Cantor's Diagonal?
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From: zuhair on 7 Feb 2010 16:44 On Feb 7, 4:29 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Feb 7, 7:48 am, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > On Feb 4, 12:35 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > And as I said already: (quoting) > > > > "And, if you interpret elements of B as subset of N (by thinking of > > > the > > > elements of B as characteristic functions, and identifying the > > > characteristic function with the corresponding subset), then what is > > > the subset h? It is the set of those elements n of N such that n is > > > not an element of g(n) (that is, h(n)=1 if and only if g(n)[n]=0).. > > > That is, the diagonal number h is *exactly* *the* *same* as the > > > diagonal set you get in the proof of Cantor's Theorem that any > > > function g:X-->P(X) is not surjective. " > > > That is still not clear to me. If anybody can further clarify that, it > > would be of great help. > > > Let me try: > > > first let me begin with the characteristic function: > > > The definition given above is: > > > If A is a subset of N, then > > Chi_A:N-->{0,1} is the characteristic function of A, where Chi_A(n) = > > 1 if n in A, and Chi_A(n)=0 if n is not in A. > > > Let A be the set of all even numbers, so A={ 2n| n e N } > > > What is the characteristic function of A? > > > Chi_A= {<0,1>,<1,0>,<2,1>,<3,0>,....} > > > Now lets replace each subset A of N > > You are already well on your way to confusing yourself. A was the set > of even numbers. Now A is an arbitrary variable standing for an > arbitrary subset of N. Which is it? > > > > > by its characteristic function > > Chi_A, and then apply the apply the general argument to it. > > > The diagonal of the general argument is defined by > > > For every f:N->P(N) > > > D_f={x | x e N & ~ x e f(x)} > > > Now lets replace P(N) by the set of all characteristic functions of > > subsets of N > > > let f:N --> {Chi_A| A subset of N} > > > Now each characteristic function is a set of ordered pairs of elements > > of N, while elements of N on the other hand are not ordered pairs of > > its elements. > > This is irrelevant to the argument. > > > so N and {Chi_A| A subset of N} are disjoint sets. > > Irrelevant. > > > Now lets apply the diagonal of the general argument to f > > > we'll have > > > For all f: D_f = N. > > No. When you apply the diagonal argument to characteristic function > (which are function from N to {0,1}, you are to get a characteristic > function. The characteristic function you get is > > D_f = { <n, 1- f(n)[n]> : n in N} The function whose value at n is 1 > if f(n)[n] is 0, and 0 if f(n)[n]=1. Remember? That's what the > argument for binary sequences was, and that is what {Chi_A|A subset of > N} is: it is the set of all binary sequences. > > > and as we know N is not a member of {Chi_A| A subset of N} > > You are confused. You aren't "translating", you are trying to feed > English words to a Russian dictionary. You are supposed to TRANSLATE > the argument, not just the inputs. > > > > > Now although N is not in the range of f, but N is not in the co-domain > > of f, > > so we cannot conclude that every function f such that > > f:N --> {Chi_A| A subset of N} is not surjective. > > > This mean that replacing each subset N with its characteristic > > function > > and applying the general argument, would not result in any proof of > > N being strictly smaller than the set of all characteristic functions > > of subsets of N. > > > This mean that the two arguments are not the same arguments. > > It means you (i) didn't understand and (ii) don't know what you are > doing and (iii) are still clueless. The arguments, however, are the > same argument. > > See: if you are going to apply the argument to binary N-functions, > then you use the definition applicable to binary N-functions. If you > are going to apply it to subsets, you use the definition that applies > to subsets. Instead, you are trying to apply the definition for > subsets to the binary N-function, and then complain that it doesn't > make sense. It is exactly as if you were trying to translate a book by > only translating every other word instead of the entire sentence, and > then claimed that translation was impossible because the result didn't > make sense. > > > so attempt of translating each arguments into the other fails > > bidirectionally. > > Your *flawed* attempt