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From: Tony Orlow on 20 Oct 2005 15:25 David Kastrup said: > Tony Orlow <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> albstorz(a)gmx.de writes: > >> > >> > Virgil wrote: > >> >> In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, > >> >> albstorz(a)gmx.de wrote: > >> >> > >> >> > David R Tribble wrote: > >> >> > > >> >> > > > >> >> > > Because I can prove it (and it's a very old proof). A powerset of > >> >> > > a nonempty set contains more elements that the set. Can you prove > >> >> > > otherwise? > >> >> > > >> >> > This argument is stupid. Is there any magic in the powerfunction? > >> >> > >> >> "Proofs" are not stupid until they can be refuted. The proof that for an > >> >> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. > >> > > >> > > >> > Even if you think that the powersets of finite and infinite sets > >> > have both a greater cardinality than their starting sets, you would > >> > not really think it depends on the same cause in both cases. > >> > > >> > You must proof it independently for finite and for infinite sets. In > >> > this sense the argument is stupid. > >> > >> Uh, no. The proof depends merely on the fact that some value has to > >> be either a member of a set, or not. And if some value is in one set, > >> bur not another, then those two sets are different. > > Oh! So, N is a different size from N/{1}? > > Wrong. N is a different set from N\{1}, but that does not mean that > the sets have different cardinality. To illustrate this for the > really dense people with a simpler example: 2 is in {2} and not in > {1}, and that means that the set {2} is different from the set {1}, > even though both sets have the same cardinality, namely 1. What about {1}/{1,2,3}? I think that has a negative number of elements, don't you David? I knew you'd agree! I bet you can help me with imaginary sets too. Perhaps we can collaborate? That'll be great! > > Whenever one thinks that your stupidity has reached bottom, you still > manage to come up with something even more harebrained. So, you like to eat hare brains? Do you have them with cabbage and potatoes? Or do you just eat them with corn? > > -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 15:29 Virgil said: > In article <1129721243.222130.296540(a)o13g2000cwo.googlegroups.com>, > albstorz(a)gmx.de wrote: > > > William Hughes wrote: > > > albst...(a)gmx.de wrote: > > > > David R Tribble wrote: > > > > > > > > > > > > > > Because I can prove it (and it's a very old proof). A powerset of > > > > > a nonempty set contains more elements that the set. Can you prove > > > > > otherwise? > > > > > > > > This argument is stupid. Is there any magic in the powerfunction? A > > > > hidden megabooster for transcendental overflow? What is the very > > > > special aspect of the powerfunction to be so magic? > > > > Why should all operations with transfinite numbers lead to results with > > > > the same "level" of infinity, but only powerfunction beams up to the > > > > next level? > > > > Is not true: a^2 = a*a? 2a = a+a? > > > > > > Yes, but the powerfunction does not look like a^2 but 2^a. > > > > > > > Is the powerfunction something other than a very shortcut for multiple > > > > additions? > > > > > > Yes. You cannot represent 2^x as multiple additions. > > > > > > Since we talk about x e {1,2,3,4,5,...}, why not? > > Lets see you give a general representation of 2^x as an addition, then! > > > > > > > Words to live by. Start by noting that a finite set has a > > > "number of elements" that can be described by a natural number > > > while an infinite set (e.g. the set of natural numbers) does > > > not have a "number of elements" that can be described by a > > > natural number. However, some things are true for both > > > finite and infinite sets. e.g. the fact that there is no > > > bijection between X and P(X). > > > > > > -William Hughes > > > > > > No. > > Then lets see one of your alleged bijections from some set X to its > power set P(X). > > The only valid disproof of a proof of non-existence is an example of > what is alleged to be non-existent. Hmm, so what was the problem with my bijection again? What rule did I break. No one has been able to tell me, no matter how many times I beg for assistance in understanding this deep and mysterious subject. > > If anyone counterclaims agains the proof of no surjection from and set X > to its power set P(X), that person owes us an example of such a mapping. And if anyone wants to claim their proof stands in the face of an obvious counterexample, that person owes us an explanation of how that counterexample does not apply. So, what of it, Virgie? What rule does my bijection between *N and P(*N) break? Take your time.... > -- Smiles, Tony
