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. Can I make a suggestion? Could AW and/or JM please supply a valid example of any applied set intersection where all of one set is a subset of the other, i.e. provide a simple problem and then show how set theory correctly illustrates this problem where the illustration shows one set as a subset of the other. Could you please point out why you think the intersection was mathematically valid in the example provided. Thanks for your help, John Edser Independent Researcher PO Box 266 Church Pt NSW 2105 Australia [email protected]

On Wed, 4 Feb 2004 06:18:30 +0000 (UTC), "John Edser" <[email protected]> wrote: >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. > >Can I make a suggestion? > >Could AW and/or JM please supply a valid example of any applied set intersection where all of one >set is a subset of the other, i.e. provide a simple problem and then show how set theory correctly >illustrates this problem where the illustration shows one set as a subset of the other. Could you >please point out why you think the intersection was mathematically valid in the example provided. > I find it extremely difficult to understand just why John Edser has so much trouble with elementary set theory. There is no big conceptual difference between set union and set intersection as he claims. There is no such thing as a "set container". Consider the sets S1 = { a, b, c } and S2 = { a, b, d }. Then the intersection of S1 and S2 is simply the set { a, b } and the union of S1 and S2 is the set {a, b, c, d }. Taking the union or taking the intersection do absolutely nothing to the original sets. They still exist unchanged. The intersection and the union are brand new sets, not modifications of the original. Now consider the case requested where one set is a proper subset of the other: S1 = {a, b, c } and S3 = {a, b}. Now the intersection of S1 and S3 is {a, b} which is equal to S3 while the union of S1 and S3 is {a, b, c} which is equal to S1. The fact that the union and the intersection happen to be equal to (contain the same elements) as one of the originals has absolutely no signficance on whether the original sets are somehow modified -- they are not. What is the big deal? Why has there been a thread about set intersection for all these many months?

>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. >Can I make a suggestion? Could AW and/or JM please supply a valid example of any applied set >intersection where all of one set is a subset of the other, i.e. provide a simple problem and then >show how set theory correctly illustrates this problem where the illustration shows one set as a >subset of the other. Could you please point out why you think the intersection was mathematically >valid in the example provided. RN:- I find it extremely difficult to understand just why John Edser has so much trouble with elementary set theory. There is no big conceptual difference between set union and set intersection as he claims. There is no such thing as a "set container". Consider the sets S1 = { a, b, c } and S2 = { a, b, d }. 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. 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. ________________________________________ The “set container” refers to the edge of _separate_ sets, that is all. Unless sets can be represented as separate then set theory cannot even exist. When representing independent set of _fitness_ it is of upmost importance to represent a barrier between independent sets because they are all competing against each other within the defined universal set: one Darwinian population. Clearly, if all fitness sets are competing but you only represent them as the one universal set of fitness by merging all of them using set union (this is the normal Neo Darwinian way of representing absolute fitness) then only zero competition can now exist because only one set exists and this set cannot compete (be compared) against itself. However if you represent them as intersections they can all be compared to each other as independent sets of fitness because were NOT merged. RN:- Then the intersection of S1 and S2 is simply the set { a, b } and the union of S1 and S2 is the set {a, b, c, d }. Taking the union or taking the intersection do absolutely nothing to the original sets. They still exist unchanged. The intersection and the union are brand new sets, not modifications of the original. JE:- Please explain the logical _difference_ between set union and set intersection. If they are logically the same then it must be possible for one set to become a subset of the other using set union, but it isn’t. If I compare sets of fitness I do nothing to each set yet. However a comparison can represented using set intersection because the sets are _not_ merged, they _remain_ separate. I can’t compare independent sets of fitness using set union because the sets are lost when they are merged and merging them provides no information re: their relative size difference. Set union is not logically the same as set intersection. Adding to a total is set union. Comparing totals is set intersection. Making and comparing totals are not logically the same thing. RN:- Now consider the case requested where one set is a proper subset of the other: S1 = {a, b, c } and S3 = {a, b}. Now the intersection of S1 and S3 is {a, b} which is equal to S3 while the union of S1 and S3 is {a, b, c} which is equal to S1. The fact that the union and the intersection happen to be equal to (contain the same elements) as one of the originals has absolutely no signficance on whether the original sets are somehow modified -- they are not. What is the big deal? Why has there been a thread about set intersection for all these many months? JE:- Nobody can agree re: exactly HOW independent sets of fitness can be represented within set theory to illustrate one selective event. Since one selective event is just a default comparison of two independent fitness totals, then this must be able to be illustrated using set theory. Please provide what was requested: an application of set theory, OUTSIDE OF MATHEMATICS, that allows one set to become a subset of the other as a valid set theory illustration of that specific problem, using set intersection. Many Thanks, John Edser Independent Researcher PO Box 266 Church Pt NSW 2105 Australia [email protected]