failed, because it was wrongheaded. Purposely > so, I suspect, because I find it hard to believe that *anyone* could > be so clueless as you have presented yourself to be for so long. > > > However there is another way of looking at matters, we can translate > > from maps to subsets of N using these characteristic functions *after* > > diagonalization has been made, and not prior to diagonalization as in > > the attempts above. > > > So from the general argument translate each subset A of N to its > > characteristic function i.e. to Chi_A , and we also translate the > > Diagonal D_f={x:xeN & ~ x e f(n)} > > to its characteristic function Chi_(D_f), and then we prove that > > Chi_(D_f) is the same diagonal we get from the Diagonal argument, > > which must be the case. > > You think? Perhaps translating the entire thing instead of just > piecemeal might yield a translation that actually makes sense? Miracle > of miracles! > > > But the problem is that this is not always the case, for example using > > the general argument we can have the empty set as the diagonal of each > > function f:N-->P(N) > > were n e f(n) for every n e N. So the Diagonal of these functions > > would be the empty set. > > > Now what would be the Characteristic function of { }, it must be > > {<{ }, 1> }, but > > this set is not a map from N to {0,1}. > > No, silly. > > Again: if A is *any* subset, then Chi_A(n) = 0 if n not in A, and 1 if > n is in A. > > So if A is the empty set, then Chi_A(n) =0 for all n. > > So the characteristic function of the empty set AS A SUBSET OF N is > > { <n,0> : n in N}. > > Characteristic functions of SUBSETS don't depend only on the set you > are looking at, they depend on the ambient set. They are defined on > the AMBIENT set, and are defined in terms of the subspace. In this > argument, you are looking at the characteristic functions of the > subsets of X, so they always have domain X. The "characteristic > function of the empty set" AS A SUBSET OF N is different from the > characteristic function of the empty set as a subset of any set Y with > Y=/=N. > > > Now lets go to the other direction, i.e. lets attempt to translate > > from > > the Diagonal argument to the General argument. > > > So we'll translate each map and the diagonal as well, to its > > corresponding > > subset of N using inverse characteristic functions, now the diagonal > > would be > > the diagonal of argument 1. > > > So it seems from the above, unless i am mistaken (which might be the > > most likely case), that the Diagonal argument is a sub-argument of > > the general argument. > > The "diagonal argument" IS the general argument. They are the same > argument, if you aren't purposely dishonest or obtuse. No there is no purposeful dishonesty as you suspect or anything like that, I will look to your responses above an try to understand them step by step and report back my doubts if there is any, If I see that you managed to prove your claim, then I will confess it, as I confessed that the diagonal argument do prove that the set N is strictly subnumerous to the set of binary maps without any well ordering needed, and also I confess that I had the wrong impression about this argument, as I confessed that your statement that I was confusing the illustration for the argument itself, as I confessed that exactly the same argument can be generalized to any set X. In the same line if I am convinced by your illustrations above, I will state it. There is no dishonesty here. > > -- > Arturo Magidin
From: Arturo Magidin on 7 Feb 2010 16:59 On Feb 7, 3:44 pm, zuhair <zaljo...(a)gmail.com> wrote: > > The "diagonal argument" IS the general argument. They are the same > > argument, if you aren't purposely dishonest or obtuse. > > No there is no purposeful dishonesty as you suspect or anything like > that, I will look to your responses above an try to understand them > step by step and report back my doubts if there is any, If I see that > you managed to prove your claim, To *your satisfaction*. The "claim" (in fact a trivial observation) has been established long before me and it is clear and patently obvious to plenty of people in this thread alone. The problem is that *you* don't get it, so kindly put the approrpiate qualifier there. It's not whether or not I can "prove [my] claim", it is whether or not I manage to hammer it through *your* thick skull. -- Arturo Magidin
