Cantor's Diagonal?
in [Math]
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From: FF on 7 Feb 2010 14:48 On 7 Feb 2010 18:06:51 GMT, Ilmari Karonen wrote: >> Exist y (P(y)) -> For all A exist x for all y ( y e x <-> ( y e A & >> P(y) ) ) >> > Um... wait. Fix some (non-empty, non-all-inclusive) set B, and let > P(y) <=> y /e B (where /e means "is not a member of"). Then, in your > modified axiom of separation for this P, what is x when A = B? So maybe it should read For all A: Ey(y e A & P(y)) -> ExAy(y e x <-> y e A & P(y)) ? FF
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
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 |