An uncountable countable set
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From: Tony Orlow on 20 Aug 2006 09:32 Virgil wrote: > In article <ec8hlf$qq8$1(a)ruby.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > >> If the formula applies to an infinite number of finites, then does the >> sum of the aleph_0 finite naturals equal (aleph_0^2+aleph_0)/2? > > Only if (aleph_0^2+aleph_0)/2 = aleph_0. Which it does, in standard theory. It is not until you get to 2^aleph_0 that you are able to distinguish aleph_0 from a larger infinity. > > But even then, one must have some prior definition of addition of > infinite cardinalities, and some prior definition multiplication of > infinite cardinalities (or at least squarings of them), and some prior > definition of division infinite cardinalities by finite cardinalities > (or at least halving of infinite cardinalities), none of which TO has > provided, so his "formula" is all nonsense. > We have the prior definitions of all those operations with regards to variables in general. All that's additionally required is allowing the variables to assume infinite values. > > >> I think that what WM is pointing out here is that aleph_0 does not >> behave like a normal number > > Only a nutcase would suppose that an infinite cardinal need behave > entirely like a finite cardinal anyway. Transfinite cardinals and limit ordinals don't behave like numbers or quantities at all, but it's not nutty to expect numbers to act like numbers. It's nutty to allow them to act like potions and incantations. Nothing breaks when infinite numbers are treated like numbers. If you think something does, then it's only transfinite set that suffers, but that appears rather expendable, anyway. > >> Given an injection f: N -> P(N), why does the set >> M(f) = {n in N | n !in f(n)} >> not exist? I don't know why that looks like my question - it's not. > > It does. But for a non-surjective f: N -> P(N), which all functions from > N to P(N) must be, there will not be any x in N such that f(x) = M(f) Yeah, that would the the lst natural. We all know THAT doesn't exist. Or, is it aleph_0? Same thing anyway. ;) TO
From: Tony Orlow on 20 Aug 2006 09:43 Virgil wrote: > In article <ec8hsd$r5t$1(a)ruby.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > >> Dik T. Winter wrote: >>> In article <ec7gek$frg$1(a)ruby.cit.cornell.edu> aeo6(a)cornnell.edu writes: >>> > Inductively provable: Any set of consecutive naturals starting at 1 >>> > always has its largest element equal to size of the set. >>> >>> What is the size of the empty set? What is the largest element of the >>> empty set? >> The empty set is not a set of consecutive naturals starting at 1, >> because it does not include 1 as an element, and therefore it does not >> stand as a counterexample. However, even in your von Neumann ordinals, >> the empty set is 0, and contains zero elements, and could be considered >> the 0th in this sequence of sets. You fail to address the inductive >> proof that the non-empty sets of this series, of which the naturals is >> the limit, always have this property. > > TO fails to note that induction does not apply. Virgil somehow fails to see that it does. > The form of induction is: > Let F(x) be a predicate whose argument, x, is a member of a minimal > inductive set S.* > If F(x) is true for the first member of S > and > if whenever F(x) is true then F(successor(x)) is also true. > then > the principle of induction says that F(x) is true for each member of S. Right. Consider S to be the set of all sets of consecutive naturals starting at 1. Is N a member of this set? You bet it is. Is this an inductive set? Indeed, there is a first, {1}, and each has a successor which is itself union the successor of its size. So what is F(x) for each of the members of S? F(x) is size(x)=max(x), which is trivially provable. Since N is a member of this inductively defined set, size(N)=max(N). QED, Doll. > > It does not say anything about what is true for the set S itself, only > for its members. S is the set of all such sets. N is a member of S. So, it says something about N. > > In particular it does not say that F(S) is true. > No, it says that F(N) is true, since N e S. > > * Note a minimal inductive set is a non-empty set, S, together with a > "successor" function s:S -> S such that > (1) there is a unique member of S which is not in the image of s. > (2) s is injective. > (3) If T is a subset of S such that > (a) T contians the unique member of S not in the image of s and > (b) for all x in S, if x \in T then f(x) \in T > then T = S > > Note that the set of von Neumann naturals, N, with s(x) = x \/{x} and > the non-image element {}, are a minimal inductive set under the above > definition. Yes, and it's inductively provable likewise that the size of a successor ordinal is one greater than its max element. That would make aleph_0 1 greater than the max finite natural. The only way around this is the declaration of the limit ordinals, but that's just a philosophical monkey wrench. > > There is no equivalent model in which the ordinal of each element is > equal to the element itself as that would require sets to be members of > themselves, which is impossible for ordinals. Don't confuse max with size. Relate them formulaically. Have a nice day! TO
From: Tony Orlow on 20 Aug 2006 09:46 Virgil wrote: > In article <ec8if0$rtd$1(a)ruby.cit.cornell.edu>, > Tony Orlow <aeo6(a)cornell.edu> wrote: > > >> Oh, how I love being compared to JSH! Shall I begin ranting about the >> conspiracy against me, and how you will all suffer for your mathematical >> sins against my superior intellectual achievements? No, I'll save that >> for when I'm actually schizophrenic, thanks. ;) > > Tomorrow? >> Indeed, I agree with WM's logic concerning the identity relationship >> between element count and value in the naturals. > > Except that ordinal numbers cannot work that way, > as it would require each ordinal to be a member of itself, > which is not possible for ordinals. Ordinals and cardinals are schlock. They're not part of any theory I ascribe to. The identity relationship between element count and value in the naturals is far more elementary and fundamental, and they are not compatible with this foundation. So, breaking them is really not much of a concern for me. Thanks for the reminder anyway. TO
From: mueckenh on 20 Aug 2006 09:46 Franziska Neugebauer schrieb: > > It is wrong to say that L is in |N > > It is _false_. "L is in omega" is _not_ _true_. Therefore it is wrong. > > > and it is wrong to say that L is larger than any n e |N. > > _Nowhere_ did *I* write something like "L is larger than any n e N". > This is probably your own wording. > > > Actual infinity does not exist! > > I don't discuss Cantorisms. Do you discuss whether the cardinal number of omega is a number larger than any natural number? > > > I told you that already several times. > > You persistently try to posit Cantorisms. > > > But refuse to understand what "to exist" means. > > An entity a exists if |= E a. It is that easy to establish? Then I hasten to write |= E agreement between us that transfinite set theory is nonsense. Regards, WM
From: mueckenh on 20 Aug 2006 09:49
Franziska Neugebauer schrieb: > > So you claim my proof is wrong. Where exactly? The result. > > There are not infinitely many differences, if we consider the finity > > of each number. > > Proof? Consider a staircase: Without a stair surpassing the height H the staircase cannot surpass height H. Without a stair surpassing every finite height the staircase cannot surpass every finite height, i.e., cannot have infinitely many steps of height 1. (surpassing every finite = infinite). But the length surpasses every finite length (according to Cantor). > > > There are infinitely many differences, if we consider infinitely many > > numbers. > > There _are_ infinitely many numbers n e omega and therefore there are > "infinitely many differences of 1" (as proven). If you climb 10 meters, then you arrive at height 10 m. But you climb infinitely many meters without arriving at an infinite height? > > > Therefore one of the assumptions is wrong: Either there are not > > infinitely many numbers or there is at least one number which is > > infinite by size. > > There are infinitely many numbers n e omega _and_ there is not a single > number which is "infinite by size". This is fact Is infinitely not a number in your opinion? Then infinity cannot be surpassed. Then omega + 1 is purest nonsense. Under this condition I agree with you. You are back to the pre-Cantorism. > >> Do you agree that P ~ omega? > > > > P has the same cardinality as |N. > > So P ~ omega. > > > But either there are not infinitely many numbers or there is at least > > one number which is infinite by size. > > Why (please prove!)? Do you want to ignore facts? You must not think that facts are facts because you call them facts. Fact is: if omega is a number, then omega differences of height 1 result in the total size omega, because omega * 1 = omega. Isn't it a fact? Regards, WM |