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From: Daryl McCullough on 31 Oct 2005 15:40 Tony Orlow says... > >Randy Poe said: >> If there exists y such there is no x with f(x) = y, then >> you provably do not have a bijection. By definition. >That is true if y is well defined, but the entire set, with no >identified end, is not a well defined set in the sense that >you could ever map anything to it besides an element that also >has no end, like ......1111. That proof is a "no >last element" proof by contradiction, and as such, is invalid for proving >anything else besides that there is no last element. Look, we have the following facts: 1. If x maps to the entire set (using your mapping), then x would be the largest element. 2. There is no largest element. 3. Therefore, nothing maps to the entire set. 4. Therefore, the mapping is not a bijection. >Which even maps to the last finite natural? There is no largest finite natural, and there is also no largest finite even number. So there is no problem. In your case, however, there is no largest element, but there *is* a largest subset (namely, the entire set). That's the difference. You have a set *N with no largest element. You have a second set P(*N) that *does* have a largest element. How can you have an order-preserving bijection between a set with no largest element and a set with a largest element? >There is no last finite natural in the unending set? Huh! >There is no last element in the unending set of subsets >of the unending set of naturals, either. But there is a largest subset, namely the entire set. Does any element map to this largest element? >> Do you have a specific objection to my map? > >Absolutely. It's the same objection Albrecht has to your >infinite set of finite naturals: it's finite. >As a finite set, unbounded or not, it cannot map every >element of an infinite set like *N uniquely. Then show what element of *N does not correspond to any subset of N. -- Daryl McCullough Ithaca, NY
From: Virgil on 31 Oct 2005 15:05 In article <MPG.1dcff6fff44c324198a59a(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dc9588314461aa298a572(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > Virgil said: > > > > > > If it did suffice, TO would be able to give the argument that > > > > is mapped into the set of all elements not mapped into their > > > > images. > > > > > The set of all elements not mapped to sets which contain them is > > > the set of all elements, since none map to sets which contain > > > them. > > > > A perfectly legitimate mapping from any set, S, to its power set, > > P(S), is the function that maps each element to the singleton set > > which contains it, f(x) = {x}. For this function, {x in S: x not in > > f(x)} is the empty set, even though TO in his perpetual state of > > confusion claims it must be the whole of P(S). > Are you able to stick to the bijection I offered? Distraction and > irrelevance are not valid modes of argument. > > > > Since the composition of {x in S: x not in f(x)} depends critically > > on the nature of the function f, all one can say about such sets, > > in general, is that for each f from any set S to its power set > > P(S), there {x in S:x not in f(x)} exists as a subset of S and > > member of P(S). > In the bijection I offered, the natural bijection between an ordered > set and its power set, every element maps to a set which contains a > greatest element no greater than log2 (order-wise) of that number. > > > > > The natural which maps to it is represented by an unending string > > > of bits. > > > > According to TO's description of "TOnaturals", they are all > > infinite strings of bits. > Yes, but not all have infinite numbers of significant bits. TO did not say that they had to be significant, and it is quite true that all of his strings are alleged to have infinitely many bits. . > This has > all bits equal to 1, with no end. Where does it fail? I do not see how having a string with infinitely many 1 bits provides a string which maps to {x in S:x not in f(x)} > > > > > > > > > This is the number which maps to the entire set. :D > > > > But why does TO insist that the set of naturals that map to sets > > not > > containing themselves contains all naturals, How does he know that > > no natural maps to any set containing itself? Such a wild claim > > requires proof, or at least some sort of evidence beyond the mere > > claim itself. > I already explained before, and now above. The singleton {x} maps to > 2^x. It may map FROM 2^x, but it certainly does not map TO anything at all under a mapping FROM S TO P(S). > it will be 2^(x+1)-2^x elements before we get the next element, > x+1, as a singleton set. It should be obvious to you that no element > can ever be a member of the set to which it maps. I gave a listing to > illustrate this. Demanding rigorous proof of the obvious is a waste > of time. With your experience, i am sure you can whip up a proof of > this in five minutes. It is not my respponsibility to "whip up" proofs of TO's allegations. If he cannot, or will not, do it, his claims may quite properly be rejected. > > > > If TO can prove this, there is a lot about his mapping from *N to > > P(*N) that he is carefully not telling us. > Like what? That is precisely what we would like to know! > > > > > > But it's not a bijection between *N and P(*N). It's not a > > > > > > bijection between *N and P(N), or even between N and P(N). > > > > > > It is a bijection between N and some of the members of > > > > > > P(N). Obviously that's not good enough. > > > > > > > > > That's not what it is. I have declared it to be between *N > > > > > and P(*N). > > > > > > > > To declare something does not make it so, at least outside of > > > > TOmatics. > > > > > It's my bijection, after all. > > > > > > While it is undoubtedly yours, whether it is a bijection is a matter > > of contention. > > > > > I can declare it to be whatever I wish. > > > > Such a declaration is invalid outside TOmatics without evidence to > > back it up, and so far, the evidence all runs counter to that > > claim. > Like what? Like the fact that , while for any f:S ->P(S), the set {x in S:x not in f(x)} is a well definied member of P(S), allegeing that there is any s in S such that f(s) = {x in S:x not in f(x)} is self-contradictory. > > > > > > > > Outside of TOmatics, declarations require proofs before they > > > > need be accepted. > > > > > > > Not when