Cantor Confusion
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From: Carsten Schultz on 2 Feb 2007 20:12 mueckenh(a)rz.fh-augsburg.de schrieb: > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. From that it follows that there are exactly four sets with four elements, since these are the elements of 4. It also follows that there is only one set with four elements, namely four. So 4=1. You should write a book about this. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Dik T. Winter on 2 Feb 2007 21:07 In article <1170413330.341310.269660(a)p10g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 2 Feb., 02:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170348341.624257.130...(a)p10g2000cwp.googlegroups.com> mueck.= .... > > If 4 is the set of all existing (whatever that does mean) sets with 4 > > elements and 5 is the set of all existing sets with 5 elements, we find > > immediately that 4 is *not* a subset of 5. With this definition of > > numbers as sets subsetting does not give what you wish. > > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > elements. That can not be true. 4 can not be all kinds of different things at once, at least not in mathematics. With this kind of definition 4 contains 4. > > > Therefore we have to investigate whether such a representation exists. > > > Note that I use the definition of a number by Peano in order to look > > > for the existence of that number. > > > > What does the latter sentence *mean*? But indeed, for most numbers some > > representations do not exist, while other representations do exist. The > > existence of a number is independent of the existence of a representation. > > This point of view has lead to the present mess-math. > > > For instance, for most rational numbers a decimal representation does not > > exist. > > Correct, for instance for 1/7. And for computable numbers some representation does exist. > > > The definition of a number is given as non-circular as possible: p. 3: > > > 1) 1 ist eine nat=FCrliche Zahl. > > > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. > > > > in N is wrong here; N is not defined. > > N is already in (1), because (1) is identical to "1 is in N". N is not *defined*. Whether you use it in (1) is irrelevant, what is relevant is that it is not defined. > > A more proper version is: > > 2) Jede nat�rliche Zahl a had einen bestimmten Nachfolger a', und > > dass ist auch ein nat�rliche Zahl. > > What is every natural number if N is not defined? This is a recursive definition of natural numbers. By (1) we have one natural number, by (2), from that single natural number we get a lot of other natural numbers. > > > 3) Es gibt keine Zahl mit dem Nachfolger 1. > > > 4) Aus a' = b' folgt a = b. > > > 5) Jede Menge M von nat=FCrlichen Zahlen, welche die Zahl 1 und zu jed= > er > > > Zahl a in M auch den Nachfolger a' enth=E4lt, enth=E4lt alle nat=FC= > rlichen > > > Zahlen. > > > > Yes, I know Peano pretty well. But this is *not* a circular definition. = > I > > wonder why you think it is circular. > > I did not say it is circular. I said "as non-circular as possible". If > it is not circular in your opinion, then be happy. (In fact every > definition is circular, because every language is. You cannot explain > anything without already using some unexplained wording. If you want > to explaine the unexplained by the words already explained, you get > circular. That is unavoidable. Cp. N being undefined but appearing in > the Peano axioms. But that is not the point here.) This is getting phylosofical. In mathematics a definition is circular if in the definition one of the deciding features is the term you want to define. So a definition as you gave: 3 is the set of all sets of 3 elements is a perfect example of a circular definition. The reason is that it states precisely nothing about what 3 is. It is not better than the definition: 3 is 3. In order to know whether a particular set fits in 3 you have to know what 3 is, and to know that you have to know whether that particular set does fit. > > > p. 130: > > > > > > 1) 1 in M. > > > 2) If n in M then n + 1 in M. > > I prefer the successor of n, rather than n + 1 here. > > I think that everybody able to read and understand these lines will > know what "+ 1" means while the successor is not immediately clear. > (The successor of n coul be n+2 or 2*n or 10*n or ....) Yes, indeed, and that is the crux. Abstraction. When you use the '+ 1' notation you are already assuming the existence of addition. When you do not, you can properly define addition (and all other operations) using the Peano axioms. You are *defining* the natural numbers, presumably without any knowledge about what natural numbers even are. And I may note that with '2*n' you can get a set that is isomorphic (with respect to all operations) to the natural numbers, only the naming is different. Consider the set of powers of 2, (your 2*n case) define: a '+' b = 2^[ log_2(a) + log_2(b) + 1 ] a '*' b = 2^[ log_2(a) * log_2(b) + log_2(a) + log_2(b) ] Call this set K. You may verify that the rings: R(N, +, *) and R(K, '+', '*') are isomorphic. > > And *in this context* the definition of 3 is: > > succ(succ(1)) > > no set at all. And