"John Edser" <[email protected]> wrote in message news:<[email protected]>... > >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. I think that I have only created confusion by raising the possibility of sets that contain other sets as elements. 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, Homo 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 Homo 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. >[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. Isn't set B is the *union* of the sets A and B? > 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. > ________________________________________ > 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. > The ?set container? refers to the edge of _separate_ sets, that is all. Unless sets can be > represented as separate then set theory cannot even exist. When representing independent set of > _fitness_ it is of upmost importance to represent a barrier between independent sets because they > are all competing against each other within the defined universal set: one Darwinian population. > Clearly, if all fitness sets are competing but you only represent them as the one universal set of > fitness by merging all of them using set union (this is the normal Neo Darwinian way of > representing absolute fitness) then only zero competition can now exist because only one set > exists and this set cannot compete (be compared) against itself. However if you represent them as > intersections they can all be compared to each other as independent sets of fitness because were > NOT merged. > It is clear that one thing I am missing is an idea of what John means by "independence". I am also missing how genes are getting passed to the next generation. I am missing how all of this relates (and I am pretty sure it does relate) to John's agruments against the likelihood of "group selection". > RN:- Then the intersection of S1 and S2 is simply the set { a, b } and the union of S1 and S2 is > the set {a, b, c, d }. Taking the union or taking the intersection do absolutely nothing to the > original sets. They still exist unchanged. The intersection and the union are brand new sets, not > modifications of the original. > > JE:- Please explain the logical _difference_ between set union and set intersection. If they are > logically the same then it must be possible for one set to become a subset of the other using set > union, but it isn?t. Of course it is. Take any two sets A and B. Define C as the union of A and B. Then A is a subset of C. Define D as the intersection of A and B. Then A will probably not be a subset of D, though it will be if A was a subset of B to begin with. A will be the same as D, and thus a subset of D, though not a "proper" subset of D. Or, were you talking about whether A is a subset of B? If so, then it either is a subset, or it is not, and any formation of "new" sets by intersection or union isn't going to change that. The logical difference between union and intersection is as described earlier in my post - intersection uses "BOTH_AND_" logic and union uses "EITHER_OR_" logic. I am beginning to guess that you see "intersection" as something that happens when one set is moved on top of another - like dinner plates. You see the set as changed, somehow, by this process - or at least you see the relationship between the sets as changed. Set A was not a subset of B, but then you intersect them, and now A IS a subset of B. Is that what you are thinking? > If I compare sets of fitness I do nothing to each set yet. However a comparison can represented > using set intersection because the sets are _not_ merged, they _remain_ separate. I can?t compare > independent sets of fitness using set union because the sets are lost when they are merged and > merging them provides no information re: their relative size difference. Set union is not > logically the same as set intersection. Adding to a total is set union. Comparing totals is set > intersection. Making and comparing totals are not logically the same thing. > A suggestion - stop talking about "sets" and try to describe a "selection event" using some other metaphor. I might suggest stacks of coins - one coin per fitness element. It is clear that your ideas about "sets" don't correspond to those of the people you are talking to. But maybe we can see what you are saying if you use a different illustration.