you are defining a bijection. Unless TO has a different definition of bijection than anyone else, while one can define a map, and even that requires meeting some standards, one must then prove that that map is a bijection. > > > > To declare that something is a bijection is to claim it has certain > > properties. If it does not, in fact, have those properties, then no > > declaration will make it into what it is not. > > > > Unless TO has a function from the complete, entire, finished *N to > > the complete, entire, finished P(*N), he does not have a relevant > > function at all. > You tell me how *N finishes, and I'll give you the natural which maps > to it. Fair enough? What is *N precisely that it can exist as a set but be so ambiguous as a set as to have the membership question ever unanswerable? One cannot have a set for which the membership question is unanswerable. Whatever one might have under those conditions, it is not a set. TO keeps conflating processes, which can be endless, with sets, which must be entirely determinate. > > "No largest natural! (gong) Huyah huyah Ommmm....." TO at his most literate > > > > And unless such a function,f:*N -> P(*N), once defined and > > "complete", can be shown to map some member of *N to {x in *N: x > > not in f(x)}, he does not have what can be called a bijection > > outside TOmatics, regardless of what TO chooses to call it inside > > TOmatics. > Except that there is no point at which the bijection fails, since > both *N and P (*N) are unending. Then neither N* not P(*N) are sets and there is no such thing as a function from one incomplete object to another incomplete object, much less a bijection, except in TOmatics, where everything is both possible and impossible. > > > > > Like which, for instance? > > > > > > > > The set of all members of *N not in the set they map to. > > > > > That's the entire set. But if any element maps to T = {x in *N:x not in f(x)}, then it cannot be in T and we cannot then have *N = T, as TO insists we must have. > > > > TO keeps saying that, but never gives any reason why, even though > > examples have been given showing that it can even be the empty set. > See above, keep in mind my general bijection, and read, for god's > sake. No element is in the set to which it maps. Therefore, the set > of all elements not in the subsets to which they map includes every > element in the entire unending set. But if T = *N then, to have a bijection, one must have some element mapping to T = *N which is, of necessity, a member of *N, and we have a contradiction due to the assumption that a bijection can exist. > > > > Considering TO's track record with unsupported statements, and > > considering that there are examples showing that TO's claim need > > not be true, we must regard any such TO claim as false, at least > > outside TOmatics, until proven otherwise. > Considering that Virgil can't even stay on topic, I don't know why I > even bother responding to his posts. My topic is perpetually, that the illogic of TOmatics is invalid in standard mathematics. > > > > > > > > > > > > > > > > > So fill in the blanks, and show us those other mappings. > > > > > > Otherwise your "bijection" is incomplete and is not, in > > > > > > fact, a bijection between *N and P(*N). > > > > > > > > > WHich other mappings do you want? What would satisfy you? > > > > > > > > A mapping which maps some member of *N to the set of all > > > > members of *N not in the set they map to. > > > ...111111111111111111111111111111111111111111111111111111111111111 > > > 1111 1, with oo^oo^oo^oo^oo......... 1's. Now blow out the > > > candles, Virgil. > > > > How are we to know whether the above abomination does or does not > > map to the set of all members of *N not in the set they map to ? > It's got a 1 for every element in the set. I already explained > (multiple times, now) that my bijection does not have any elements > that are included in the sets to which they map, so the number you > are asking for is the natural which maps to the entire unending set. Then you are saying that T = {x in *N:x not in f(x)} = *N, which is a self-contradiction. > > TO's claims, being untrustworthy, require supporting evidence if > > they are to be acceptable outside of TOmatics. And what happens > > inside is irrelevant to sane mathematics anyway. > Sanity, like evil, is often relative. The sanity of the one, TO is the insanity of the many,,everyone else.
From: Virgil on 31 Oct 2005 16:32 In article <MPG.1dcfff7f104b00a498a59e(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > albstorz(a)gmx.de said: > > You argue: there is no infinite natural number since the peano axioms > > don't allow an infinite natural number. > > That's right. I agree with you. > > Alas, Albrecht, it is not true. It is outside of TOmania!.
From: Virgil on 31 Oct 2005 16:44 In article <MPG.1dd0002f705e945e98a59f(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Of course, you cling to the notion that inductive proof only works > for finite naturals, a rather circular and self-supporting assertion, > that makes it conveniently unavailable for the topic at hand. The inductive axiom in the Peano axioms say that if you have a statement about some natural which is true about the first natural and whenever it is true for some natural it is also true for the successor of that natural, then it is true for every natural. If we define sets being finite or infinite by the Dedekind criterion, then we can show that the set of naturals less that or equal to a given natural is always a finite set but the set of all naturals is not a finite set. If our definition of a finite natural is then that the set of naturals less than tor equal to it is a finite set, we can prove inductively that every naturals is finite. This is all that is needed for standard mathematics. To the extent that TO assumes conditions contrary to these, he is making up a dreamworld that has no relevance to standard mathematics now, nor will it have in the future.
From: Virgil on 31 Oct 2005 16:46
In article <MPG.1dd0080cc699f44298a5a0(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > All this proves is that there is no size of the set, as Albrecht has been > saying, since the set size is equal to its largest element, which doesn't > exist. Great proof. Try again. This proves nothing about the infinitude of > the > set. By the Dedekind definition, which is the only relevant one, it does. |