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From: Virgil on 19 Oct 2005 14:03 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. 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.
From: Virgil on 19 Oct 2005 14:07 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).
From: stephen on 19 Oct 2005 14:08 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? Presumably that is the "last" infinite even number. And which 0 represents 0:010........10101? It is an odd number afterall, so one of those 0's must represent it. Stephen
From: Virgil on 19 Oct 2005 14:12 In article <1129722002.134872.315200(a)g49g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > 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. > > > Regards > AS As the argument which proves that there is no surjection from a set to its power set in no way uses the finitenesss or infiniteness of either the set or its power set why are separate proofs required? To comply with this silly demand, one could simply repeat the same proof twice, and it will serve once for finite cases and the other for infinite cases.
From: Virgil on 19 Oct 2005 14:23
In article <1129725511.849831.257090(a)g47g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > William Hughes wrote: > > albstorz(a)gmx.de wrote: > > > > <snip> > > > > > > > > If we accept the uncountability as a form of infinity, this leads to > > > the paradoxon that the natural numbers are not countable. > > > > No, the natural numbers are countable precisely because > > they do count themselves. > > This is exact my argument: there are uncountable many natural numbers > (nothing other means infinity many) but the natural numbers are shurely > countable since they count themself. > You are not able to recognise a paradoxon if you see it. > The only "paradox" here is AS using "countable" versus "uncountable" in two mutually contradictory senses in one sentence. In standard terminology 'countable' does not require finiteness. 'Countable' merely requires that there be a surjection from the set of naturals to the set in question, or, equivalently, an injection from the set in question to the set of naturals. And the set of naturals has both. To be uncountable, a set must be not countable by the definition above, so for AS to claim that the set of naturals is both countable and uncountable is HIS error, not anyone else's. |