the existence of 3 follows from the first set of > > axioms you gave. > > The existence does not follow from axioms. Axioms state something in > an arbitrary way. They can even state the famous pink elephant. The non-existence of a pink elephant in the realm of the axiom that states that there is a pink elephant is not mathematics but philosophy. Axioms state what things exist (or do not exist) in their realm. In geometry there is the parallel axiom (that you also use as example for the axiom of choice in your book). In two forms of geometry there exists a line through a point parallel to a given line not through that point. In one form of geometry such a line does not exist. (Exist meaning exist mathematically here.) The axiom states existence or non-existence, and that is all. > Whether it exists remains to be investigated. Your existence is not a mathematical existence. > > Your definition is circular, this one is not. > > I do not give a definition but I look for the existence of the number > already defined. So when I ask you for a definition you do not give a definition? Yes, I have been thinking that all along. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: cbrown on 2 Feb 2007 21:34 On Feb 2, 1:49 pm, David Marcus <DavidMar...(a)alumdotmit.edu> wrote: > Dave Seaman wrote: > > On Fri, 02 Feb 2007 20:29:03 GMT, Andy Smith wrote: > > > Andy Smith <A...(a)phoenixsystems.co.uk> writes > > > (snip everything else) > > > > At root I think my problem comes down to achieving a suitably Zen-like > > > perspective on the following apparently incompatible statements: > > > > 1) The real line is made up of an ordered and infinite set of points, > > > and is connected. > > > > 2) No point on the real line has an adjacent point. > > > I don't understand why you think those two statements are incompatible. > > If any point on the real line actually *had* an adjacent point, then the > > line would be disconnected precisely at the gap between those two points. > > Hence, connectedness is incompatible with the existence of adjacent > > points. > > I suppose he is thinking of points as having size, e.g., like little > marbles. Of course, they aren't like that. > Just to screw with Andy's head... IIRC, Rudy Rucker in "White Light" talks about a book with an uncountable number of pages. I think it was an Encyclopedia of Curves, containing an image of every curve in R^2, spread across two adjacent pages, and of course drawn in lovingly accurate detail. The pages were numbered; but to number the pages, they used an infinitely long string. This was written at the bootom of the page, with the first line in 1pt type, the second line in 1/2 pt type, and so on with each succesive line being written in a type size 1/2 the size of the previous line. You can certainly open this book to any particular page; but the pages are so thin that no matter how hard you try, you always end up flipping a bunch of pages, not a single page. (Which is only to say.: Pages cannot be planes, any more than marbles can be points!). Cheers - Chas
From: cbrown on 2 Feb 2007 21:44 On Feb 2, 5:12 pm, Carsten Schultz <cars...(a)codimi.de> wrote: > mueck...(a)rz.fh-augsburg.de schrieb: > > > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > > elements. > > From that it follows that there are exactly four sets with four > elements, since these are the elements of 4. It also follows that there > is only one set with four elements, namely four. So 4=1. You should > write a book about this. > Dik is even reviewing it, I think. Cheers - Chas
From: Dik T. Winter on 2 Feb 2007 22:12
In article <1170413742.648825.136900(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > Frankly, I do not see much difference to: The union of finite paths > > > contains an infinite path. > > > > But that is true. > > So you say: > The union of finite trees contains an infinite tree. > The union of finite chains contains an infinite chain. > The union of finite paths contains an infinite path. Indeed. > > > What I state again, again and again, but what you > > misread each and every time is: "the union of sets of finite paths > > does not contain an infinite path". > > So you say the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > contains the infinite path p(oo) = {0,0,0,...}. > > But the union of finite sets of finite paths > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > does not contain the union of finite paths > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... Again, indeed. > I would say: The union of all finite sets of finite path contains all > finite paths. Right. As I have stated all the time. And so the set of paths in the complete tree is *not* a subset of the union of all finite sets of finite paths, something you have stated repeatedly. > The union of these finite paths can be consructed (it is > a set of nodes) and is realized in the binary tree by its nodes. It > contains all infinite paths of the infinite tree. > > The tree contains the union of all finite sets of finite path as well > as the union of these paths. Right. But you *use* that the union of all finite sets of finite paths *contains* the union of these paths. > > > And it is not contained in any of the finite trees. Why is it said to > > > be contained in the union of all finite trees? > > > > Because an infinite path is a union of paths. And when you construct > > unions of *sets* of paths you are not constructing unions of paths. