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From: Virgil on 22 Oct 2005 14:54 In article <1129995274.931879.33490(a)o13g2000cwo.googlegroups.com>, albstorz(a)gmx.de wrote: > David R Tribble wrote: > > Albrecht Storz wrote: > > > You can biject all constructable subsets of P(N) to N, but not the > > > unconstructable. That's an argument against the diagonal proof, since, > > > if a set will be constructable in form of an antidiagonal of a list, > > > it's a constructable set and therefore it is element of the set of the > > > sets which are bijectable with N. > > > > Then that constructable set will be in the original list, and the > > diagonal set will be different from it, and it can't be a set > > constructed from the diagonal. > > > > But that just proves that you can't construct a set from the > > diagonal of the list of sets while allowing that set to also be > > in the list. No such set exists in the list. Therefore, the > > constructed set must be a set that does not exist in the mapping, > > and therefore the mapping does not map every possible set in > > the list. > > > And you are shure the nonconstructable exists? WE are sure [sic] that the 'diagonal' is just as constructible as the list itself is. > And what meaning did > they have to you? What "unconstructables' is AS referring to? > You like them. Are they nice. What kind of properties > they have (elese than be unconstructable)? And what do you do with them > then knowing they are? As we have no idea what ASA is nattering on about, perhaps he would explicate. > > Regards > AS
From: Virgil on 22 Oct 2005 14:58 In article <1129995923.625067.33080(a)g14g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > David R Tribble wrote: > > William Hughes wrote: > > >> Yes every number is a set. No, not every set is a number. For example > > >> {peach, apple, plum, fiddle} is a set but not a number. > > > > > > > Albrecht Storz wrote: > > >> Wrong. It's a number. The number is this aspect of the set which it > > >> have in common with the set {diddle, daddle, doddle, duddle}. > > >> To say, a > > >> set is a number or a set has a number is a slight difference which has > > >> no effect in this considerations. > > > > Albrecht, if every number is a set, which set is 3? > > > > What number is the set {a, {b}, {{c}}, {{{d}}}, ...}? > > > > If sets are numbers, then adding sets should obey the same rules as > > adding numbers, right? What is {1,2,3} + {0,2,4}? Is it 3+3 = 6? > > > 3 is every set with 3 members. In what sense? If 3 were "every set with 3 members" then 3 would not have 3 members. > Shurely the sets are different. And they > have different properties. But in the aspect of size of membership they > are exact same. Natural numbers are the essence of the sets concerning > their size. Not in NBG. > > David,try to be logic and honest to yourself. > It's not a play to win or loose. It's a play to understand or to stay > weak in mind. AS seems to have the "stay weak in mind" part down pat, but needs more practice with "understand". > You might belief in a one century old tale or in a more than a thousand > years old known thruth. > > Regards > AS
From: albstorz on 22 Oct 2005 15:47 David Kastrup wrote: > albstorz(a)gmx.de writes: > > > Tony Orlow wrote: > >> albstorz(a)gmx.de said: > >> > > >> > 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. > > >> 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. > > > > No Tony. > > > > If you think so, you should also argue, that since the cartesian > > product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the > > finite case is always of graeter cardinality than the starting sets, it > > should be also in the infinite case. But it isn't. > > You are arguing with the guy who thinks that the cardinality of the > set of even naturals is less than that of the naturals. > > The amusing thing is that although Tony is talking rubbish here, > powersets _do_ have a larger cardinality than the original set, since > any presumed mapping f:S->P(S) can't cover {x in S| x not in f(x)} > since that set is different from all f(x) for all x in S. This proof needs the expression "x in S| x not in f(x)" has to be meaningful. But it isn't in all the proofs about infinite sets. The antidiagonal is an unicorn. That's all. The proof of the existence of unicorns don't bring them in existence. They don't exist in spite of any proof about it. The other way makes a shoe out of this rubbish: If a proof leads to the result, that unicorns exist, you have proofed your proof (respectively your assumptions) to be wrong. Understand this, and you are out of cantorian fog. It will fall like scales from your eyes: all will be clear and easy. > > So while you win this particular argument against Tony (not too hard > to do), you are still wrong in the main. The above proof does not > depend on powers, calculation, finiteness or infiniteness. > > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum Powers of integers could be brake down to multiples of integers which could be brake down to sums of integers. Why should power be able to give a result which is unreachable by multiplication or addition? That's magic math you depend on. Regards AS
