> >JM:- ...C' contains only two things, not four. Those two things are both sets, and each of them

> >contains two of the original objects. The set "container" does not "dissolve" when a set "moves

> >inside" another set.

> >JE:- The set container inside, does dissolve when sets are joined by set union. This is a

> >critical difference between union and intersection. AW severely criticised me when I suggested

> >that sets merged by set union lose their integrity because of this fact. Set union is a non

> >reversible logic but set intersection is a reversible logic. Understanding this difference is

> >critical for any application of set theory to evolutionary theory. If independent sets of fitness

> >are merged by set union they lose their fitness independence but this is 100% preserved when

> >fitness sets intersect.

JM:- I think that I have only created confusion by raising the possibility of sets that contain

other sets as elements.

JE:- Why? Numbers are sets of sets and you can’t get more basic than that.

JM:- Sets that contain other sets as *elements* are not common in applied set theory, though it is

common in the mathematician's *pure* set theory. John has perhaps misunderstood what I said and has

interpreted it in terms of sets containing other sets as *subsets*. That is common in both pure and

applied set theory. Let me try to give an example of what I am saying from biology.

First, we will consider sets of species. We might define the set Mammalia to be the set of all

species that are mammals. So, **** sapiens is a member or element of the set Mammalia. But John

Edser is not an element of Mammalia (as defined here). John is an organism, not a species. If we

think of species as sets of organisms, then we might say that John is an element of the set ****

sapiens, and thus an element of an element of Mammalia.

Now, suppose that we define C as the set of all carnivorous species, and V as the set of all

viviparous species. The intersection of C and V will be a set of species (not a set of organisms!)

that are BOTH carnivorous AND viviparous. The union of C and V will contain even more species,

including Equus equus - which is viviparous, but not carnivorous. The union is the set of species

that are EITHER carnivorous OR viviparous. The sets C and V continue to exist - no damage was done

to them by either intersection nor union formation.

JE:- It does not matter if you define one set element as another set, or not. The elements remain

exactly as you _define_ them; no more and no less. If Mammalia are the set of all species that are

mammals then myself as an individual mammal, can only remain TOTALLY INVISIBLE within a set union of

things defined as one species of mammal. Within any set union each element is now regarded to be as

_exactly_ the same type of set element, i.e. each set element _remains_ separate but you cannot tell

them apart. Dare I say it(?) they are _equivalent_. It does mot matter if they have many other

things that are different! They are EQUIVALENT BY DEFINITION. If you define C as the set of all

carnivorous species, and V as the set of all viviparous species then their set union is the set of

species that are EITHER carnivorous OR viviparous. However, as far as each set element within this

set union is concerned they are EQUIVALENT, i.e. you can’t anymore, say which is, and which is not,

carnivorous OR viviparous by ONLY inspecting the set elements within that set union using the

_definition_ of each set element. You have to _reverse_ that set union to find this out. Within set

union the sets C and V DO NOT continue to exist. The fact that you can easily identify their set

elements within the set union because you are not just a dumb machine does not alter anything re:

the logic. The union defines all set elements as equivalent by definition. When you say you can

still tell the difference between carnivorous OR viviparous mammals within a set union of both you

are simply reversing the set union in your minds eye. A dumb machine, or nature, cannot do this.

Note that if you intersect these sets then what is in the intersection are set elements (mammal

species) from both sets, i.e. mammals that are both carnivorous and viviparous. Thus you can still

identify mammal species that are only viviparous, only carnivorous and mammal species that are both.

ONLY here do both sets C and V continue to exist within the logic of a dumb machine, via the

definition of what each set element is defined to be and NOT _outside_ of that definition.

If A = B then A cannot be distinguished from B when both exist.

>[snip]

>

> JE:- We were attempting to represent a _biological_ problem using set theory, specifically:

> illustrate one selective event. I represent this event as the total intersection between a minimum

> of 2 independent parental fitness sets, say: A = 3 B = 4. Two overlapping Venn circles are drawn

> where the 3 areas from left to right contain:

>

> ,3,1

>

> Set A has all of its fitness elements in the intersection. Set B has only 3 of them.

> Thus: Set A is a subset of set B so nature only has just one set to select: set B.

JM:- Isn't set B is the *union* of the sets A and B?

JE:- No. The union is just a single set of 7. Here the independent fitness sets have been merged

within one total (their identity has now been lost by definition) and not compared (their identity

retained by definition). An intersection is a comparison. Set union is a total.