On Sat, 7 Feb 2004 05:52:54 +0000 (UTC), "John Edser" <[email protected]> wrote: > > >>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. >>Can I make a suggestion? Could AW and/or JM please supply a valid example of any applied set >>intersection where all of one set is a subset of the other, i.e. provide a simple problem and then >>show how set theory correctly illustrates this problem where the illustration shows one set as a >>subset of the other. Could you please point out why you think the intersection was mathematically >>valid in the example provided. > >RN:- I find it extremely difficult to understand just why John Edser has so much trouble with >elementary set theory. There is no big conceptual difference between set union and set intersection >as he claims. There is no such thing as a "set container". Consider the sets S1 = { a, b, c } and >S2 = { a, b, d }. > > >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. 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. >________________________________________ > >The “set container” refers to the edge of _separate_ sets, that is all. Unless sets can be >represented as separate then set theory cannot even exist. When representing independent set of >_fitness_ it is of upmost importance to represent a barrier between independent sets because they >are all competing against each other within the defined universal set: one Darwinian population. >Clearly, if all fitness sets are competing but you only represent them as the one universal set of >fitness by merging all of them using set union (this is the normal Neo Darwinian way of >representing absolute fitness) then only zero competition can now exist because only one set exists >and this set cannot compete (be compared) against itself. However if you represent them as >intersections they can all be compared to each other as independent sets of fitness because were >NOT merged. > >RN:- Then the intersection of S1 and S2 is simply the set { a, b } and the union of S1 and S2 is >the set {a, b, c, d }. Taking the union or taking the intersection do absolutely nothing to the >original sets. They still exist unchanged. The intersection and the union are brand new sets, not >modifications of the original. > >JE:- Please explain the logical _difference_ between set union and set intersection. If they are >logically the same then it must be possible for one set to become a subset of the other using set >union, but it isn’t. If I compare sets of fitness I do nothing to each set yet. However a >comparison can represented using set intersection because the sets are _not_ merged, they _remain_ >separate. I can’t compare independent sets of fitness using set union because the sets are lost >when they are merged and merging them provides no information re: their relative size difference. >Set union is not logically the same as set intersection. Adding to a total is set union. Comparing >totals is set intersection. Making and comparing totals are not logically the same thing. > >RN:- Now consider the case requested where one set is a proper subset of the other: S1 = {a, b, c } >and S3 = {a, b}. Now the intersection of S1 and S3 is {a, b} which is equal to S3 while the union >of S1 and S3 is {a, b, c} which is equal to S1. The fact that the union and the intersection happen >to be equal to (contain the same elements) as one of the originals has absolutely no signficance on >whether the original sets are somehow modified -- they are not. What is the big deal? Why has there >been a thread about set intersection for all these many months? > >JE:- Nobody can agree re: exactly HOW independent sets of fitness can be represented within set >theory to illustrate one selective event. Since one selective event is just a default comparison of >two independent fitness totals, then this must be able to be illustrated using set theory. > >Please provide what was requested: an application of set theory, OUTSIDE OF MATHEMATICS, that >allows one set to become a subset of the other as a valid set theory illustration of that specific >problem, using set intersection. > Frankly, I have no interest in getting involved in your dispute over how you represent "independent sets of fitness". It is simply that you are either misinterpreting or misusing the notions of set theory. Not everything in the work must necessarily be able to be illustrated using set theory. Especially, if you define the problem in a totally inappropriate way, set theory is inappropriate or improperly used. Of course I didn't say that set intersection and set union are identical. I said they shared many features including the fact that neither of them has anything at all to do with the notion of a "set container", whatever that may be. On both cases, you act on the elements of two separate sets to produce a new set containing elements selected from the two original sets. The difference is in which elements you select. The similarity is that an operation on two sets yields a new set Nowhere does the notion of boundaries arise. When you compute set union, the sets are NOT merged. The sets are NOT lost. Set theory is part of mathematics. You are simply abusing mathematics by improperly defining sets to try to model your notion of fitness.

