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From: Tony Orlow on 26 Oct 2005 10:10 Virgil said: > In article <MPG.1dc855247151905798a563(a)newsstand.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > David R Tribble said: > > > Virgil said: > > > >> If each subset of *N is to be represented by an infinite binary sequence > > > >> of digits with 1 in some position representing the presence of a member > > > >> *N and 0 representing its absence, then one element sets must be > > > >> represented by strings with one 1 in them. > > > > > > > > > > Tony Orlow wrote: > > > > This makes sense, but doesn't jibe with what you said before, as far as I > > > > can > > > > tell. Yes, each singleton set in P(*N) will map to a natural whose binary > > > > representation has a single bit. This only includes N out of 2^N subsets. > > > > > > > > > > Your scheme listed these mappings: > > > f(0) = {} > > > f(1) = {0} > > > f(2) = {1} > > > f(3) = {0,1} > > > f(4) = {2} > > > f(5) = {0,2} > > > f(6) = {1,2} > > > f(7) = {0,1,2} > > > f(8) = {3} > > > ... > > > > > > This list defines only the mappings of finite naturals in *N to > > > the finite subsets in P(*N). Your list does not show any infinite > > > naturals, nor does it show any infinite subsets, nor does it show any > > > subsets containing infinite naturals. > > I was asked about that and gave examples for the sets of odds and evens, > > which > > map to 2N/3 and N/3. There are other examples. It should suffice to say the > > same patter continues without end. > > As there is no "pattern" sufficient to go to the aleged end, it does not > suffice. > > If it did suffice, TO would be able to give the argument that is mapped > into the set of all elements not mapped into their images. The set of all elements not mapped to sets which contain them is the set of all elements, since none map to sets which contain them. The natural which maps to it is represented by an unending string of bits. It never ends. There is one for every natural. For every natural, there is one after it. For every bit, there is one after it. This is the number which maps to the entire set. :D > > > > > > Your list shows an infinite number of finite naturals mapping to an > > > equally infinite set of finite subsets. That's easy. > > > > There is no infinite set of finite naturals, as Albrecht has tried > > valiantly to illustrate. > > The set of finite naturals is Dedekind infinite, which is quite infinite > enough outside TOmatics. Sufficiently imponderable? Good enough. > > > > > > But it's not > > > a bijection between *N and P(*N). It's not a bijection between > > > *N and P(N), or even between N and P(N). It is a bijection between > > > N and some of the members of P(N). Obviously that's not good enough. > > > That's not what it is. I have declared it to be between *N and P(*N). > > To declare something does not make it so, at least outside of TOmatics. It's my bijection, after all. I can declare it to be whatever I wish. Go ahead and rain on my parade, but don't move the police barriers. > > Outside of TOmatics, declarations require proofs before they need be > accepted. Not when you are defining a bijection. What do you prove when you say the evens are mapped with f(x)=2x? Buzz off! > > > It includes such numbers in *N as 0:01.....0101 for the evens and > > 0:1010...1010 for the odds. You have no reason to think this set is > > limited to finite values. > And we have o reason to think it a bijection until TO can find some > member of *N which maps to the set of all members of *N not in the set > they map to. ..........................1111111111111111111111111111111111111111!!!!!!!!!!!!!! > > > > > Even if we were to grant you that the list somehow includes infinite > > > naturals in *N mapping to subsets, you would still be missing a lot > > > of subsets in P(*N). > > > Like which, for instance? > > The set of all members of *N not in the set they map to. That's the entire set. See above. Any more? > > > > > > > So fill in the blanks, and show us those other mappings. Otherwise > > > your "bijection" is incomplete and is not, in fact, a bijection > > > between *N and P(*N). > > > WHich other mappings do you want? What would satisfy you? > > A mapping which maps some member of *N to the set of all members of *N > not in the set they map to. ....11111111111111111111111111111111111111111111111111111111111111111111, with oo^oo^oo^oo^oo......... 1's. Now blow out the candles, Virgil. > -- Smiles, Tony
From: Tony Orlow on 26 Oct 2005 10:17 William Hughes said: > > Tony Orlow wrote: > > David R Tribble said: > > > Tony Orlow wrote: > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > Virgil said: > > > >> {x in S:x not in f(x)} > > > > > > > > > > Tony Orlow wrote: > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > Your mapping does not include any natural that maps to the entire set > > > *N, which it must do in order to be called a bijection, since *N is one > > > of the members of P(*N). Show us that one single mapping, please. > > It would obviously be the natural with an infinite unending strings of 1's, > > right? Ah, but you want to know, how MANY 1's? > > No, this is irrelevent. No natural has a binary representation that > consists only of 1's. I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit the bill. Why do you say I cannot declare that the 1's go on indefinitely? We are talking about *N, the set including infinite naturals, with infinite numbers of bits. When will I run out of bits for my subset elements? Remember, there is no last element, as you remind me again, below. > > > > > > > > But none of us can figure out where you keep pulling this "last > > > element" gibberish from. Just let it go, Tony. > > The entire set includes the last element, > > > > Tut tut. You have not been doing your exercises. > Please repeat 150 times after lunch: > > Some sets do not have a last element. > > - William Hughes > > -- Smiles, Tony