From: albstorz on 22 Oct 2005 15:54 David Kastrup wrote: > albstorz(a)gmx.de writes: > > > David R Tribble 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? > >> > > >> > >> Albrecht Storz wrote: > >> > 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? [...] > >> > > >> > I'm very sensible about this because this argument is found in very > >> > much books although it's total meaningless. > >> > > >> > (Weak minds might be impressed by the big numbers which are easily > >> > produced by powerfunction.) > >> > > >> > What in finity holds may not (or do not) hold in infinity. > >> > >> So what is your proof? > > > > The Gedanken in the context is, if finite powersets are always > > greater than there sarting sets, is there any connectivity to the > > idea that infinite powersets are greater than there starting sets. > > There is no connectivity. > > You are babbling. The standard proof about the larger cardinality of > powersets makes no use of finiteness, infiniteness or, in fact, the > power function at all. > > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum To someone, who don't know what is the issue, argueing seems to be babbling. The question was: is there any meaning about the fact, that finite powersets are greater than their starting sets concerning the size of infinite sets? This and only just this point was under discussion. Regards AS
From: David Kastrup on 22 Oct 2005 15:56
albstorz(a)gmx.de writes: > David Kastrup wrote: >> albstorz(a)gmx.de writes: >> >> > Tony Orlow wrote: >> >> albstorz(a)gmx.de said: >> >> > >> >> > 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. >> >> >> 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. >> > >> > No Tony. >> > >> > If you think so, you should also argue, that since the cartesian >> > product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in >> > the finite case is always of graeter cardinality than the >> > starting sets, it should be also in the infinite case. But it >> > isn't. >> >> You are arguing with the guy who thinks that the cardinality of the >> set of even naturals is less than that of the naturals. >> >> The amusing thing is that although Tony is talking rubbish here, >> powersets _do_ have a larger cardinality than the original set, >> since any presumed mapping f:S->P(S) can't cover {x in S| x not in >> f(x)} since that set is different from all f(x) for all x in S. > > This proof needs the expression "x in S| x not in f(x)" has to be > meaningful. So where is the problem with its meaning for you? S was _assumed_ to be a set. And f(x) was _assumed_ to be defined as a mapping function from S, so f(x) is defined for all x in S. And since f(x) was assumed to be a member of P(S), then f(x) is again a set. So why can't I just take all x in S that have the property of not being an element of f(x)? > But it isn't in all the proofs about infinite sets. Irrelevant. We are talking about the above proof. Of course, for every truth you'll be able to find millions of invalid proofs. But that does not change a truth, if there is just a single valid proof. So what is wrong with the above proof? > The antidiagonal is an unicorn. That's all. The proof of the > existence of unicorns don't bring them in existence. You are babbling nonsense because you are out of arguments. > They don't exist in spite of any proof about it. The other way > makes a shoe out of this rubbish: If a proof leads to the result, > that unicorns exist, you have proofed your proof (respectively your > assumptions) to be wrong. Understand this, and you are out of > cantorian fog. It will fall like scales from your eyes: all will be > clear and easy. Sorry, but I don't have the required amount of alcoholic beverages and mind-altering drugs in the house. >> So while you win this particular argument against Tony (not too >> hard to do), you are still wrong in the main. The above proof does >> not depend on powers, calculation, finiteness or infiniteness. > > Powers of integers could be brake down to multiples of integers > which could be brake down to sums of integers. Why should power be > able to give a result which is unreachable by multiplication or > addition? That's magic math you depend on. You are babbling nonsense. I repeat: the above proof does not depend on powers, calculation, finiteness or infiniteness. No "magic math" is involved, and no powers are involved. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |