An uncountable countable set
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From: Virgil on 2 Sep 2006 16:40 In article <1157227963.905905.309230(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David R Tribble schrieb: > > > Dik T. Winter schrieb: > > >> But indeed. If the ordinality of a set is infinite (w or larger), then > > >> so is its cardinality (aleph-0 or larger). On the other hand, if the > > >> cardinality of a set is infinite (aleph-0 or larger), and if it can be > > >> well-ordered, so is its ordinality after well-ordering (w or larger). > > > > > > > mueckenh wrote: > > > As long as the set of natural numbers includes only finite numbers, its > > > cardinal number is also finite. > > > > What is that cardinal number? Do you have a name for it? > > The set is potentially infinite. It has no cardinal number in the sense > of set theory. Does "Mueckenh" claim that the set of all naturals is incapable of being bijected with any other set? That is what "Mueckenh" is implying when he claims that the set of all naturals does not have any cardinality in the sense of set theory.
From: Virgil on 2 Sep 2006 16:43 In article <1157228031.247138.146740(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1156363517.395289.207370(a)i42g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Tony Orlow schrieb: > > > > > > > > > > > Ordinals and cardinals are necessities if we want to talk about > > > > > set "order" and "size" in any kind of logical, well-defined way. > > > > > > That is wrong. We can talk about finite sets and about infinite sets > > > which all have the same magnitude by the measure of intercession > > > instead of bijection. > > > > "Intercession"? What sort of measure is that? > > > > > > > > > > > > What do you call the "size" of a countable set with no end? > > > > > You don't have a name for it, do you? > > > > > > > > No, I find focused concentration on the Twilight Zone, the "boundary" > > > > between finite and infinite, to be a rather fruitless exercise. There is > > > > no such distinct boundary. > > > > > > > > > > > Because there is no boundary at all. All sets are finite, but some are > > > without end. > > > > Since "infinite" commonly means endless, that is a bit tricky. > > > > > > > > > > Those are called potentially infinite. All of them have > > > the same magnitude if measured by intercession. > > > > > > Definition: "A and B intercede": An order can be defined such that: if > > > b_1 and b_2 are elements of set B, then a at least one element a of set > > > A lies between them in this order, and if a_1 and a_2 are elements of > > > set A, then at least one element b of set B lies between them in this > > > order. > > > > > > > > Example: Rational and irrational numbers intercede in the natural order > > > by magnitude. > > > > But when one well orders the rationals, one apparently changes the size > > of the set without adding or removing a single member, since it can now > > be made to intercede with the naturals. > > > > I find any measure of set size that depends on order to the extent that > > reordering without adding or deleting any members changes the size, to > > be ridiculous. > > All infinite sets (of finite numbers) intercede each other. There is > only one equivalence class. It is not depending on order. "Mueckenh" 's claim implies that any infinite set must surject onto its power set. But that has been shown to be false.
From: Dik T. Winter on 2 Sep 2006 21:56 In article <1157193775.450075.276010(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > Quantifier dyslexia is the assumption that there is a number that > > > counts all natural numbers, i.e. which is larger than all natural > > > numbers. > > > > That is not quantifier dyslexia. That is an error by Cantor I already > > did comment on. That is, when he writes that to count all natural > > numbers you need numbers of the second cardinality (or was it the first? > > Die erste Zahlenklasse (I) ist die Menge der endlichen ganzen Zahlen 1, > 2, 3, ..., auf sie folgt die zweite Zahlenklasse (II), bestehend aus > gewissen in bestimmter Sukzession einander folgenden unendlichen ganzen > Zahlen; erst nachdem die zweite Zahlenklasse definiert ist, kommt man > zur dritten, dann zur vierten usw. > > > I disremember what he called aleph-0), he was wrong. To count all natural > > numbers you need only natural numbers. > > Great! So we have complete consensus now. Excuse me that I did overlook > your previous statement concerning this. I suspect not, because I think you are misreading about the error I wrote about. > > On the other hand, the "size" > > of the set of natural numbers is not a natural number. > > It is not a number at all. Well, we mathematicians are willing to call things numbers whenever we feel like it. The term is not sacred. Cayley numbers, cardinal numbers, p-adic numbers, cardinal numbers, and you can go on. Whenever you want to talk particularly about some kind of numbers you should state what kind you are talking about. But I have stated this before. But if you wish, provide a definition of "number". As far as I know, there is not one in mathematics. > > Note, you may critique Cantor's set theory, but that was set theory in > > its infancy. It still did contain inconsistencies and errors. Since > > that time quite a bit has been developed and corrected. > > In particular by intermingling the different meanings of infinity. In particular by just separating the meanings of potential and actual infinity. > > Right. Because 0.111... has only finite digit positions we have: > > (1) *each* digit position can be indexed > > (2) *each* digit position can be covered > > but not > > (3) the number can be covered. > > Your "each" means in symbols of logic: "A = (for) all". > The number is nothing than all of its digit positions. > Therefore your statement is a self contradiction. Where? In logical terms (A meaning "for all" and E meaning "there is"): (1) A{p = digit position} E{q = list item} {such that q indexes p} (2) A{p = digit position} E{q = list item} {such that q covers p} but not (3) E{q = list item} A{p = digit position} {such that q covers p} this is simply false, just as: (4) E{q = list item} A{p = digit position} {such that q indexes p} is false. Again, your interpretation of "for all" is false. It does *not* mean "for all at once", it means "for each one individually". If you disagree, and think they mean the same, in your opinion also the statement: (4) there is a number that indexes all digit positions. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 2 Sep 2006 22:16 In article <1157193952.357765.296410(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1156842504.966630.131290(a)h48g2000cwc.googlegroups.com> muecke= > nh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > [ talking about {1, 2, 3, ..., w} ] > > > > > > Let us remove omega from that set. What is the resulting > > > > cardinality > > > > > > Not an actually infinite one, because without omega there are only > > > finite natural numbers. > > > > But what then? > > It is a potentially infinite set, as described by Peano and by all the > mathematicians before Cantor. Yes. I asked you about the cardinality, not what it was. What is the cardinality of that set? > > > > As there is a bijection between that set and the set including > > > > omega, their cardinalities are the same. > > > > > > No. Without omega there is no actual infinity. > > > > But there is a bijection. And two sets that are in bijection with each > > other have the same cardinality. You may think that is a stupid > > definition, but nevertheless, it is a definition that works. > > Look at my last post concerning the final proof that 0.111... with > actually infinitely many digits does not exist. That is not an answer to the question I asked. I ask about the cardinality of a set and you come back with something completely unrelated. (But see my rebuttal of that article.) > > > You should know that the n-th natural number is defined by the sum of > > > the number n-1 plus 1 (Peano). This can be retraced to the sum of n > > > 1's. What is there to be defined? If there are infinitely many naturals > > > (and if infinity is a number), then there is at least one infinite sum > > > of 1's (one from each natural). > > > > Assuming that that sum terminates. Which it does not as there is no last > > natural number. Not even set theory makes that a terminating sum. And > > this is regardless of whether infinity is a number or not. > > Cantor defined aleph_0 in this manner: "Da aus jedem einzelnen Elemente > m, wenn man von seiner Beschaffenheit absieht, eine "Eins" wird, so ist > die Kardinalzahl [von M] selbst eine bestimmte aus lauter Einsen > zusammengesetzte Menge, die als intellektuelles Abbild oder Projektion > der gegebenen Menge M in unserm Geiste Existenz hat." But if you think > he was wrong, there is no need to discuss this topic. Where is he talking about addition? "when man von seiner Beschaffenheit absieht"; I would translate this as "when you disregard all properties". And so, in the transformation the ability to add is lost. And so you get a set (the current, proper, term is multi-set, I believe) consisting of only ones. > > > > > > OK, I kind of understand. I do not state that infinity is a number > > > > aleph_0. There exists a host of infinities. And aleph_0 is the > > > > infinity that gives the equivalence class of sets that are in > > > > bijection with the set of natural numbers; the canonical set with > > > > cardinality infinity. > > > > > > If aleph_0 > n for n e N, and if there are even numbers x > aleph_0, > > > then aleph_0 is a number. Cantor stated that. If you do not do so, then > > > we are in agreement in that point. > > > > You may call it a number, or not. I prefer to call it a cardinal number. > > I prefer not to use the word "number" singly, unless there is context > > that shows what kind of numbers I am talking about. So I may be talking > > about the numbers in the ring Q(sqrt(-3)), or even about the integers in > > that ring. Or about the 5-adic numbers, or the Cayley numbers, but > > always with context. I think that in his early papers he indeed called > > them "Zahlen", but in later papers he did call them "M?chtigkeit". If > > you read his papers you may have found that he changes terminology > > between papers. Understandable, because it was an early development. > > On the other hand, you will see letters by Kronecker where he calls > > the elements of Q(sqrt(-3)) or somesuch "Zahlen" (while he already > > much earlier tried to remove all non-integers from proofs). > > My point is only that: aleph_0 is not in trichotomy with natural > numbers and other cardinals. Prove it. > > What is the cardinality of the set of all finite natural numbers? > > It has none. Why not? Cardinality has been defined so it applies to *each* set. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 2 Sep 2006 23:04
