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From: Virgil on 31 Oct 2005 17:54 In article <MPG.1dd0241af280198e98a5ad(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In order for a mapping to be a bijection, it must "get to" every point! > > > > Or perhaps any old mapping can be called a bijection in the strange > > world of TOmatics. > > > > Outside of TOmaics, a function from a set S to a set T is a bijection > > ONLY IF for every s in S, f(s) is a unique member of T, AND for every t > > in T, there is a unique s in S such that f(s) = t. > > > > TO's alleged "bijections" do not meet these criteria, so that, outside > > TOmatics, they are not bijections at all. > This bijection certainly meets those criteria, for every element you can name > in P(*N), besides the very last member, the entire set *N. None of your > bijections for unending sets ever name a mapping for a last element. There is > no such thing. TO claims that he has a mapping f:*N -> P(*N) such that, for some n in *N, f(n) = T, where T = {x in *N: x not in f(x)}. But consider whether, for TO's function, f, and that n, one finds n in T or n not in T. If n is in T then n is in f(n), so that n is not in {x in *N: x not in f(x)} = T. If n not in T then n not in f(n), so n is in {x in *N: x not in f(x)} = T Thus TO's assumption requires a contradiction > Wake up, > Virgil. The splinter you perceive in my eye is but a reflection of the log in > yours. That TO's claim implies a contradiction is plain enough to see despite any eye trouble. But TO's claims often imply contradictions, which is what makes his TOmatic delusions so useless to serious mathematics.
From: Virgil on 31 Oct 2005 17:56 In article <MPG.1dd02b7e915c80de98a5ae(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Randy Poe said: > > > > Tony Orlow wrote: > > > David R Tribble said: > > > > We are forced to conclude that there is no natural s that maps > > > > to *N, and that therefore your mapping scheme is not a > > > > bijection between *N and P(*N). > > > You are not forced to conclude that this prevents a bijection. > > > The entire infinite set would be the final subset enumerated, but > > > the set does not end, nor does the set of its subsets, so the > > > bijection will never, ever get to this point. You used "TOmatics" > > > fairly well. What if we apply it to the evens? Whatever n you > > > choose, you have 2n, but 2n is a natural too, so you have to map > > > it to 4n, etc. > > > > The objection above is that there exists at least one element of > > P(*N) which is not equal to f(s) for any s in *N. That objection > > says nothing about "last". > Ahem. Is the entire set *N not the last element in the ordered set > P(*N)? In my biojection it is, and it's the entire set is generally > considered the last element of the power set, the null set being the > first. So, at least in the case of my bijection, this is exactly what > that missing mapping translates to, and it is probably provably > always the case. > > > > The same objection can not be raised about f:N->E, f(x) = 2x. Every > > element in E is mapped by some element in N. Every single one. > > Nobody but you says you have to consider "the last" in order to > > justify the claim that this statement is true for every single > > member of the set of evens. > But you are saying that for my bijection, effectively. COntradiction > abound by the unspoken assumptions of such last elements in this > transfinite mess. > > > > The surjection part of my simple proof of the bijectivity of > > f(x)=2x had this simple structure: Let y be any element of the > > codomain. Then there exists x such that f(x) = y. > > > > If you wanted to establish surjectivity of your map f:*N->P(*N) you > > need the same structure: Let y be any element of P(*N). Then (TO > > can show) there exists x in *N such that f(*N) = y. > That bijection is established by the mutual bijection between the > infinite binary strings and these two sets. If Randy can just "say" > that *N is the same as the set of all infinite binary strings, then > why can't I? And, certainly, if I am representing the power set as a > set of binary strings, there is a bit for every one of the infinite > number of naturals in *N, so it is also the set of infinite binary > strings. So, what is your objection? What is left to be proven? TO claims that he has a mapping f:*N -> P(*N) such that, for some n in *N, f(n) = T, where T = {x in *N: x not in f(x)}. But consider whether one has n in T or n not in T. If n is in T then n is in f(n), so that n is not in {x in *N: x not in f(x)} = T. If n not in T then n not in f(n), so n is in T = {x in *N: x not in f(x)} = T. Thus TO's assumption requires a contradiction.
From: Virgil on 31 Oct 2005 18:02 In article <MPG.1dd02c4d8cdfd07498a5af(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > > Albrecht Storz wrote: > > > >> My argumentation is very easy: > > > >> Every nat. number represents a set. If you look at the first 100 nat. > > > >> numbers, the 100th nat. number "100" represents the set {1, ... , 100}. > > > >> As this holds for every nat. number, if there are infinite nat. numbers > > > >> there must be a infiniteth nat. number representing this set. > > > > Albrecht is arguing that if every A is a B, then it must be the case > > that every B is also an A. > No, dumbass, he is arguing that if A=B, then if A is finite, B is finite, and > if A is infinite B is infinite, which is perfectly true, despite your whining. I read it as saying that if a lot (but not all) of the bounded sets of naturals has some property, then all unbounded sets of naturals must also have that property. This is patently false for the property of being bounded, so can be false for other properties as well. And is false for the property being implied. > > > > That every natural represents a set does not imply that every set > > represents a natural. > > > > Albrecht would fail any logic class with such a foolish claim.
From: Virgil on 31 Oct 2005 18:04 In article <MPG.1dd02c90fdd2aaa198a5b0(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Well, Albrecht, perhaps you should click on the link below and see the Well > Ordering of the Real Numbers. I think it's pretty snazzy! Enjoy! Snazzy does not imply valid.
From: Virgil on 31 Oct 2005 18:06
In article <MPG.1dd02f849136f06398a5b1(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > albstorz(a)gmx.de said: The blind leading the blind! |