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From: Virgil on 31 Oct 2005 16:55 In article <MPG.1dd00d647e86874798a5a1(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > And yet, the bijection between 2 unending sets does not require you > to be able to name the end of the sets, does it? TO is confusing order properties of ordered sets with membership properties of arbitrary sets. In order to have a set at all, the issue of membership must be determinate, but this has nothing to do with any order propereties of an ordered set. And ordered set can be endless, in the sense of not having any last member according to the imposed order, but set must be complete to be a set as all BEFORE it can be an ordered set. > Look again at the bijection I offered. Your mapping between *N and > P(N) is not valid in Bigulosity Theory. On the contrary, in "Bigulosity Theory" anything, and its negation, is valid. > > > > Actually, in a sane world, the fact that your mapping > > produces a bijection from *N to P(N), using up all the > > elements of *N, would be enough to convince you that > > the enumeration doesn't go on to map the rest of P(*N). > > But no, somehow you can believe both that your list > > is bijection from *N to P(N), and the VERY SAME LIST > > magically stretches when you want it to, to become a > > bijection from *N to P(*N). > It is the position of standard set theorists that infinite sets just > stretch to infinity without end, and that this is a justification for > declaring that any bijection means equality. WRONG! Sets are not elastic. They are static. > I am simply pointing out that, given this line of thinking, Given TO's line of thinking, chaos rules.
From: Virgil on 31 Oct 2005 16:57 In article <MPG.1dd00db4e999cda698a5a2(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Robert J. Kolker said: > > Tony Orlow wrote: > > >>betters indicates strongly that you will never learn to be one. > > > > > > You have no clue what I'm doing here, do you Bob? > > > > Yes I do. You are a troll. > > > > > > I know what they are, but they don't make sense. > > > > They don't make sense to YOU. But what does? > > > > Bob Kolker > > > > > Things well beyind you, apparently, Bob. You don't even believe in mind, so > it's not surprising. Concluding that TO does not have a mind is merely a simple deduction from available evidence.
From: Virgil on 31 Oct 2005 17:02 In article <MPG.1dd00ff01017385998a5a3(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > It is not a requirement of an infinite bijection that the end > be declared, which is precisely what you are doing in the standard > proof of the unbijectibility of the power set with the set. It is a requirement for bijection (everywhere except in TOmatics) that every element of the domain have a uniquely corresponding member of the codomain and every element of the codomain have a uniquely corresponding member of the domain. Incompleteness of correspondence does not produce bijections, or even mappings.
From: Virgil on 31 Oct 2005 17:08 In article <MPG.1dd01e629b1b79cd98a5a8(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dcab744b1856f9198a57e(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > Randy Poe said: > > > > > > Since f:N->E is both injective and surjective, it is bijective. > > > > > Okay, thanks, how about a proof of bijection between the natural > > > numbers and the binary strings? > > > > Which set of naturals and which set of binary strings? In TOmatics there > > are at least two of each. > > > > For the standard bijection between the Dedekind infinite set of finite > > naturals and the Dedekind infinite set of finite binary strings, map > > each finite natural to its finite binary string representation and voila! > > What do you mean "voila!"? If that's not sufficient for my bijection, then > it's > not for yours. Prove it is a bijection. Show me how it's done, Stud. f(1) = 1 and f(succ(n)) = f(n)+1, where the right hand sides are in standard binary arithmetic. By induction, it is done! Or does TO need more help with binary arithkmetic? > > > > Since the Dedekind infinite set of finite natural is the only set of > > naturals outside of TOmatics, that bijectin will have to suffice, at > > least everywhere outside Tomatics > Prove it is a bijection. Show me how it's done. See above.
From: Virgil on 31 Oct 2005 17:14
In article <MPG.1dd01ecab8eb559298a5a9(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > Could it be that TO is wrong? Again? > It's not at all obvious. In riogorous math you have to prove the bijection is > valid. Are you not up to the rigors of your own field of study? In the Dedekind infinite set of naturals, for each natural, n, there is a successor, succ(n). For each finite binary string, s, there is an algorithm for producing s+1. By induction, f(first) = 1, f(succ(n)) = f(n)+1, does it. |