RN:- Of course I didn't say that set intersection and set union are identical. I said they shared many features including the fact that neither of them has anything at all to do with the notion of a "set container", whatever that may be. JE:- _______________________________________________________ Firstly and most importantly, do you agree that any addition or subtraction can be illustrated using set theory? _______________________________________________________ How do you signify the separation of the sets? RN:- On both cases, you act on the elements of two separate sets to produce a new set containing elements selected from the two original sets. JE:- If this is the case, how is a set union different to set intersection is this respect, i.e. how is a set union different to a set intersection when “you act on the elements of two separate sets to produce a new set containing elements selected from the two original sets” RN:- The difference is in which elements you select. The similarity is that an operation on two sets yields a new set Nowhere does the notion of boundaries arise. If “nowhere does the notion of boundaries arise” how were the sets known to be separate in the 1st place and how do we know the boundaries of the new set/sets created in each case? If “the only difference is in which elements you select” how is the entire intersection of two sets different to their union? If you would provide a simple example in real life, i.e. an illustration of set theory outside of pure mathematics (as originally requested) of both: 1) The union of two sets. 2) The entire intersection of two sets this would help me/us understand the error we are accused of making. Regards, John Edser Independent Researcher PO Box 266 Church Pt NSW 2105 Australia [email protected]

"John Edser" <[email protected]> wrote in message news:<[email protected]>... > If you would provide a simple example in real life, i.e. an illustration of set theory outside of > pure mathematics (as originally requested) of both: > > 1) The union of two sets. > > 2) The entire intersection of two sets > > this would help me/us understand the error we are accused of making. A set is a conceptual collection of *distinguishible* things. They are creations of the human (or some other intelligent) mind. For example, we might define the set P to be the set of all placental species. We might define M to be the set of all mammalian species. The primary question related to sets is whether some thing is a member of the set or not. For example, platypus is a member of M but not a member of P. Wolf is a member of both sets P and M. Rattlesnake is in neither. Given two sets, you may define the union as the set which contains everything that is in either the first set or the second set or both. Define set Q to be the union of P and M. Platypus is a member of Q, and so is wolf. Wolf is not a member twice - no set can have duplicate members. Given two sets you may also define the intersection as the set which contains everything that is in both the first set and the second. Define set Z to be the intersection of P and M. Wolf is a member of Z. Platypus is not a member of Z. Two sets are said to be equal if they have the same members. One set is a subset of another if everything in the first set is also in the second. Notice the following about the sets P, M, Q, and Z defined above: P is a subset of M. Q is equal to M. Z is equal to P. Someone who has two bags containing *indistinguishible* pennies, five pennies in one bag and eight pennies in another probably should not be using set theory to compare the counts, to add or subtract the counts, etc. Doing those comparisons, additions, and subtractions are perfectly fine and perfectly simple things to do, but it just shouldn't be done using set theory. Set theory is for *distinguishible* things. Set theory is non-physical. A thing can be (in fact, it has to be) a member of a huge number of different sets. But nothing can be a member twice in the same set.

> >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, Homo 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 Homo 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]

