Cantor Confusion
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From: Virgil on 29 Nov 2006 13:47 In article <456DC03A.1000206(a)et.uni-magdeburg.de>, Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: > On 11/29/2006 5:24 PM, mueckenh(a)rz.fh-augsburg.de wrote: > We have to accept thart there are sets which are capable of > > growing, as Fraenkel et al. expess it. > > While there are several books by Fraenkel: > > Einleitung in die Mengenlehre 2nd ed. 1923, 3rd ed. 1946, > Gesammelte Abhandlungen Cantor, Dedekind 1932 > Das Leben Georg Cantors 1932 > Abstract set theory 1961 > Lebenskreise - Aus den Erinnerungen eines j�dischen Mathematikers 1967 > > perhaps there is only one book by Fraenkel et al.: > Foundations of Set Theory 1958. Correct? > > > Then we have finite sets without > > a largest element. > > Why not with a growing largest element? > > Regards, > Eckard Growing elements would allow claims like: "2 + 2 = 5 for large enough values of 2".
From: William Hughes on 29 Nov 2006 14:53 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > The two statements: > > > > > > > > The elements of N satsify property X > > > > and > > > > The set N satisfies property X > > > > > > > > are independent.. Induction proves that the elements of N are finite. > > > > Induction does not show that N is finite. > > > > > > Of course it does not. Induction is a sensible method. By induction > > > only sensible assertions can be proved. The *set generated by > > > induction* is always finite. > > > > > > > Using your interpretation of "set generated by induction" > > > > all sets of natural numbers are finite. > > if all sets of natural numbers can be generated by induction, > > Do you deny this assertion? > > > > So all we have to do to prove that all sets of natural numbers > > are finite is to assume that all sets of natural numbers > > can be generated by induction. > > > > No set of natural numbers without a largest element can > > be generated by induction. > > Do you deny this assertion? > > > > So what we have to do to prove that all sets of natural numbers > > are finite is to assume that there is no set of natural numbers > > without a largest element. > > No. We have to accept thart there are sets which are capable of > growing, as Fraenkel et al. expess it. Then we have finite sets without > a largest element. Then we describe reality correctly. (Leave Fraenkel et al. out of it. Finite sets without a largest element are your idea). O.K. Let A be a finite set without a largest element. What is the cardinal number of A. Clearly it cannot be any finite cardinal number? Recall, you want to do more than just state that A does not have a cardinal number. You want to show that assuming that A has a cardinal number leads to a contradiction. - William Hughes - William Hughes
From: mueckenh on 29 Nov 2006 16:04 Dik T. Winter schrieb: > In article <1164637973.064418.67510(a)45g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > Nothing more than opinion. You think there are no such representations, > > > so everything that states that there are such representations is wrong. > > > > I proved that there are no such representations. If they were (the > > paths in the infinite tree, for instance), then we had a contradiction. > > You are, with your tree, wrong. Please try to understand it first! Up to now you didn't. > In your tree (where the nodes represent > bits), it can be shown that all nodes represent a number: the collection > of bits found when going from the root to the current bit. So the nodes > represent all rational numbers in [0,1) where the denominator is a power > of two. You state that the paths represent numbers. Let us analyse that, > and begin with the finite paths that terminate at some finite edge. In my tree there are no finite (terminating) paths! >When > you consider such paths, you can assign numbers to each final edge: the > same number as the node were it comes from (nodes represent bits, so when > you come at an edge you can only add the bit of the last node you passed, > you can not look in the future). So your paths (whether finite or infinite) > all represent a number that is also represented by a node. In your terminology every digit 3 in 0.333... represents a number. The first 3 represents the number 0.3, the second 3 represents the number 0.33 and the 10th 3 represents the number 0.3333333333. What does this interpretation have to do with the existence of the decimal representations of the number 1/3, namely 0.333... in a list of real numbers? ========================== > In your infinite tree, all edges come fome a node > at finite distance, and so all edges together only represent the numbers > with a finite binary expansion. So all decimal representations represent finite decimal expansions? Or what is the difference? > > Then 1/3 is not in Cantor's list. > It is. Can you imagine to represent the numbers in Cantor's list by paths? > Again, see above. As there is no node at an infinite distance, there is > also not an edge that comes from such a node, and so all edges represent > a number with a finite binary expansion. As there is no digit "3" at an infinite distance from the "0." in 0.333..., there is no infinite decimal expansion. What is the difference to a representations by paths? ================ EIT: > Eh? N is the set of natural numbers. The quantifiers can be exchanged > when N is finite. Not when something completely different is finite. We consider only finite lines. Within every line the quantifiers can be exchanged. Regards, WM