From: Randy Poe on 26 Oct 2005 10:28 Tony Orlow wrote: > Virgil said: > > To declare something does not make it so, at least outside of TOmatics. > It's my bijection, after all. I can declare it to be whatever I wish. You can't declare it to be a bijection unless you prove it to be so. > Go ahead > and rain on my parade, but don't move the police barriers. > > > > Outside of TOmatics, declarations require proofs before they need be > > accepted. > Not when you are defining a bijection. Yes, when you are defining a bijection. > What do you prove when you say the evens are mapped with f(x)=2x? That this map is bijective. It isn't assumed. It has to be proved. All of us who offer it as a bijection are fully prepared to back up that claim with a proof. We don't claim it is "obvious" or "needs no proof" or "is a bijection because we put it on the table with a label THIS IS A BIJECTION." It's a map. If you aren't sure it's bijective, then demand a proof, just as we demand of you. To wit: Let x be any natural. Then 2x is divisible by 2 and is a member of the set of evens. Thus f:N->E Let y be any even, i.e. y = 2x for some natural. Therefore f(x) = y, i.e., every even is the image of some natural number. Thus f is surjective. Let f(x) = f(x') = y for some x, x' in N. Then 2x = 2x' = y, or (2x-2x') = 0, i.e. x=x'. Since f(x)=f(x') => x = x', then f is injective. Since f:N->E is both injective and surjective, it is bijective. ----------------- Conspicuously absent here is any TO-ish mention of a "last" or "largest" element. By letting x or y be arbitrary elements, the reasoning applies to every element of the respective sets. - Randy
From: William Hughes on 26 Oct 2005 11:01 Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > David R Tribble said: > > > > Tony Orlow wrote: > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > Virgil said: > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > Tony Orlow wrote: > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > of the members of P(*N). Show us that one single mapping, please. > > > It would obviously be the natural with an infinite unending strings of 1's, > > > right? Ah, but you want to know, how MANY 1's? > > > > No, this is irrelevent. No natural has a binary representation that > > consists only of 1's. > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > the bill. Why do you say I cannot declare that the 1's go on indefinitely? We > are talking about *N, the set including infinite naturals, with infinite > numbers of bits. When will I run out of bits for my subset elements? Remember, > there is no last element, as you remind me again, below. You do not "run out of bits". The reason that no natural number can have a binary representation composed only of ones is that every natural has a successor which also has a binary representation. Suppose that a natural n had a binary representation that is composed only of ones. What could the representation of n+1 be? It cannot consist only of ones (this is the binary representation of n). It cannot have a 0, as a binary representation containing a 0 is the binary representation of a number smaller than the number with a binary representation consisting only of ones. Thus n+1 cannot have a binary representation. Contradiction. -William Hughes
From: imaginatorium on 26 Oct 2005 11:57
Tony Orlow wrote: > William Hughes said: > > > > Tony Orlow wrote: > > > David R Tribble said: > > > > Tony Orlow wrote: > > > > >> So, which element of the power set does not have a natural mapped to it? > > > > > > > > > > > > > Virgil said: > > > > >> {x in S:x not in f(x)} > > > > > > > > > > > > > Tony Orlow wrote: > > > > > Oh yeah, the entire set, last element and all. What element was that again? > > > > > > > > Your mapping does not include any natural that maps to the entire set > > > > *N, which it must do in order to be called a bijection, since *N is one > > > > of the members of P(*N). Show us that one single mapping, please. > > > It would obviously be the natural with an infinite unending strings of 1's, > > > right? Ah, but you want to know, how MANY 1's? > > > > No, this is irrelevent. No natural has a binary representation that > > consists only of 1's. > I assume you mean in the infinite string, otherwise 1, 11, 111, etc would fit > the bill. Why do you say I cannot declare that the 1's go on indefinitely? Oh, because mathematics does not work by "declaring that P", where P is some proposition you'd like to be true. I wonder if you've noticed that (with the possible exception of other cranks) you are the only one who goes about "declaring" this and that, or "applying" values to things you want to "measure". Brian Chandler http://imaginatorium.org |