Cantor's Diagonal?
in [Math]
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From: Arturo Magidin on 7 Feb 2010 14:15 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: > > > On Feb 4, 7:08 am, zuhair <zaljo...(a)gmail.com> wrote: > > > > On Feb 4, 1:04 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Feb 3, 11:55 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > Now I *SEE* this so called *Diagonal* argument. > > > > > Given your comments below, NO, you don't. You still don't understand > > > > it and you are just fooling yourself. As usual. > > > > > > It seems to me that a modification of this argument can actually work > > > > > for every well orderable set, however I don't know if a modification > > > > > of this argument can be made general enough to prove that the power of > > > > > every non well orderable set is bigger than it. > > > > > There is no "modification" needed. The argument IS EXACTLY THE SAME > > > > ARGUMENT that shows that no set is bijectable with its power set. > > > > > > The other proof > > > > > There is no "other" proof. THEY ARE THE SAME ARGUMENT. THEY ARE THE > > > > SAME PROOF. > > > > > > does > > > > > that for all sets, so it seems to be more general than the diagonal > > > > > argument, > > > > > THEY ARE THE EXACT SAME ARGUMENT. > > > > How in hell they are *exactly* the same argument, can you tell me. > > > I already did, you ignorant boffoon. > > > Binary sequences N-->{0,1} correspond to subsets, by identifying a > > subset with its characteristic function. 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. > > > 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. Mike Terry has done so. > > 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 by its characteristic function > Chi_A, and then apply the apply the general argument to it. You are being either disingenious, or willfully obtuse. As to your continued insistence that the argument "may fail in other set theories", you are being purposely dishonest. Naturally, a proof that works in theory T need not work in theory T'. As I pointed out, the diagonal argument can be "tweaked" (aka modified slightly) to work in set theories that do not have empty sets if you do two modifications: one is to modify the statement so that it only applies to sets that are not singletons; the second modification is that the argument needs to be 'tweaked' (suitably modified) so that the constructed function/set is not the zero function/emtpy. This is a process similar to the one used to avoid problems of dual representations when presenting the argument as proving no bijection exists between the natural numbers and the real numbers in the interval [0,1). -- Arturo Magidin
From: zuhair on 7 Feb 2010 14:20 > You are being either disingenious, or willfully obtuse. same to be said of you also. > > As to your continued insistence that the argument "may fail in other > set theories", you are being purposely dishonest. Naturally, a proof > that works in theory T need not work in theory T'. As I pointed out, > the diagonal argument can be "tweaked" (aka modified slightly) to work > in set theories that do not have empty sets if you do two > modifications: one is to modify the statement so that it only applies > to sets that are not singletons; the second modification is that the > argument needs to be 'tweaked' (suitably modified) so that the > constructed function/set is not the zero function/emtpy. This is a > process similar to the one used to avoid problems of dual > representations when presenting the argument as proving no bijection > exists between the natural numbers and the real numbers in the > interval [0,1). you willfully chose to speak about the less important part of my post, as I already acknowledged to Jesse, you didn't reply to the part of my questions about the argument in ZF itself, which is the part that matters here, and this is something that I also name as dishonesty, you are simply running away. > > -- > Arturo Magidin
From: zuhair on 7 Feb 2010 14:32 On Feb 7, 10:52 am, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "zuhair" <zaljo...(a)gmail.com> wrote in message > > news:2a51fa70-59c3-48a4-a694-608f8dcf5d67(a)m16g2000yqc.googlegroups.com... > > > > > > > On Feb 4, 12:35 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > On Feb 4, 7:08 am, zuhair <zaljo...(a)gmail.com> wrote: > > > > > On Feb 4, 1:04 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > > On Feb 3, 11:55 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > Now I *SEE* this so called *Diagonal* argument. > > > > > > Given your comments below, NO, you don't. You still don't understand > > > > > it and you are just fooling yourself. As usual. > > > > > > > It seems to me that a modification of this argument can actually > work > > > > > > for every well orderable set, however I don't know if a > modification > > > > > > of this argument can be made general enough to prove that the > power of > > > > > > every non well orderable set is bigger than it. > > > > > > There is no "modification" needed. The argument IS EXACTLY THE SAME > > > > > ARGUMENT that shows that no set is bijectable with its power set. > > > > > > > The other proof > > > > > > There is no "other" proof. THEY ARE THE SAME ARGUMENT. THEY ARE THE > > > > > SAME PROOF. > > > > > > > does > > > > > > that for all sets, so it seems to be more general than the > diagonal > > > > > > argument, > > > > > > THEY ARE THE EXACT SAME ARGUMENT. > > > > > How in hell they are *exactly* the same argument, can you tell me. > > > > I already did, you ignorant boffoon. > > > > Binary sequences N-->{0,1} correspond to subsets, by identifying a > > > subset with its characteristic function. 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. > > > > 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 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. > > > so N and {Chi_A| A subset of N} are disjoint sets. > > > Now lets apply the diagonal of the general argument to f > > > we'll have > > > For all f: D_f = N. > > I'm sure you are being *deliberately* obtuse here. > > The point that has been made is that: > 1) there is a natural bijection between P(N) and {0,1}^N > 2) the proof that P(N) is not countable (special case of X and P(X) not > being equinumerous) and the proof that {0,1}^N is not countable (by > considering binary sequences) use "exactly the same argument", in view of > the natural bijection. > > I.e. if we are given > f: N-->{0,1}^N > this "naturally" corresponds to a > g: N-->P(N) > and the diagonal argument (for showing X and P(X) are not equinumerous) > gives us a G in P(N) such that: > Ax in N: g(x) != G, so g is not surjective. > > The set G corresponds naturally to an F in {0,1}^N, and clearly we have > Ax in N: f(x) != F, showing that f is not surjective.. > > Furthermore, when you look at the Cantor proof that f is not surjective > (looking at binary sequences), you see that it is constructing the exact > same F as we would get above by mapping back and forth between P(N) and > {0,1}^N as required in the obvious manner. In fact, considering the > naturalness of the bijection between P(N) and {0,1}^N, most people would > have no trouble asserting that the proofs are using "exactly the same > argument". > > Anyone who understands the general powerset proof, and who is aware of the > natural bijection between P(N) and {0,1}^N, would have no difficulty proving > the binary-sequence proof by applying "the same argument". > > Note, this doesn't mean that the proofs are line by line identical - that's > misinterpreting the meaning of the phrase "exactly the same argument" - > argument and proof are not synonyms! > > So... you are (deliberately I suspect) misapplying the "general argument" to > f. You should not be applying the powerset-proof argument directly against > the ordered pairs making up the functions Chi_A etc., but rather, mapping > back and forth between P(N) and {0,1} as required to apply the argument.... > > > > [snip consequences of mistaken application of general argument] > > > > > This mean that the two arguments are not the same arguments. > > No it doesn't. > > It is true that whether the two arguments are "exactly the same" is a matter > of common sense interpretation, and a matter of having sufficient insight > into the structure of the two proofs to apply common sense. I.e. the > question is not purely a mathematical question, so nobody can prove to you > that the arguments are EXACTLY the same if you don't have the common sense / > insight to see it... > > Mike. That's not enough. You left it to personal interpretation which is variable.But that only proves my stand point that they are not EXACTLY the same argument as Arturo way saying, they are SIMILAR, but not EXACTLY the same, you only managed to prove my point actually, EXACTLY is EXACTLY, it means they can be translated to each other in both directions, which is not the case, of course it is clear that the two arguments are very similar to each other as if they are twin arguments, but they are not EXACTLY similar, sorry Mike, in that you are wrong here, and only insisting against the obvious answer, sorry. Zuhair
From: Arturo Magidin on 7 Feb 2010 14:34 On Feb 7, 1:20 pm, zuhair <zaljo...(a)gmail.com> wrote: > > You are being either disingenious, or willfully obtuse. > > same to be said of you also. Only if you want to project your dishonesty on others. > > As to your continued insistence that the argument "may fail in other > > set theories", you are being purposely dishonest. Naturally, a proof > > that works in theory T need not work in theory T'. As I pointed out, > > the diagonal argument can be "tweaked" (aka modified slightly) to work > > in set theories that do not have empty sets if you do two > > modifications: one is to modify the statement so that it only applies > > to sets that are not singletons; the second modification is that the > > argument needs to be 'tweaked' (suitably modified) so that the > > constructed function/set is not the zero function/emtpy. This is a > > process similar to the one used to avoid problems of dual > > representations when presenting the argument as proving no bijection > > exists between the natural numbers and the real numbers in the > > interval [0,1). > > you willfully chose to speak about the less important part of my post, The "important" part of your post was your willful misunderstanding, which was addressed by Mike Terry in detail. I pointed you to that response. > as I already acknowledged to Jesse, you didn't reply to the part of > my questions about the argument in ZF itself, which is the part that > matters here, and this is something that I also name as dishonesty, > you are simply running away. You are being dishonest and you are projecting your lies on others. As I said clearly: "Mike Terry has done so." You know: in that part that you DELETED before claiming I was not answering you. The only person running away is you, as always. You are not just a willful ignoramus, Zuhair, you are not just dishonest, you are also rather incompetent as a liar. -- Arturo Magidin
From: zuhair on 7 Feb 2010 14:34
On Feb 7, 2:15 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: > > > > On Feb 4, 7:08 am, zuhair <zaljo...(a)gmail.com> wrote: > > > > > On Feb 4, 1:04 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > > On Feb 3, 11:55 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > Now I *SEE* this so called *Diagonal* argument. > > > > > > Given your comments below, NO, you don't. You still don't understand > > > > > it and you are just fooling yourself. As usual. > > > > > > > It seems to me that a modification of this argument can actually work > > > > > > for every well orderable set, however I don't know if a modification > > > > > > of this argument can be made general enough to prove that the power of > > > > > > every non well orderable set is bigger than it. > > > > > > There is no "modification" needed. The argument IS EXACTLY THE SAME > > > > > ARGUMENT that shows that no set is bijectable with its power set. > > > > > > > The other proof > > > > > > There is no "other" proof. THEY ARE THE SAME ARGUMENT. THEY ARE THE > > > > > SAME PROOF. > > > > > > > does > > > > > > that for all sets, so it seems to be more general than the diagonal > > > > > > argument, > > > > > > THEY ARE THE EXACT SAME ARGUMENT. > > > > > How in hell they are *exactly* the same argument, can you tell me. > > > > I already did, you ignorant boffoon. > > > > Binary sequences N-->{0,1} correspond to subsets, by identifying a > > > subset with its characteristic function. 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. > > > > 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. > > Mike Terry has done so. > > > > > > > > > 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 by its characteristic function > > Chi_A, and then apply the apply the general argument to it. > > You are being either disingenious, or willfully obtuse. > > As to your continued insistence that the argument "may fail in other > set theories", you are being purposely dishonest. Naturally, a proof > that works in theory T need not work in theory T'. As I pointed out, > the diagonal argument can be "tweaked" (aka modified slightly) to work > in set theories that do not have empty sets if you do two > modifications: one is to modify the statement so that it only applies > to sets that are not singletons; the second modification is that the > argument needs to be 'tweaked' (suitably modified) so that the > constructed function/set is not the zero function/emtpy. This is a > process similar to the one used to avoid problems of dual > representations when presenting the argument as proving no bijection > exists between the natural numbers and the real numbers in the > interval [0,1). > > -- > Arturo Magidin On second look, I see that you've mentioned Mike's reply, I overlooked that, but still Mike's response is actually a negative one, sorry, I thought there is a formal way of proving to arguments to be EXACTLY the same, or IDENTICAL, it seem as Mike's said, the grounds are only personal. Zuhair |