.
> 
> 
>>>JE:-
>>>Yes.
> 
> 
>>BOH:-
>>So, if there are no intersections, where does this leave your claim that 
>>intersecting sets are fundamental to understanding selection?
> 
> 
>>JE:-
>>Mad Hatter nonsense. 
> 
> 
> BOH:-
> Ah thank you.
> 
> JE:-
> You are welcome. Pay my respects to the
> white rabbit; he looks unwell...
> 
> Sets can be supposed to be intersected or not
> intersected, at will. OK?
> 
No!  We've already agreed that intersection has a definition, so sets 
can only be intersected if there are elements that are members of both sets.

> My proposition re: natural selection supposes 
> they are always 100% intersected within the
> same population. 

But this can only come about if the offspring are products of all hte 
parents, by the definition of an intersection.

I can make any supposition
> I like as to the level of intersection as long
> as the view is incorporated within a testable
> theory of nature; which it is. OK?
> 
OK,
> 
>>JE:-
>>Sets can be validly
>>_assumed_ to be intersected if and only if 
>>they contain equivalent set elements. I _assume_
>>sets of total parental fitness to be 100% 
>>automatically intersected within one population.
> 
> 
> BOH:-
> But you've already agreed that they aren't intersected.  You're 
> contradicting yourself.
> 
> JE:-
> Please attend to your poor comprehension 
> skills. You stated: "So, if no sets are intersected, 
> then there are no intersections". I did NOT propose
> that there was "no intersection" of all parental
> absolute fitness sets within the same population,
> I proposed the exact OPPOSITE: all such sets are 100%
> intersected. OK?
> 
You're not following the argument:
1. You have agreed that elements in an intersection of two sets are 
those which are members of both sets.
2. You have defined fitness sets as the offspring produced by a parent.
3. You have also agreed that we are not considering sexual species, so 
that each offspring (=fitness element) only has one parent.

 From this, we can conclude that no fitness element has 2 parents. 
Hence, the intersection is empty.

You can state that you have 100% intersections all day, but until you 
show how this is compatible with the logic outlined above, it is 
impossible for me to agree with you.

> 
> 
>>>>JE:-
>>>>If and only if, each set element within every
>>>>absolute fitness set, i.e. each fitness element, 
>>>>is _eqivalent_within one population THEN
>>>>each absolute fitness set CAN be intersected. 
> 
> 
>>>BOH:-
>>>Right, they can be intersected, but they only will be intersected if 
>>>there is an element that is a member of both sets.  i.e. if it has two 
>>>parents.
> 
> 
>>>JE:-
>>>Ridiculous. Nothing in mathematics 
>>>restricts set intersection to sexual
>>>products.
> 
> 
>>BOH:-
>>Huh?  No mention of "sexual products" was made or implied.
> 
>  
> 
>>JE:-
>>You wrote: " ..they only will be intersected if 
>>there is an element that is a member of both sets.  
>>i.e. if it has two parents." Where in biology
>>does nature provide offspring (the defined fitness 
>>element) that requires two parents that does not 
>>involve the production of sexual products from 
>>each parent?
> 
> 
> BOH:-
> You tell me - you're the one insisting that they are intersected. 
> 
> JE:-
> You were invalidly insisting that in order
> to be "a member of both sets", i.e. within a
> mathematical intersection, the offspring must
> have two parents. Mathematics does not dictate
> that sets of offspring must be sexually reproduced 
> to be intersected. 

No, but it does insist that they must have two parents.  Strange, I 
thought we had already agreed about this.

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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