"John Edser" <[email protected]> wrote in message news:<[email protected]>... > > 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. > 5. In any case, we have a bunch of sets of next generation organisms - one set for each element of > the original parental set. > 6. The sets of next generation organisms become "blurry" in some way so that they contain > indistinguishable "fitness elements" rather than distinguishable organisms. > 7. 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. > 8. 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. JM:- OK, 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 > 1) Both sets remain intact. > 2) 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? JM:- I now (for the first time) understand you. I agree that what you are trying to do is perfectly simple math, and that Nature can do the relevant operations without a brain in her head. However, we disagree whether what you are doing is set theory. It is something else - equally elementary and fundamental as set theory - but it is NOT set theory IMHO. To avoid confusion, I would like to suggest that you use different terminology for what you are doing. I am not aware of any standard mathematical terminology, so we will invent some. Here is the suggested new terminology and the correspondence with the set theory terminology that you have been using. set => bag element => token union => merger intersection => alignment A set is a collection of distinguishable elements. A bag is a collection of indistinguishable tokens. (At least they are indistinguishable to the bag operations of merger and alignment as described below). An element may be in several sets. A token can be in only one bag (one at a time). Sets live outside time - they do not change in any way when new sets are defined by intersection or union. Bags live in time - they can be created or destroyed, and the contents may change. The union of two sets is the collection of any elements that were in either of the two sets. The merger of two bags is the creation of a new bag containing all of the tokens from the two bags, along with the destruction of the two bags themselves. The intersection of two sets is the collection of elements that are in both of the two bags. The alignment of two bags is a one-to-one matching of the elements of the two bags that makes it possible to determine which of the two bags had the most elements. Neither of the two bags being compared is changed in any way by alignment. There is an additional operation called "tokenizing" which consists of creating a new bag whose token count exactly matches the number of elements in a set. For example, if I have a set of goats, I can tokenize it to create a bag of tokens that matches my set of goats. The operations of alignment and tokenizing may seem similar to counting, but they are simpler in one respect: To count, I need to know the names of a lot of numbers - one two three ... But to do tokenizing or alignment, I don't need to know the names of numbers - all I have to do is match one-to-one. I promised, at one point, that once I understood what a selective event was, I would tell you how to describe it using set theory. Well, I hope you won't hold me to that promise. I am sure that one can model bag theory in set theory, but it won't be pretty. Bag theory is simple enough on its own - trying to model it in set theory will only make it ugly. I'm pretty sure that AW and BOH will also understand your explanation here, if you point it out to them. All that remains is to use bag theory to explain what you are talking about when you use the words "selective event". Thanks John.

> JE:- The proposition I am proposing is just very simple. I have a minimum of two separate totals > to compare. That is 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 ones (stones). When 3 ones join they form one set of 3 called set A. When, in a > different place, 4 ones 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 doesn'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 > 1) Both sets remain intact. > 2) 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. > > Note 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? JM:- I now (for the first time) understand you. I agree that what you are trying to do is perfectly simple math, and that Nature can do the relevant operations without a brain in her head. However, we disagree whether what you are doing is set theory. It is something else - equally elementary and fundamental as set theory - but it is NOT set theory IMHO. JE:- ______________________________ To create each total that represents a single separate set, all I needed to do was add up. To compare the two totals with each other, I only needed to subtract one total from the other. It seems absurd to me for anybody to claim that just adding up to create two separate totals and then subtracting them cannot _validly_ and _easily_ be represented using _standard_ set theory, i.e. just using Venn Circles. _____________________________ Why can't a Venn circle represent a bag of stones, where each set element within each circle is a token, i.e. a number that can represent _anything_ that is defined to be countable and within a set? Why can't the contents of the two bags of stones being poured into a pot be a valid union of two Venn Circles and all their respective elements? Why can't the comparison of the two lines of stones be validly represented as the overlapping area between two Venn circles, i.e. the intersect of two Venn circles? Another illustration: If I represent two separate Venn circles A and B, _both_ with 4 set elements each such that it is defined as valid to join or intersect them totally, how would you represent: 1) The set union of A and B, 2) The total set intersection of A and B, just using a simple Venn circle illustration? Here are my solutions: 3) The set union of A and B: One circle with 8 elements. 4) The set intersection of A and B: Two overlapping circles with the Venn circle signature: ,4,0 which means: = zero elements in set A 4 = 4 elements in the intersection = zero elements in set B Proving: both sets are exactly the same size. I admit that my representation of the simple comparison of two totals may be incorrect. However I can see no other way to represent this event as a valid illustration of standard set theory. Obviously, I will never agree that my proposal _cannot_ be represented using a standard form of set theory. The proof that it must be able to be so represented, sits before everybody. _______________________________________________ Two totals _can_ be made and they _can_ be compared using simple arithmetic so this event _must_ be able to be represented using just a _standard_ form of set theory. So rather obviously, I am _forced_ to repeat my previous request, again. Would any reader PLEASE provide a corrected set theory configuration that validly represents the following simple event: 5) Two separate totals. 6) Their comparison. _____________________________________________ >snip explanation that I did not understand< JM:- I promised, at one point, that once I understood what a selective event was, I would tell you how to describe it using set theory. Well, I hope you won't hold me to that promise. JE:- I have no other choice but to hold you to your promise. JM:- I am sure that one can model bag theory in set theory, but it won't be pretty. Bag theory is simple enough on its own - trying to model it in set theory will only make it ugly. JE:- Lets find out. JM:- I'm pretty sure that AW and BOH will also understand your explanation here, if you point it out to them. JE:- If either chooses to respond and attempt to illustrate such a simple event using standard set theory, I would be most grateful! My guess is that neither will respond. Respectfully Yours, John Edser Independent Researcher PO Box 266 Church Pt NSW 2105 Australia [email protected]

