Cantor's Diagonal?
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From: FF on 7 Feb 2010 14:52 On Sun, 7 Feb 2010 10:13:49 -0800 (PST), FredJeffries wrote: >>> Well, I *think* that Frege didn't believe in the empty set. >>> >> Nope. Actually, he was one of the first to work with it. >> > Didn't R L Moore espouse the non-existence of the empty set? Kanamori: The empty set, the singleton, and the ordered pair (2003) http://www.math.ucla.edu/~asl/bsl/0903/0903-001.ps FF
From: zuhair on 7 Feb 2010 15:10 On Feb 7, 2:51 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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. No that is not true, I tried, they are not the same, they don't translate line by line as you said, you are making a claim without any evidence, just talk. Zuhair > > > better use more humble wording if the matter is so. > > Better try actually engaging your BRAIN next time, dum-dum. > > -- > Arturo Magidin
From: Arturo Magidin on 7 Feb 2010 15:30 On Feb 7, 2:10 pm, zuhair <zaljo...(a)gmail.com> wrote: > On Feb 7, 2:51 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > 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. > > No that is not true, I tried, Your attempt was to apply one argument direclty to the wrong function. Your attempt was flawed due to your obtuse misunderstanding. >they are not the same, they don't > translate line by line as you said, you are making a claim without any > evidence, just talk. I posted the translation, dum-dum. The only "just talking" is you. You should know by now not to lie. Let T be the correspondence from binary N-sequences to subsets of N. Given a binary N-sequence (a function f:N-->{0,1}), T(f) is the set T(f) = {n in N : f(n) = 1}. Let S be the inverse correspondence, from subsets of N to binary N- sequences. Given a subset A of N, S(A) is the function binary N- sequences which I denoted Chi_A, S(A)[n] = 1 if and only if n in A. The argument for sequences: 1. Let f:N-->{0,1}^N. 2. Define g:N-->{0,1} by g(n) = 0 if and only if f(n)[n] = 1. 3. For all n, g=/=f(n), since g(n) =/= f(n)[n]. 4. Therefore, f is not onto. QED The argument for subsets: 1'. Let F:N-->P(N). 2'. Define G in P(N) by n not in G if and only if n in F(n). 3'. For all n, G=/=F(n), since "n in G" has the opposite truth value as "n in F(n)". 4'. Therefore, F is not onto. QED. Translation from N-sequences to subsets: set F(n) = T(f(n)). Set G = T(g). Then: 1T. Let f:N-{0,1}^N; this defines F:N-->P(N). This is 1'. So 1' = 1T. 2T. n not in T(g) if and only if [now using the translation] g(n) = 0 if and only if f(n)[n]=1 if and only if n in T(f(n)) n in F(n). What is T(g)? It is G from 2'. So 2T is 2'. 3T. For all n, T(g)=/= T(f(n)) for all n. Since T(g)=G, and T(f(n)) = F(N), 3T = 3' 4T. Therefore, f (and so F) is not surjective. This is 4'. Translation going the other way. given F:N-->P(N), this gives f:N-- >{0,1}^N via f(n)=S(F(n)). 1S Let F:N-->P(N). Let f:N--{0,1}^N be given by f(n) = S(F(n)). This is the same as 1. 2S. S(G):N-->{0,1} is defined by S(G)=0 if and only if S(F(n))[n]=1. That is S(G) is exactly the same as the function g from 2, since S(F(n))[n]=f(n)[n]. 3S For all n, S(G)(n) =/= S(F(n))[n], so S(G) =/= S(F(n)). Since S(G)=g and S(F(n)) = f(n), this is exactly the same as 3. 4S Thus, F (and therefore S(F)) is not onto. The functions you have and construct correspond EXACTLY ( not "similarly", not "sort of", not "subjectively", not "personally", but EXACTLY) to the subsets you have and construct. If you replace N with X throughout (considering binary X-sequences and subsets of X), you get the corresponding proofs for arbitrary sets X and either their power sets, or the set of all binary X-sequences. Lots of empty words, but they are all coming fom the empty head above your shoulders. -- Arturo Magidin
From: William Hughes on 7 Feb 2010 15:47 On Feb 7, 3:48 pm, zuhair <zaljo...(a)gmail.com> wrote: > when one use a terminology like *Exactly* the > same, or *identical*, one would expect the existence of a rigorous > formal proof of this claim, Nope, you would use the term "exactly" when the proofs are so similar, that the only way to deny they are the same is to be obtuse. In this case both arguments start with a mapping and lead to the same set which is not in the image of the mapping. The term exactly seems more than appropriate. -William Hughes
From: Arturo Magidin on 7 Feb 2010 16:29
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. -- Arturo Magidin |