From: stephen on 20 Oct 2005 15:25 Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: >> Tony Orlow <aeo6(a)cornell.edu> wrote: >> > David R Tribble said: >> >> Tony Orlow wrote: >> >> >> >> >> >> > What do you want me to try, anyway, and infinite mapping, >> >> > element-by-element? A bijection's a bijection, right? >> >> >> >> Yes, that would be nice. Please show us your bijection. >> > f(0) = ...000 = {} >> > f(1) = ...001 = {0} >> > f(2) = ...010 = {1} >> > f(3) = ...011 = {0,1} >> > f(4) = ...100 = {2} >> > f(5) = ...101 = {0,2} >> > f(6) = ...110 = {1,2} >> > f(7) = ...111 = {0,1,2} >> >> If we define >> w= { x : x not in f(x) } >> then we get >> w = {0, 1, 2, 3, ...... } >> >> So for what y does f(y) = { 0, 1, 2, 3, ..... }? >> >> And is y in f(y)? >> >> Let me guess. You will claim that F(N) = { 0, 1, 2, 3, ..... }. >> So is N in { 0, 1, 2, 3, ..... }? If it is, then N is not in >> w, and F(N) does not equal w. If it is not, then N is in w, >> and F(N) does not equal w. >> >> <snip> >> >> > But I have constructed a bijection between the two using an intermediate binary >> > representation. What is the specific rule I have broken concerning the >> > construction of bijections. If I haven't broken any such rule, then is it true >> > that a bijection between two sets means that have the same size, or even >> > cardinality? >> >> No you have not. >> >> Stephen >> > Notice above that no natural maps to a subset which contains it, so w is all of > N. Imagining any completed w leads to a contradiction, since the natural that > would map to it is always bigger than every natural in that set. That means that imagining any completed bijection also leads to a contradiction. w has just as much existence as your bijection. If the bijection is complete, so is w. > That's okay > though, because for every natural, there's a larger one. If x exists, 2^x > exists. The sets are infinite, so the bijection continues, even though there is > a difference between the values which are in the subsets and the values which > denote the subsets. It is not a bijection. N is a subset of N, and is therefore in the powerset of N, and if no natural maps to N, then there is no bijection. Stephen
From: Tony Orlow on 20 Oct 2005 15:30 Virgil said: > In article <1129721500.027651.21470(a)f14g2000cwb.googlegroups.com>, > albstorz(a)gmx.de wrote: > > > Virgil wrote: > > > In article <1129684564.158859.119300(a)o13g2000cwo.googlegroups.com>, > > > albstorz(a)gmx.de wrote: > > > > > So biject two non-well-orderable sets. > > > > > > > > > With or without an axiom of choice? > > > > > > I think it is more interesting for the most people without AC, since AC > > is not widely loved by the mathematics. > > > > But you may show both if you want. > > > > Regards > > AS > > That shows how little AS knows about things, since given the axiom of > choice, there is no such thing as a non-well-orderable set (though > admittedly there are some sets known that have yet to be well-ordered). > Gee, like what, Virgil? Let's have a well-ordering party, for those not-yet- well-ordered sets. -- Smiles, Tony
From: Tony Orlow on 20 Oct 2005 15:37
stephen(a)nomail.com said: > Tony Orlow <aeo6(a)cornell.edu> wrote: > > Look, it's certainly not my position that the pwoer set is the same sie as the > > set. It's clearly not. I just see a bijection between them, which only bolsters > > my argument that bijection alone is not sufficient to equate the sizes of two > > sets. > > > When it comes to the evens (let's start with 0), the value 0:010.......1010101 > > represents such a subset, and is essentially binary N/3. > > So which number does that first 1 represent? log2(N-1). Perhaps we should say we have 2^N subsets and strings of N digits. That's what I should have said. Then that number is 2^N/3, and the first 1 in it represents N-2, since the bits are numbered from 0 through N-1. > Presumably that is > the "last" infinite even number. The last one before N. There is no ultimate last even, as we all know, silly canilly. > And which 0 represents > 0:010........10101? It is an odd number afterall, so one of those > 0's must represent it. No, that number is 2^N/3, and would be represented by the 2^N/3th bit, which is not in the first N bits, right? This is the heart of your power set proof. > > Stephen > -- Smiles, Tony |