From: mueckenh on 29 Nov 2006 16:12 Dik T. Winter schrieb: > In article <1164637800.628277.302340(a)14g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > You wrote: > > > > > > Cantor wrote: "It is remarkable that removing a countable set > > > > > > leaves the plaine *being connected*." But it is not remarkable > > > > > > that removing an uncountable set leaves the plaine being > > > > > > connected? What a foolish assertion. > ... > > > > Yes, but Cantor considered only point sets which are dense. > > > > > > You did not state that above. > > > > I wrote about algebraic and transcendental numbers. So being dense is > > implied and need not be mentioned. > > Perhaps. But that was not clear from what you wrote. And you even > let the examples where the inner portion of a circle were removed go, > while the apparently do *not* satisfy what Cantor considers! You cannot expect that I carefully read and answer every sentence of Mr. Bader. It is only by accident that I saw and answered his posting (because he seems to be very interested in my papers and to study them in great detail). I usually ignore him. > > > > But under *those* conditions there there are still uncountable sets that > > > leave the plane connected and also sets that leave the plane disconnected. > > > > Of course, but Cantor did not recognize it. And as far as I know nobody > > recognized it before I did. At least nobody mentioned it. > > Of course nobody mentions it, because it is highly unremarkable! Well, after I showed it. Cantor's proof cannot be used to find an uncountable set which leaves the plane connected. > Remove > all irrational points and you still have a connected space. Remove in > addition a line and you have a disconnected space. What is remarkable > about that? What *is* remarkable is that whatever countable dense set > you remove, the result is still connected. And the property of being > dense is not even needed for both results. Just this shows that Cantor, who insists on it, was not aware of the fact that uncountable sets leave the plane connected. Regards, WM
From: mueckenh on 29 Nov 2006 16:21
Virgil schrieb: > For one, any open connected subset of the reals can be proved > uncountable by the same Cantor first proof argument, but disconnected > subsets may not. Cantor does not talk of connected or disconnected, but only of an arbitrary sequence "beliebige Reihe reeller Zahlgrößen" > > > > In addition, two sequences of transcendental numbers can converge to a > > rational number, such that we have the same situation as described > > above. > > The set of transcendentals is not a connected set in R, so violates the > hypotheses of the proof. There is no such hypothesis at all. Where did you think to read that? The *result* of the proof is applied to the transcendental numbers. "kombiniert man die Inhalte dieser beiden Paragraphen, so ist damit ein neuer Beweis des zuerst von Liouville bewiesenen Satzes gegeben, daß es in jedem vorgegebenen Intervalle unendlich viele transzendente, d. h. nicht algebraische reelle Zahlen gibt." > > > > Therefore both sets, Q ???and T (transcendental numbers), have the same > > status with respect to *this* uncountability proof. > > Both are totally disconnected as subsets of the reals (between any two > members there is a non-member) so that the hypotheses of the proof are > violated. There is no such hypothesis at all. The proof applied to these subsets cannot distinguish there cardinality. That is what I mentioned. > > > And we are not > > able, *based on this very proof*, to distinguish between them. > > Why should we, as the proof does not apply to either. As I mentioned, the proof yields the uncountability of the transcendental numbers, when the countability of the algebraic numbers is established. When also the countability of the nonalgebraic numbers is established, the proof shows a contradiction. > > > On the other hand, the proof can show the uncountability of a countable > > set. If, for instance, the alternating harmonic sequence > > (-1)^n/ n --> 0 > > is taken as sequence (1), yielding the intervals (-1 , 1/2), (-1/3 , > > 1/4), ... we find that > > its limit 0 does not belong to the sequence, although the set of > > numbers involved is obviously > > denumerable. > > That doesn't work it unless you show that NO sequence can be made > including all of the values of S = {0} union { (-1)^n/n: n in N} > > But consider the sequence 0, -1, 1/2, -1/3, 1/4, -1/5, ... which does > contain all of the members of S. Consider the sequence in Cantors proof which starts at position 0 with the common limit. > > > > The alternating harmonic sequence does not, of course, contain all real > > numbers, but this simple example demonstrates that Cantor's first proof > > is not conclusive. > > > Wrong! The theorem speaks of ANY sequence, not just hand picked > sequences. Take any sequence of rational numbers selected from their order by magnitude. > > Cantor's proof can be used to show the uncountability of the rational > > numbers. In Q there are sequences which converge to rational numbers. > > But there are also sequences which do not, and it is the existence of > those sequences that do not which scuttles WM's misrepresentation. > Recall in the proof, that the analysis must hold for ANY sequence which > is alleged to count every member! if it fails for any, it fails entirely. Therefore Cantor's proof fails for those sequences which carry their limit as a first element. Therefore Cantor's second proof fails for the list 0.0 0.1 0.11 0.111 .... when replacing 0 by 1. Therefore the real numbers are countable, proven by the paths of the binary tree. ================== > Ey in S Ax in S ((x != y) --> (y > x)) is false in every ordered set, > S, which does not have a largest member, such as N or Q or R. But it is not true in a set of elements of one line of the EIT which always has a largest member. Regards, WM |