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From: Robert J. Kolker on 19 Nov 2005 13:12 zuhair wrote: > > > So propositional functions are something not intelligible?huh! So what does P(x) = x. Proposition functions have as a range truth values. Bob Kolker
From: Robert Low on 19 Nov 2005 17:48 zuhair wrote: > Robert Low wrote: >>zuhair wrote: >>>Robert Low wrote: >>>>zuhair wrote: >>>>>Similar condition happens with >>>>>P(x) = Singlton x = {x} >>>>Assuming that this made sense, >>>>>Now P(P(x))= Singlton {x} = {{x}} which is also a singlton >>>>This would still be wrong. P({x})={ {}, {x} } >>>>which has two elements. >>>>In general, if a set A has n(>=0) elements, P(A) has 2^n >>>>elements. >>>You are confusing P(x) for power set of x , which is not what I am >>>refering to , here P means propositional function and not Power. >>Oh, I'm sorry. I thought you were writing something intelligible >>but incorrect. I see I was wrong about the first part of >>that. > So propositional functions are something not intelligible?huh! Oh, propositional functions are intelligible. It's what you write about them that isn't. > How ignorant you are . I'm fortunate enough to live a life of sheer bliss.
From: zuhair on 20 Nov 2005 03:35 Robert Low wrote: > zuhair wrote: > > Robert Low wrote: > >>zuhair wrote: > >>>Robert Low wrote: > >>>>zuhair wrote: > >>>>>Similar condition happens with > >>>>>P(x) = Singlton x = {x} > >>>>Assuming that this made sense, > >>>>>Now P(P(x))= Singlton {x} = {{x}} which is also a singlton > >>>>This would still be wrong. P({x})={ {}, {x} } > >>>>which has two elements. > >>>>In general, if a set A has n(>=0) elements, P(A) has 2^n > >>>>elements. > >>>You are confusing P(x) for power set of x , which is not what I am > >>>refering to , here P means propositional function and not Power. > >>Oh, I'm sorry. I thought you were writing something intelligible > >>but incorrect. I see I was wrong about the first part of > >>that. > > So propositional functions are something not intelligible?huh! > > Oh, propositional functions are intelligible. It's what > you write about them that isn't. > > > How ignorant you are . > > I'm fortunate enough to live a life of sheer bliss. Well let me confess, I am also ignorant of these functions and how they should be wrote. but from what I know when we say x is a man , x is a variable which can include any SUBJECT which has man as its PREDICATE and thus makes a correct proposition. I write that P(x)= man , which means which x has man as its predicate? the answers to that question will make the set MAN " Mn" so in reality Mn = { x| x=man } = P(x) So the predicate of every member in the set "P (x)", is what is common between all the set members , so it is the set itself, so:- Mn = P(x) Now P(x) = x , means which x has itself x as a predicate Now every thing can have itself as a predicate since any x is x. because everything is identical to itself and this is also implied in saying A is a subset of A. Therefor if we make a set U of all x that fulfilles P(x) = x U={ x| P(x)=x } ( P(x)=x means P(x)=x is True , but I don't wright "is True" because this is understood) Then U is the set of all sets. However as I said in reality P(x) = y is in reality the set itself. so for the above example U= P(x) Now P(x) itself is also identical to itself. So P( P(x) ) = P (x) is True. so it should be a member of U. but since P(x) = U Then U is a member of U. so U = { U } And this goes for ever since P ( P(x) ) = P ( P (x) ) which leads to U being a member of U being a member of U. or U = { { U } } At the end the result will be a kind of U =...........{{{{{ U }}}}}................ An unbounded set, of coarse such a set is a proper class since it cannot be contained in a bigger set than it. Now that U is in reality not identical to itself. so U is not a subset of U. In reality U is a proper subset of U , equivalently U is a proper superset of U. So U is uncatchable. Now what I wanted to say is that the natural numbers set can be formed in away similar to U above. see P(x) = {x} is a singlton set. now if x = 1,2 then it doesn't fulfill P above. only single things can be put inside two brackets to form a singlton set. But {x} itself also can be put between two brackets to form a singlton set { {x} }. so P ( P (x) ) = { {x} } is a singlton set. IS TRUE. Now we have the same as with U . so:- ................P(P(P(x))).......... = .................{{{x}}}.................. Now lets define N as the set above with x = { } so N = ...............P(P(P({}))).......... = ................