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. What definition of union do you use? Ah, I am seeing now. The axiom of sumsets. But I am *not* talking about a single set, I am talking about *multiple* sets. To rephrase your wording so that they are applicable: > The union of sets of sets of elements is not a set of elements, but a set > of sets of elements. To know why the axiom of sumsets is also the axiom of union, you have to consider the following: given two sets A and B by the axiom of pairing there exists a set {A, B} by the axiom of union there exists a set U{A, B} consisting of the elements of A and B. That set is normally called the union of A and B. I allow that the terminology is a bit confusing. But U applied to a single set of sets yields the set of the elements. But U applied to multiple sets does not. And you are applying it to multiple sets, so it does not. > The union of one ot more sets of paths, as I defined it, can be a > path. But is not in the union of the sets of paths. Pray follow the logic. > > > Because there cannot be more than infinitely many natural numbers. All > > > those infinitely many numbers, however, are allegedly finite. > > > > And there are infinitely many such. And they are, indeed, all finite. > > The path lengths however do not map to natural numbers, which you assert. > > That is valid for finite paths, but not for infinite paths. > > The finite natural numbers map on the finite pathlengths. > The finite segments of natural numbers map on the finite pathlengths. > > If there is an infinite pathlength, then there must be an infinite > number as well as an infinite segment of natural numbers. Why must there be an infinite number? But rest assured, there may be an infinite segment (depending on how you define segment). > You say, the former does not exist in N while the latter does exist in > the form of the whole set N. > > That is a contradiction. You see that there is no infinite pathlength > possible without an infinite pathlength. This is gibberish. I only state that infinite paths have infinite pathlength, and that there is no natural number that maps to. What is the contradiction? > > Let's, for > > once, go to a proper definition (assuming all are ordered sets): > > given a path {n_1, n_2, ..., n_k} then the path length is > > |{n_1, n_2, ..., n_k}| = k. > > Clearly for finite paths the path length is finite, and so a natural > > number. However, when we construct the path: > > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > > we get as path: > > {n_1, n_2, n_3, n_4, ...} > > the path length in this case is *not* a natural number. It is the > > cardinality of N. > > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. No, I do not see it. You state that the pathlength is a number. Let me grant that. So omega is a number. But I do not see why N should contain omega, because by definition it is the set of *natural* numbers, and omega is not a natural number. It all boils down to the same story. Let A be the set of natural numbers plus omege. A is infinite (I think you agree). Remove omega from A. What do you have now? I state that you now have N. Apparently you disagree. Why? I also state that because A is infinite, N is also infinite. But for some reasons you state that N is finite. For what reasons? > > > But you said that there is an infinite pathlength corresponding to an > > > infinite set of paths. > > > > But that set of paths is *not* the union of the P(i). > > Above you said that the set of all paths p(n) contains the infinite > path p(oo). Yes. But not that the union of the P(i) contains an infinite path. > > The union of sets of finite trees would neither contain, nor establish, > > an infinite tree. It would not contain one, because none of the > > constituent sets contains an infinite tree. It would not establish > > one, because the union is a set of trees, not a tree. > > The union of a set of sets of elements is a set of elements, not a set > of sets of elements. But I am not talking about the union of a set. I am talking about the union of set*s*. Lets analyse. Given two sets A and B: by pairing {A, B} exists by union U{A, B} exists. The definition of A U B is U{A, B}. > The union of a set of trees, as I defined it, is a tree. I was talking about the union of set*s* of trees. > > A tree is a > > set of nodes, a set of trees is a set of sets of nodes. > > And the union of two or more sets of nodes is a set of nodes. Right. > Therefore the union of a set of trees, as I defined it, is a tree. And I never did contradict this. > > > > > > Wrong. Infinite paths do not have a path-length that is a > > > > > > natural number. > > > > > > > > > > Pathlengths ARE natural numbers. > > > > > > > > For finite paths. > > > > > > In all cases, by definition. > > > > *What* definition? > > Definition: Map the pathlength x on the number x. Fine, I do the mapping. How do I get that x is a *natural* number? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |