Cantor's Diagonal?
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From: zuhair on 7 Feb 2010 14:37 On Feb 7, 2:34 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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 Sorry Arturo, as regards your statement of the two arguments being *EXACTLY* the same, you and Mike are both only saying it on personal grounds, there is no formal prove of this claim, so you have no grounds to stand on, I thought that there is something more to that matter. Zuhair
From: Arturo Magidin on 7 Feb 2010 14:45 On Feb 7, 1:32 pm, zuhair <zaljo...(a)gmail.com> wrote: > 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. It's not a question of "interpretation", it's a question of applying the correspondence. > But that only proves my stand point that they are not EXACTLY > the same argument No , it proves you are still clueless. The arguments (not the proofs, the ARGUMENTS) are exactly the same in both cases. > as Arturo way saying, they are SIMILAR, but not > EXACTLY the same, They are exactly the same. > you only managed to prove my point actually, You are too incompetent to realize that your "point" was utterly demolished. > EXACTLY > is EXACTLY, it means they can be translated to each other in both > directions, Which is EXACTLY the case. If you translate the arguments using the correspondence between P(N) and {0,1}^N, the arguments are translated into one another. > which is not the case, Actually, it is EXACTLY 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, Yes, they are. Translate the construction from one to the other using the correspondence, and at each step you are doing the exact same thing. > sorry Mike, in that you > are wrong here, and only insisting against the obvious answer, sorry. Arrogance and incompetence, all in the same package. Very efficient of you. Again: function N-->{0,1}^N correspond to subsets of N via viewing them as characteristic function; subsets of N correspond to function N-->{0,1} via characteristic functions. 1. Let f:N--> {0,1}^N be any function 1'. Let f:N-->P(N) be any function. 2. Define g:N-->{0,1} as follows: 2'. Define G in P(N) as follows. 3. g(n) = 1 if f(n)[n] = 0, and g(n)=0 if f(n)[n]=1. 3' n is in G if n if not in f(n), and n is not in G if n is not in f(n). 4. Then g is a function from N to {0,1}. 4'. Then G is a subset of N. 5. For any n in N, g =/= f(n) since f(n)[n] =/= g(n). 5'. For any n in N, G=/=f(n), since the truth value of "n in G" is different from the truth value of "n in f(n)". 6. Therefore, f is not onto. 6' Therefore, f is not onto. 7. QED 7' QED The "translations" go either way, and they correspond EXACTLY. It is EXACTLY THE SAME ARGUMENT. -- Arturo Magidin
From: zuhair on 7 Feb 2010 14:48 On Feb 7, 2:37 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Feb 7, 1:34 pm, zuhair <zaljo...(a)gmail.com> wrote: > > > > > > > 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). > > > On second look, I see that you've mentioned Mike's reply, I overlooked > > that, > > Which of course, means that your claims that I was "running away" or > being dishonest were... dishonest. > > 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, > > If you are too dumb to know the difference between "argument" and > "proof", as seems to be the case, then of course you still have > trouble. Small wonder that half a decade later, you are still just as > ignorant and as confused as you were when you started. > > > it seem as Mike's said, the grounds are only > > personal. > > No, they aren't "personal". But they do require some honest effort on > your part, which of course means they are worthless to you, since you > are incapable of either honesty or effort. > > -- > Arturo Magidin If there is no formal proof of identity of arguments as Mike mentioned in his reply, and it seems that you agree with him on that, then why one would use such big words like *EXACTLY* the same, or *Identical*, why not use more humble words like *SIMILAR* for example, that would be more appropriate, when one use a terminology like *Exactly* the same, or *identical*, one would expect the existence of a rigorous formal proof of this claim, and not something that is an open ground to personal interpretation, or something that needs some kind of *insight* that might vary among people? better use more humble wording if the matter is so. Zuhair
From: Arturo Magidin on 7 Feb 2010 14:48 On Feb 7, 1:37 pm, zuhair <zaljo...(a)gmail.com> wrote: > On Feb 7, 2:34 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > 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 > > Sorry Arturo, as regards your statement of the two arguments being > *EXACTLY* the same, you and Mike are both only saying it on personal > grounds, > there is no formal prove of this claim, There is no formal proof that you can thin yourself out of a paper bag, either. > so you have no grounds to > stand on, I have plenty of ground to stand on. The fact that YOU DON'T GET IT doen't make it disappear, it just makes you just as dumb as you were when you started lo so many years ago. > I thought that there is something more to that matter. That would require you to actually bother to think, and it is clear you are still unwilling to do so. -- Arturo Magidin
From: Arturo Magidin on 7 Feb 2010 14:51
On Feb 7, 1:48 pm, zuhair <zaljo...(a)gmail.com> wrote: > If there is no formal proof of identity of arguments as Mike mentioned > in his reply, and it seems that you agree with him on that, then why > one would use such big words like *EXACTLY* the same, or *Identical*, > why not use more humble words like *SIMILAR* for example, that would > be more appropriate, when one use a terminology like *Exactly* the > same, or *identical*, one would expect the existence of a rigorous > formal proof of this claim, and not something that is an open ground > to personal interpretation, or something that needs some kind of > *insight* that might vary among people? They are EXACTLY THE SAME ARGUMENT. The only requirement is that you apply the correspondence between subsets and binary sequences. If you think of subsets of N as being "the same as" binary sequences, then the arguments translate exactly to one another line by line. They are, in fact, EXACTLY THE SAME. It is not a matter of humbleness, it is a question of your continued obtuseness. > better use more humble wording if the matter is so. Better try actually engaging your BRAIN next time, dum-dum. -- Arturo Magidin |