"John Edser" <[email protected]> wrote in message news:<[email protected]>... > [snip] I admit that my representation of the simple comparison of two totals may be incorrect. > However I can see no other way to represent this event as a valid illustration of standard set > theory. Obviously, I will never agree that my proposal _cannot_ be represented using a standard > form of set theory. The proof that it must be able to be so represented, sits before everybody. > _______________________________________________ > Two totals _can_ be made and they _can_ be compared using simple arithmetic so this event _must_ > be able to be represented using just a _standard_ form of set theory. So rather obviously, I am > _forced_ to repeat my previous request, again. Would any reader PLEASE provide a corrected set > theory configuration that validly represents the following simple event: > 1) Two separate totals. > 2) Their comparison. > _____________________________________________ > > >snip explanation that I did not understand< JM:- Reread it. I am sure you will understand it if you are willing to give up your incorrect preconceptions about set theory. JE:- I'm sorry, it made no sense to me. JM:- I am not saying that the simple events you describe cannot be usefully visualized using diagrams that look a lot like Venn diagrams. I am simply saying that the things that those Venn circles will be representing will not be sets, as everyone else understands the term "set". JE:- OK Lets assume I am wrong. > JM:- I promised, at one point, that once I understood what a selective event was, I would tell you > how to describe it using set theory. Well, I hope you won't hold me to that promise. > JE:- I have no other choice but to hold you to your promise. > JM:- I am sure that one can model bag theory in set theory, but it won't be pretty. Bag theory is > simple enough on its own - trying to model it in set theory will only make it ugly. > JE:- Lets find out. JM:- OK. But the prerequisite was that I understand what a selective event is, in your formulation. In my earlier posting, which you clipped, I provided a fairly long list of the parts that I understood, the parts that I think I understood, and the parts that I must not understand at all because they make no sense to me. Also, as AW points out, one of the supposed benefits of using your Venn circles to visualize the process is that the correct "orders of selection" are somehow automatic. I particularly want to understand this. It seems to have something to do with your arguments against group selection. If so, I definitely don't want to lose this feature of your thinking in my promised attempt to map Edser "set theory" to the set theory that everyone else uses. JE:- OK. I am suggesting that Darwinian selection is just the default comparison of a minimum of two totals. -------------------------------------------- How do you represent the default comparison of two totals using standard set theory? set A = 10 Set B = 7 We know that set A is bigger than set B. How is this proven using set theory? --------------------------------------------- > JM:- I'm pretty sure that AW and BOH will also understand your explanation here, if you point it > out to them. > JE:- If either chooses to respond and attempt to illustrate such a simple event using standard set > theory, I would be most grateful! My guess is that neither will respond. JM:- You seem to be a pretty skillful guesser! JE:- Isn't it obvious they are attempting to hide something? Respectfully, John Edser Independent Researcher PO Box 266 Church Pt NSW 2105 Australia [email protected]

"John Edser" <[email protected]> wrote > > JE:- If either chooses to respond and attempt to illustrate such a simple > > event using standard set theory, I would be most grateful! My guess is that > > neither will respond. > > JM:- You seem to be a pretty skillful guesser! > > JE:- Isn't it obvious they are attempting to hide something? http://tinyurl.com/yqyqs Jim