From: zuhair on 7 Feb 2010 17:01 On Feb 7, 4:29 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Feb 7, 7:48 am, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > On Feb 4, 12:35 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > And as I said already: (quoting) > > > > "And, if you interpret elements of B as subset of N (by thinking of > > > the > > > elements of B as characteristic functions, and identifying the > > > characteristic function with the corresponding subset), then what is > > > the subset h? It is the set of those elements n of N such that n is > > > not an element of g(n) (that is, h(n)=1 if and only if g(n)[n]=0).. > > > That is, the diagonal number h is *exactly* *the* *same* as the > > > diagonal set you get in the proof of Cantor's Theorem that any > > > function g:X-->P(X) is not surjective. " > > > That is still not clear to me. If anybody can further clarify that, it > > would be of great help. > > > Let me try: > > > first let me begin with the characteristic function: > > > The definition given above is: > > > If A is a subset of N, then > > Chi_A:N-->{0,1} is the characteristic function of A, where Chi_A(n) = > > 1 if n in A, and Chi_A(n)=0 if n is not in A. > > > Let A be the set of all even numbers, so A={ 2n| n e N } > > > What is the characteristic function of A? > > > Chi_A= {<0,1>,<1,0>,<2,1>,<3,0>,....} > > > Now lets replace each subset A of N > > You are already well on your way to confusing yourself. A was the set > of even numbers. Now A is an arbitrary variable standing for an > arbitrary subset of N. Which is it? > > > > > by its characteristic function > > Chi_A, and then apply the apply the general argument to it. > > > The diagonal of the general argument is defined by > > > For every f:N->P(N) > > > D_f={x | x e N & ~ x e f(x)} > > > Now lets replace P(N) by the set of all characteristic functions of > > subsets of N > > > let f:N --> {Chi_A| A subset of N} > > > Now each characteristic function is a set of ordered pairs of elements > > of N, while elements of N on the other hand are not ordered pairs of > > its elements. > > This is irrelevant to the argument. > > > so N and {Chi_A| A subset of N} are disjoint sets. > > Irrelevant. > > > Now lets apply the diagonal of the general argument to f > > > we'll have > > > For all f: D_f = N. > > No. When you apply the diagonal argument to characteristic function > (which are function from N to {0,1}, you are to get a characteristic > function. The characteristic function you get is > > D_f = { <n, 1- f(n)[n]> : n in N} The function whose value at n is 1 > if f(n)[n] is 0, and 0 if f(n)[n]=1. Remember? That's what the > argument for binary sequences was, and that is what {Chi_A|A subset of > N} is: it is the set of all binary sequences. > > > and as we know N is not a member of {Chi_A| A subset of N} > > You are confused. You aren't "translating", you are trying to feed > English words to a Russian dictionary. You are supposed to TRANSLATE > the argument, not just the inputs. > > > > > Now although N is not in the range of f, but N is not in the co-domain > > of f, > > so we cannot conclude that every function f such that > > f:N --> {Chi_A| A subset of N} is not surjective. > > > This mean that replacing each subset N with its characteristic > > function > > and applying the general argument, would not result in any proof of > > N being strictly smaller than the set of all characteristic functions > > of subsets of N. > > > This mean that the two arguments are not the same arguments. > > It means you (i) didn't understand and (ii) don't know what you are > doing and (iii) are still clueless. The arguments, however, are the > same argument. > > See: if you are going to apply the argument to binary N-functions, > then you use the definition applicable to binary N-functions. If you > are going to apply it to subsets, you use the definition that applies > to subsets. Instead, you are trying to apply the definition for > subsets to the binary N-function, and then complain that it doesn't > make sense. It is exactly as if you were trying to translate a book by > only translating every other word instead of the entire sentence, and > then claimed that translation was impossible because the result didn't > make sense. > > > so attempt of translating each arguments into the other fails > > bidirectionally. > > Your *flawed* attempt failed, because it was wrongheaded. Purposely > so, I suspect, because I find it hard to believe that *anyone* could > be so clueless as you have presented yourself to be for so long. > > > However there is another way of looking at matters, we can translate > > from maps to subsets of N using these characteristic functions *after* > > diagonalization has been made, and not prior to diagonalization as in > > the attempts above. > > > So from the general argument translate each subset A of N to its > > characteristic function i.e. to Chi_A , and we also translate the > > Diagonal D_f={x:xeN & ~ x e f(n)} > > to its characteristic function Chi_(D_f), and then we prove that > > Chi_(D_f) is the same diagonal we get from the Diagonal argument, > > which must be the case. > > You think? Perhaps translating the entire thing instead of just > piecemeal might yield a translation that actually makes sense? Miracle > of miracles! Well sometimes, yes I do! > > > But the problem is that this is not always the case, for example using > > the general argument we can have the empty set as the diagonal of each > > function f:N-->P(N) > > were n e f(n) for every n e N. So the Diagonal of these functions > > would be the empty set. > > > Now what would be the Characteristic function of { }, it must be > > {<{ }, 1> }, but > > this set is not a map from N to {0,1}. > > No, silly. Thanks for the complement. > > Again: if A is *any* subset, then Chi_A(n) = 0 if n not in A, and 1 if > n is in A. > > So if A is the empty set, then Chi_A(n) =0 for all n. > > So the characteristic function of the empty set AS A SUBSET OF N is > > { <n,0> : n in N}. yea, I missed that, that was my error, I got your point now. > > Characteristic functions of SUBSETS don't depend only on the set you > are looking at, they depend on the ambient set. They are defined on > the AMBIENT set, and are defined in terms of the subspace. In this > argument, you are looking at the characteristic functions of the > subsets of X, so they always have domain X. The "characteristic > function of the empty set" AS A SUBSET OF N is different from the > characteristic function of the empty set as a subset of any set Y with > Y=/=N. Ah, yes, true, the arguments really translates to each other, you were right as I suspected. > > > Now lets go to the other direction, i.e. lets attempt to translate > > from > > the Diagonal argument to the General argument. > > > So we'll translate each map and the diagonal as well, to its > > corresponding > > subset of N using inverse characteristic functions, now the diagonal > > would be > > the diagonal of argument 1. > > > So it seems from the above, unless i am mistaken (which might be the > > most likely case), that the Diagonal argument is a sub-argument of > > the general argument. > > The "diagonal argument" IS the general argument. They are the same > argument, if you aren't purposely dishonest or obtuse. You were right, I was wrong. > > -- > Arturo Magidin
From: zuhair on 7 Feb 2010 17:04 On Feb 7, 4:59 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Feb 7, 3:44 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > The "diagonal argument" IS the general argument. They are the same > > > argument, if you aren't purposely dishonest or obtuse. > > > No there is no purposeful dishonesty as you suspect or anything like > > that, I will look to your responses above an try to understand them > > step by step and report back my doubts if there is any, If I see that > > you managed to prove your claim, > > To *your satisfaction*. The "claim" (in fact a trivial observation) > has been established long before me and it is clear and patently > obvious to plenty of people in this thread alone. The problem is that > *you* don't get it, so kindly put the approrpiate qualifier there. > It's not whether or not I can "prove [my] claim", it is whether or not > I manage to hammer it through *your* thick skull. Yea my skull is indeed think, well it is useful, it protect my brain from dangerous hammering. > > -- > Arturo Magidin
From: Arturo Magidin on 7 Feb 2010 17:05
On Feb 7, 4:01 pm, zuhair <zaljo...(a)gmail.com> wrote: > > No, silly. > > Thanks for the complement. You get complements from the axioms of ZF, not from me. To get *compliments*, you're going to have to do better. > Ah, yes, true, the arguments really translates to each other, you were > right as I suspected. "As I suspected"? Pull the other one, it's got bells on. Half a dozen times you pontificated on how wrong and "unhumble" I had been with that claim. Do you believe yourself when you say things like that? -- Arturo Magidin |