In article <1157194212.124804.231590(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > I did already tell you which edges belong to the path 1/3. The nodes > > > can be enumerated like the edges. > > > > Yes. But you still only have terminating paths. At each level all paths > > are terminated by a node. And so 1/3 is not in the tree because there > > is no terminating edge and node. > > Then such numbers like 1/3 do not exist (in that representation). Indeed, in your tree with terminating edges, such numbers do not exist. > But > the same holds for Cantor's list. I do not see how you can jump to that conclusion. > > Oh, well. As in mathematics the reals require a construction process, so > > also your tree requires a construction process. In mathematics the reals > > are constructed from the rationals (and I know at least four methods to > > do that, that can be shown to give equivalent results). And the rationals > > are constructed from the integers. I asked you before, but you never did > > reply. Do you know how the rationals are constructed from the integers in > > mathematics? More basic, do you know how arithmetic on naturals is > > defined, based on the Peano axioms? > > More to the topic: Do you know how Cantor's diagonal is constructed > from a list of reals? And how this list is constructed? Again, the second proof was *not* about the reals. But I know how the diagonal is constructed. I have no idea how the list is constructed. But again (it appears that I have to repeat myself a lot in this discussion), there is *no* construction of the list given. The proof is along the lines of: "given *any* list, I can show such". So the proof is valid without regards to the actual construction of the list. > > > There is no process of construction. If you have difficulties to > > > comprehend that: it is the same as with Cantor's list. The tree is > > > defined once and for all. That's it. > > > > Wrong. But I am not going to explain that again. > > What is the difference between my tree and Cantor's list. There was no > explanation and there is none, because there is no difference. You give a construction process for your tree. In the case of the list there is no construction process involved. The proof is about "given *any* tree". > > > All real numbers of the interval [0, 1] are there, some of them even > > > twice. > > > > Many of them are not. Your tree only contain numbers with a finite binary > > expansion. 1/3, 1/5 and 1/7 are not among them. > > But Cantor's list only contains numbers with an infinite binary > expansion? In particular his diagonal construction. The list is not a list that is constructed, but a list that is given, or assumed. No construction involved. Whether the list contains finite binary expansions or not is completely irrelevant. The process with Cantors proof is more like the following: give me a list of infinite sequences of 0's and 1's, and I show that there is a sequence that is not in the list. Do you have problems with proofs by contradiction? > > > You seem to misunderstand the tree. If the diagonal is in Cantor's > > > list, then 1/3 is in my tree. Can you give a reason why it should not > > > (other than that then set theory is inconsistent)? > > > > I have explained already many times, but you are not willing to listen. > > Please give a reference, or better: copy and paste your explanation. I have repeated my argument above. But I think you will not read it. 1/3 is not in your tree because all edges (and hence paths) are terminating. The diagonal is not in the list because it is defined in such a way that it can not be in the list. There is no relation between the two. > > > You recently mentioned an interesting aspect: The algebraic numbers, > > > i.e., the polynoms are countable, because they are finite. And, > > > therefore, you wanted only to count the finite segments of paths in my > > > tree. I oppose, unless you agree that the 1's in 0.111... also are > > > uncountable. You see, the problem is the same: We can count finite > > > segment but not infinity. > > > > You are using a pretty strange terminology. You can count the algebraic > > numbers just because the polynomials remain finite (but there are > > infinitely many of them). You can count the finite paths because they > > are finite (but there are infinitely many of them). You can count > > the digits because each digit position is finite (but there are infinitely > > many of them). What inconsistency? > > You can count the levels of my tree just because they remain finite > (but there are infinitely many of them). Right. > The number 1/3 has only finite digit positions at finite levels of my > tree. Where should any uncountable edge appear? Depends on how you count. If you count only the edges leading to 1/3 you can do that (but never reach 1/3). If you want to do the same with 1/5 the case is similar. The problem appears when you want to count *all* edges. At each level, the number of edges is 2^n - 1 or somesuch. But also at each level we still have paths that need to be distinguished. > > > Either you agree that all the edges of my tree are countable or you > > > agree that Cantor's diagonal is uncountable. Or you state at least how > > > many infinite path in my tree you would consider to have completely > > > countable edges. > > > > This question makes not sense to me. > > The problem is the following: You assert that the digit positions of > 1/3 are countable as well as the levels of my tree (all of them), but > you deny that the edges at these levels are not all countable. As I have stated again and again, if your edges *do* terminate, the number of edges is countable, but 1/3 is not in your tree. There is no terminating edge that leads to 1/3. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |