An uncountable countable set
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From: MoeBlee on 23 Aug 2006 18:36 Tony Orlow wrote: > Set theory contradicts with: > > (1) E y e N, A x>y, x< 2*x < x^2 < 2^x (y=2) > > because: > > (2) A y e N, aleph_0>y I don't know what you intend '<' to stand for. For the domination relation? The less than relation on ordinals? I don't know what is meant by '(y=2)' in the larger formula. > and > > (3) aleph_0/2 = aleph_0 = aleph_0^2 < 2^aleph_0 > > (1) is trivially inductively provable. Do you mean (1) is a theorem of set theory, or do you mean it is provable that (1) is the negation of a theorem of set theory? > (2)and (3) are from transfinitology. What is transfinitology? What is the definition (and in what theory is this definition?) of '/' where w (omega) is in the numerator? MoeBlee
From: Tony Orlow on 23 Aug 2006 18:38 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > >> In article <ec8if0$rtd$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: > >> > Indeed, I agree with WM's logic concerning the identity relationship >> > between element count and value in the naturals. He's quite correct in >> > that regard. >> >> Well, you and he are not. The logic is flawed. > > For 1, 2, and 3, order and value are identical, even according to your > logic, I suppose. Where does the first deviation happen? Which one does > deviate first, i.e., which one does first be larger than the other? And > why? > > The cardinal number aleph_0 is infinite while the ordinal number > remains finite in order to have infinitely many finite numbers. Precisely. At what point do you increment both max value and count by adding the next natural, and having started with the two equal at every point, end up with two different values? >> And his proof is not a proof. > > O course not, because every proof which has an unpleasant result is not > a proof. I disagree!!! ;) >> > For my part, I agree that the set of finite >> > naturals is finite, though unbounded, >> >> In that case you are not using standard mathematical terminology. I >> have no idea what a finite but unbounded set is. > > That's why you cannot understand mathematics. You fall back behind > Cantor. He knew it. Since Dedekind, unbounded means infinite, but it's just not so. ROll back the film a little. > > Regards, WM >
From: MoeBlee on 23 Aug 2006 18:46 Tony Orlow wrote: > If set theory claims to have a cardinality > which fits every set, then this set stands out as the counterexample. > Any finite number of bit positions produces a finite set of strings. Any > countably infinite set of bit positions produces an uncountable set of > strings. And, there's nothing in between. This number doesn't exist. It > can't be classified. So, yes, I think you have an actual flaw within. Please say what sentence and its negation you believe are both theorems of set theory. > I have presented a system No you haven't. You've posted disconnected pieces of undefined terminology. > >> Define "truth". > > > > Formal definition is given in a formal meta-theory. Greatly simplified, > > a sentence S is true in a model M iff the evaluation function per M > > (definition of this function given courtesy of the defintion by > > recursion theorem) with the sentence as argument yields the set of all > > functions on the variables into the domain of the model. > > Maybe that's a little simplified. Not sure what you're saying. Sorry. My statement is too compressed and not pinpoint accurate given the limitations of a one paragraph answer. See Enderton's mathematical logic textbook. The full formalization is culminated in the exercise in which you are asked to make sets of functions from the variables into the universe the values of a certain recursively defined function. MoeBlee
From: Virgil on 23 Aug 2006 18:50 In article <ecig8i$ta$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Dik T. Winter wrote: > > Some clarification. > > Thanks, Dik! :) > > > > > In article <J4CnBH.DqI(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > > > In article <J472w4.Lt1(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> > > > writes: > > > > In article <1155885821.815144.187270(a)m73g2000cwd.googlegroups.com> > > > > mueckenh(a)rz.fh-augsburg.de writes: > > > > > Dik T. Winter schrieb: > > > > ... > > > > > > To be more correct. The first proof is not relying on the > > > > > > representation of reals, the second (diagonal) proof is not > > > > > > about reals at all. > > ... > > [ About Cantor's proofs:] > > I was not even completely correct in that statement. After completely > > reading his first proof and his second proof (the diagonal proof), I am > > quite sure that both his proofs were targeted at the following theorem: > > There are sets with a cardinality larger than that of the natural > > numbers. > > Yes, that would seem to be about the size of it. > > > The first part of his first proof shows that a complete ordered field > > has cardinality larger than the natural numbers. In his proof he did > > not rely an any properties of the reals other than that they form a > > complete ordered field (he uses reals to exemplify). > > Specifically, he demonstrates that there will always exist an > unaccessible real in any finite interval, given that we are only allowed > a finite (natural, that is) number of iterations. Therefore, indeed, the > number of reals in a finite interval is greater than any finite natural. > That proof is essentially valid, but it's not a proof that the reals > cannot be linearly ordered in a discrete manner. Then lets see TO prove that anyone CAN linearly order the reals "in a discrete manner", where discrete when applied to order relations means that for each element, other than an end one, has a unique nearest element on either side. > > > (The second > > part shows that the cardinality of the set of subsets of a set is > > strictly larger than that of the original set. This proof comes > > close to the proof provided by Hessenberg, but is, in my opinion, > > a bit less strict.) > > It is related to the powerset through |P(S)|=2^|S|, but he did it in > decimal, no? No!. In any case, the powerset relation boils down to the > symbolc equation N=S^L, where S is the number of logical states allowed > (there can be more than two in various systems) and where L ultimately > corresponds to the size of the root set. So, the two are related. Only for finite sets, as for infinite sets we have from TO no acceptable definition of exponentiation. > > > > > What he states in the quote provided by Mueckenheim is that that proof > > can be easily used to show that the cardinality of any set of reals in > > an interval is larger than the cardinality of the natural numbers. > > I don't disagree that the first is infinite while the second is > unbounded bt finite, and therefore smaller. As that is not what was said, TO is disagreeing, not agreeing. > Considering that the set of finite naturals is ultimately finite In what axiom system? It is not finite in any of the axiom systems that have been put forward here. > I dunno. Did I miss anything? Almost everything worth not missing..
From: Virgil on 23 Aug 2006 19:00
In article <ecigf9$1a8$1(a)ruby.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > And, it's so nice to know you've finally come around to realizing > there's a difference between 0.999... and 1 The difference between 0.999... and 1.000... as reals is the same as the difference between 2/4 and 3/6 as rationals, merely different representations of the same number. Either that or 0.999... is not a real at all, but only something that converges to a real. But then one would also have to say that no nonterminating decimals actually ARE reals, they are only things that converge to reals. Which would make 1.000... and 1.000 as different as 0.999... and 1.0. |