{{{ }}}.................. Now this N is exactly the same as the set of all natural numbers. since {} = 0 { {} } = 0,1 { { {} } } =0,1,2 { { { {} } } } =0,1,2,3 .. .. .. .............................{{{{{ }}}}}....................= 0,1,2,3,4,................. = N so N is a set that cannot be identical to itself. or N is not the subset of itself. N is always a proper subset of itself or a proper superset of itself like U. accordingly no mathematical function f(n) can catch N, For example it is well known that any finite set of N , N1can be bijected to a disjoint finite set of N , N2 were the smallest member in N2 is bigger than the biggest member in N1. Now this phenomena which only occure in infinite sets was taken to proove that a set can be bijected to some proper subset of it, by using certain bijective functions f:A ->B , were f(x)=y, f is single valued so if every y in B also belong to A then card A = card B. But knowning that infinite sets like N cannot be catched ie the union of all finite sets of set N is not N. The the above function f(x) cannot generalize its bijective rule for all N, because N is always bigger than itself. Accordingly this reflexive definition of the infinite set by being able to be injected to some proper subset of it is erronous. Because although every singlton disjoint set of N has a singlton disjoint set of some proper subset of N , yet this cannot be generalized to say that N itself is bijected to that subset , because N is always having extra members so we cannot say N has equal extention as a proper subset of it. One can adopt a non debatable definition of infinite sets like below: A is called a well ordered infinite set, if A is well ordered and if for every finite subset A1 of A there is always a finite subset of A, "A2 " , that is disjoint from A1 , wereby min A2 > max A1, if the order of members of A is increasing AND max A2 < min A1 if the order of members of A is decreasing. I think that this definition is clear and it is better than the standard definition. I think that cardinality only tells us part of the story of the infinite, and the concept of subcardinality , I have wrote in a seperate topic of Sci.math : Cardinality and injection. should be took into consideration to complete the story. Zuhair
From: boink on 20 Nov 2005 06:25 On Sun, 20 Nov 2005 00:35:50 -0800, zuhair wrote: > > Zuhair Are you on drugs?
From: MoeBlee on 20 Nov 2005 06:52
zuhair wrote: > but from what I know when we say x is a man , x is a variable which > can include any SUBJECT which has man as its PREDICATE and thus > makes a correct proposition. > > I write that P(x)= man , which means which x has man as its predicate? No, you write Px <-> x is a man. > the answers to that question will make the set MAN " Mn" > > so in reality Mn = { x| x=man } = P(x) No, the set of men = {x| Px} > So the predicate of every member in the set "P (x)", is what is common > between all the set members , so it is the set itself, so:- > > Mn = P(x) First, why are you using two letters Mn for the set of men. Let's just use one letter, M. So M = {x| Px} > Now P(x) = x , means which x has itself x as a predicate No, Px = x is a very different thing. Now you're using P as a function (or operation) symbol rather than as a predicate symbol. Choose one or the other; it can't be both. And if Px = x, then P is just the identity function. > Now every thing can have itself as a predicate since any x is x. No, what you want to say is that for all x, x=x. In that sense, for all x, the 2-place predicate symbolized by the 2-place predicate symbol '=' holds for <x x>. > because everything is identical to itself Correct. > and this is also implied in saying A is a subset of A. No, A is a subset of A is implied by the validity of the propositional form P -> P, specifically in the instance of xeA -> xeA. > Therefor if we make a set U of all x that fulfilles P(x) = x Remember, as I mentioned, you switched P to a function symbol, not a predicate symbol. Anyway, every x satisfies Px = x, if P is the identity function. So U has everything as a member. But in Z set theory, we prove there is no such set that has everything as a member. > U={ x| P(x)=x } Yes, except in Z set theory we prove that U does not exist. > P(x)=x means P(x)=x is True , but I don't wright > "is True" Actually, it should be 'Px=x' is true iff Px=x. > Then U is the set of all sets. It would be if it existed. It exists in class theories but not in Z set theory. > However as I said in reality P(x) = y is in reality the set itself. I don't know what you are trying to say with that. > so for the above example U= P(x) No, U={x| Px=x}. > Now P(x) itself is also identical to itself. Everything is identical with itself. > So P( P(x) ) = P (x) is True. True. > so it should be a member of U. What should be a member of U? Px? Yes, for all x, Px would be a member of U, if U existed. > but since P(x) = U What? Where did that come from? Px is x. But x is not necessarily U. x being identical with itself doesn't imply that x is the set of everything. I'm identical with myself; That hardly makes me the set of all sets. > Then U is a member of U. so U = { U } If U existed, then U would be a member of U. And if xe{U} -> xeU, since xeU in any case. But offhand, I don't see how it follows that if xeU, then xe{U}, since that would entail that x=U, and xeU should not entail that x=U. But still, if you assume that U exists, then the theory is inconsistent, so we could prove anything anyway. > And this goes for ever since Well your utterly senseless post does at least SEEM to go on forever... Why are you trying to do this the hard way? Why don't you just read a book on mathematical logic? > P ( P(x) ) = P ( P (x) ) which leads to U > being a member > of U being a member of U. or U = { { U } } > > At the end the result will be a kind of U =...........{{{{{ U > }}}}}................ > > An unbounded set, of coarse such a set is a proper class since it > cannot > be contained in a bigger set than it. > > Now that U is in reality not identical to itself. > > so U is not a subset of U. > > In reality U is a proper subset of U , equivalently U is a proper > superset of U. > > So U is uncatchable. > > Now what I wanted to say is that the natural numbers set can be formed > in away > similar to U above. > > see P(x) = {x} is a singlton set. > > now if x = 1,2 then it doesn't fulfill P above. > > only single things can be put inside two brackets to form a singlton > set. > > But {x} itself also can be put between two brackets to form a singlton > set > { {x} }. > > so P ( P (x) ) = { {x} } is a singlton set. IS TRUE. > > Now we have the same as with U . so:- > > ...............P(P(P(x))).......... = > ................{{{x}}}.................. > > Now lets define N as the set above with x = { } > > so N = ...............P(P(P({}))).......... = ................{{{ > }}}.................. > > Now this N is exactly the same as the set of all natural numbers. > > since {} = 0 > > { {} } = 0,1 > > { { {} } } =0,1,2 > > { { { {} } } } =0,1,2,3 > . > . > . > > ............................{{{{{ }}}}}....................= > 0,1,2,3,4,................. = N > > so N is a set that cannot be identical to itself. > > or N is not the subset of itself. > > N is always a proper subset of itself or a proper superset of itself > like U. > > accordingly no mathematical function f(n) can catch N, > > For example it is well known that any finite set of N , N1can be > bijected > > to a disjoint finite set of N , N2 were the smallest member in N2 is > bigger than > > the biggest member in N1. > > Now this phenomena which only occure in infinite sets was taken to > proove > > that a set can be bijected to some proper subset of it, by using > certain > > bijective functions f:A ->B , were f(x)=y, f is single valued so if > every y in B also belong to A then card A = card B. > > But knowning that infinite sets like N cannot be catched > > ie the union of all finite sets of set N is not N. > > The the above function f(x) cannot generalize its bijective rule for > all N, because > > N is always bigger than itself. > > Accordingly this reflexive definition of the infinite set by being able > to be injected > > to some proper subset of it is erronous. > > Because although every singlton disjoint set of N has a singlton > disjoint set > of some proper subset of N , yet this cannot be generalized to say > that N itself is bijected to that subset , because N is always having > extra members > so we cannot say N has equal extention as a proper subset of it. > > One can adopt a non debatable definition of infinite sets like below: > > > A is called a well ordered infinite set, if A is well ordered and if > for every finite subset A1 of A there is always a finite subset of A, > "A2 " , that is disjoint from A1 , wereby > min A2 > max A1, if the order of members of A is increasing AND max A2 > < min A1 if the order of members of A is decreasing. > > > I think that this definition is clear and it is better than the > standard definition. > > I think that cardinality only tells us part of the story of the > infinite, and the concept > of subcardinality , I have wrote in a seperate topic of Sci.math : > Cardinality and injection. > should be took into consideration to complete the story. > > > Zuhair MoeBlee |