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From: Virgil on 20 Oct 2005 22:49 In article <MPG.1dc1c1f5108b9b0f98a51b(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc040cc19fad5ff98a4e4(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > Did you have an objection to the bijection between *N and P(*N)? What > > > did I do wrong there, bijection-wise? > > > > You assume your result in order to prove it. > > > Virgil, you're being a lunkhead. It wasn't a proof, and I didn't assume any > conclusion. TO's inability to see what he is actually doing is one of the reasons TOmatics is so self-destructive. > I described a bijection What TO described was nowhere near to being either a complete description nor a bijection, and reasons why it is not a bijection have been repeateedly provided as well as reasons why it cannot be one. That TO will not acknowledge either does not vitiate their validity. > No one has answered that question yet. Perhaps TO is losing his memory, or what little literacy he has. > Discredit the > counterexample with established rules For whatever S and whatever f:S -> P(S), To wants to propose, TO cannot find any s in S with f(s) = {x in S:x not in f(x)}. Absent such an s, f is not a surjection and thus not a bijection either. or accept it as a refutation of the > proof. You claim to be logical? Obey logic.
From: Virgil on 20 Oct 2005 22:55 In article <MPG.1dc1c2d56cab209998a51c(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc0411e718b821798a4e5(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > Virgil said: > > > > In article <MPG.1dbf2de38cf6ec898a4da(a)newsstand.cit.cornell.edu>, > > > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > > > > > David R Tribble said: > > > > > > > > But that's an incomplete mapping, because there are not enough > > > > > > infinite > > > > > > binary strings in *N to enumerate all of the subsets of *N. Try > > > > > > it, > > > > > > if you don't believe me. > > > > > > > > > > > > > > > > > Not enough infinite binary strings? > > > > > > > > Precisely. > > > > > > > So, you need more than an infinite amount? > > > > You need "more" then you have, in the same sense that there are "more" > > reals than rationals even though there is an endless supply of both. > > > There are also more rationals than naturals Except that one can easily inject the rationals into the naturals, which assures that the 'size' of the rationals must be the same size as a subset of of the naturals. > The problem with the reals is that you > don't have a natural well-ordering of them. I can give you that, when you're > ready, which will probably never be. A valid explicit well-ordering of the reals would be of some actual mathematical interest. If TO could provide one, he might be worth a tinker's dam after all, but I am not going to bate my breath til he does!
From: Virgil on 20 Oct 2005 23:02 In article <MPG.1dc1c353824e05df98a51d(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc042163fd02d9198a4e6(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > Except for the obvious bijection between *N and P(*N). But, hey, what's > > > one > > > measly counterexample? > > > > In this case it is quite "measly", since it specifically requires that > > one assume most of the result in order to "prove" the result. > Assume what, specifically? That there is a bijection between the set of all binary strings and the set of only those binary strings having a single non-zero digit. > > But any such countable well-ordering would prove that the set of endless > > sequences of binary digits is TO-finite. > Not I am talking about actual infinitely long bitstrings, such as would > represent an actual infinite value in binary numbers. But the set of them having only a single non-zero digit IS TO-finite. And if a set of binary strings is to be represented by 1's for each member, a one element set can only have one 1.
From: Virgil on 20 Oct 2005 23:06 In article <MPG.1dc1c41a9eabe5ec98a51e(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc044c61d46b15798a4e8(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > albstorz(a)gmx.de said: > > > > > > > > > > You must proof it independently for finite and for infinite > > > > sets. In this sense the argument is stupid. > > > > > > > > > > > > Regards AS > > > > > > > > > > > Albrecht, do you accept the axiom of induction? If so, it is > > > easily provable inductively that the power set of a set of size n > > > has size 2^n, and since this is an equality property, it holds > > > for the infinite case. The power set of an infinite set is > > > infinite, but a larger infinity than the set. > > > > That is only true if 'size' means cardinality, since it is based > > entirely on injection/surjection/bijection arguments. > No, cardinality only comes to this conclusion through the clumsiest > of methods, like a bear that scratches his back by getting himself > run over by a truck. Then TO must be claiming that not only can sets of the same cardinality have different sizes, but also that sets of the same size may have different cardinalities. I should very much like to see any example of the latter situation. > > > > And TO correcting albstorz is a perfect illustration of the blind > > leading the blind. > Yes, that's how it looks, to the blind. AS TO is one of them, he can speak for them. > >
From: imaginatorium on 20 Oct 2005 23:16
Tony Orlow wrote: > imaginatorium(a)despammed.com said: > > > > Notice also how you are again unable to consider abstract sets, and > > keep mumbling about "numbers". Why on earth should the y be "larger" > > than any element in w? We've proved that y cannot exist at all, but > > there is no a priori reason any notion of "largeness" is involved. If > > we were considering the set of all finite simple groups or the set of > > all Platonic solids, there would be no "larger", but the proof applies > > exactly the same. > Did I not give what appears to be a valid bijection between the NUMBERS in *N > and the SETS in P(*N)? If "*N" is a set, in the sense of normal set theory, then no you didn't. (Am I remembering right? Is *N the set of TOnats, the set of infinitely leftward extending digits laced with colons and other decorations?) > If the bijection is not valid, please state what mistake > I made in constructing it. Well, you didn't state it in a way that anyone could derive for any element of P(*N) what element of *N was supposed to map to it. Saying 'map 0 to {0}, dot dot [defining finite cases] ... carry on into the infinite' leaves the problems to the mumbled dots at the end. > When I say the element y is "larger", perhaps what I > should say is its position is greater than the size of S, that is, it is in the > set[*], but not in the segment of the set[**] considered as S. Well, there you are. S is a set. In set theory that means that S is the set S. It contains every element that is in S, and does not contain any that are not in S. Only S is the set containing every element in S and no others. In set theory (thanks to the axiom of infinity) we consider infinite sets too: if T is an infinite set it means we cannot even in principle lists the elements at some constant readable point size on any four-sided piece of paper, however large. Even so, the elements of T are indeed all elements that belong to T. (I hope I've remembered right: we consider a mapping from S to P(S), and y is the putative element of mapping to the 'magic subset' of those elements which map to subsets not containing themselves. Yes? So y, an element of S cannot exist.) But in your version, first you're still stuck on this "numbers" thing, so you can't describe set membership without irrelevancies like "greater than". Then you have four references to sets. Can you please explain which sets they are: > When I say the element y is "larger", perhaps what I > should say is its position is greater than the size of S, This S really is the set we are talking about, no? > that is, it is in the set[*], "The set"?? Which one is that? Not S? but not in the segment of the set[**] Is this again some other (the same other?) set that isn't S? What is the relevance of this? >... considered as S. Oh no, the last "the set[**]" is "considered as S". Well, normal set theory doesn't have a defined relation "considered as". Is this the set S, or is it not? > When you ask for an > element that maps to the entire set, that assumes you have a bitstring that > that element should be equal to. How many 1's are in that bit string? Where > does it end? A bijection to the power set requires something to map onto every subset, including the set itself, which is a (non-proper [technical term]) subset. I'm a bit lost, but I expect the answer to "Where does it end?" is "It doesn't". Brian Chandler http://imaginatorium.org |