> JE:- Thus no cognition is needed to make a selection so ?automatic selection? becomes logically

> explicable. When set B is selected set A is also selected because it is a subset of B. But set A

> is only selected _after_ and not _simultaneous_ to, set B. Thus B is only sub selected after A is

> selected.

> _______________________________________

> The main dispute here is about the validity of my proposed set intersection between independent

> parental fitnesses set totals within the same Darwinian population.

> ________________________________________

>

JM:- I would claim that we don't even have a "dispute" yet, because no one except John understands

what John is saying. John has a picture in his head of how to use Venn diagrams to represent a

single "selective event". But, so far, no one else really understands what a "selective event" is!

Here is what I think I understand so far. Please, correct my misunderstandings, John.

1. Selective events happen once per generation.

2. You start with a set of organisms which I think John calls units of selection - I am not clear on

the terminology here.

3. These organisms participate in a struggle for existence and reproduction. With different degrees

of success, they bring into existence a bunch of new sets of organisms - their offspring.

then clearly each next-generation organism is a member of two sets of offspring - one for

each parent.

4. In any case, we have a bunch of sets of next generation organisms - one set for each element of

the original parental set.

5. The sets of next generation organisms become "blurry" in some way so that they contain

indistinguishable "fitness elements" rather than distinguishable organisms.

6. Visualize each of these sets of fitness elements as dinner plates with the diameter of the plate

representing the count of fitness elements contained. Stack all of the plates. The largest plate

stands out, and Nature, without a brain in her head, can recognize it.

7. And, at this point, my understanding breaks down. Why was it important for nature to be able to

recognize the "winner"? My understanding of what has to happen next is that we need to form a

union of all those sets of fitness elements so that we can reconstitute a starting set of

potential parents for the next generation of selection. I am obviously missing something here.

JE:- Forget about selection for the moment. The proposition I am proposing is just very simple. I

have a minimum of two separate totals to compare. That ALL I am proposing!

Jim and John each have a small herd of goats. Each of them carries a small bag of about the same

sized pebbles. As each herder tends his goats he has put a single pebble in his bag to represent

each goat in his herd. John has 3 pebbles in his bag representing all his goats but Jim had 4

pebbles in his bag representing all of his.

John’s bag is set A = 3

Jim’s bag is set B = 4

Each total is a simple count. In set theory terms EACH set remains separate and was created by the

set union of one’s (stones). When 3 one’s join they form one set of 3 called set A. When, in a

different place, 4 one’s join they form one set of 4 called set B.

John offers to trade goat herds. As far as Jim can see, both herds of goats are of exactly the same

quality. However both Jim and John can only count to 2 so they cannot tell which herd is the

biggest. If they just pour both bags of stones into a pot forming a pile of 7 equal stones (set

union) then this does’t help indicate which herd was the largest and they risk losing their

individual stones in the one anonymous pile. This would be a disaster because they are the only

indicator each has for the size of their own herd. Jim has a PhD from Flintstone University so he

easily solves the problem. Jim asks John to line his stones up and Jim lines his up, next to John’s.

Jim’s line of stones can be seen to be larger by one stone.

Jim now works out that John’s proposed trade is not a good deal for Jim because Jim’s herd was

larger. The lining up of the stones and their comparison is a simple set intersection that has the

Venn circle signature:-

,3,1

Draw two overlapping circles. Place 0,3 and 1, in each of the three defined areas from left to

right. Clearly

8) Both sets remain intact.

9) A total of 7 stones are being intersected.

This means:

= Just zero stones, only in John’s set. 3 = Maximum stones in the intersection (both sets) 1 = Just

1 stone, only in Jim’s set.

Where John’s set of 3 stones and Jim’s set of 4 stones remain intact as separate sets so they can be

put back into their separate bags (the set intersection reversed) without further ado.

Delete John, Jim and the stones and leave the _entire_ problem as abstract, then it can be shown

that the complete intersection of set A and B has proven A to be a subset of B. Substitute goats for

fitnesses and you have it in one.

In Darwinism, nature makes the comparison, not Jim and John and she cannot even count to 2. This

does not matter because using a complete set intersection, only one set exists for nature to select.

>snip<

Can we agree, so far?

Respectfully Yours,

John Edser Independent Researcher

PO Box 266 Church Pt NSW 2105 Australia

[email protected]