An uncountable countable set
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From: Dik T. Winter on 29 Aug 2006 18:42 In article <1156689385.395973.61570(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <virgil-B3927D.12230325082006(a)news.usenetmonster.com> Virgil <virgil(a)comcast.net> writes: .... > > What I have been able to find was that with indexing of digit WM means > > that every digit position can be indexed by a natural number. With > > covering WM means that all positions to a certain one can be indexed > > by a natural number. What WM is asserting is that when a number can > > be totally indexed, it also can be totally covered. Quantifier dislexia > > disguised. > > 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? I disremember what he called aleph-0), he was wrong. To count all natural numbers you need only natural numbers. On the other hand, the "size" of the set of natural numbers is not a natural number. 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. > My arguing holds for the unary numbers of my list. If you think not, > then give an example of a finite number n which indexes the digit > position n but does not cover all positions m =< n. > > Here is my example: The number 3 = 0.111 in my unary expression indexes > the digit position 3 of any unary number and it covers the digit > positions 1, 2, and 3 of any unary number. > > The same can be proven for any finite number n. Yes, I never ever did argue otherwise. So why do you keep on asserting that? > In 0.111... there are allegedly only finite digit positions. Hence this > proof is valid for every position of this number. Every position can be > indexed, and every position can be covered. This is dictated by logic: Yes, I never argued otherwise. So why do you keep on asserting that? > Position n can be indexed ==> every position m < n can be covered. > Not every position m can be covered ==> not every position n > m can be > indexed. 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. Because (3) would mean that there is a number that covers *all* digits, and as that number also indexes a digit position, there is a last finite digit position. But that is not the case. (And this concisely shows that for finite n, the unary representation *can* be covered.) You are repeatedly confusing (2) with (3). They are not the same. -- 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 29 Aug 2006 19:03 In article <1156842504.966630.131290(a)h48g2000cwc.googlegroups.com> mueckenh(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? > > 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. > > > > > Your example shows very clear that either both LUBs have to be included > > > - or none. In my opinion we choose none. But whatever you choose, you > > > cannot avoid to see the symmetry, can you? > > > > I see the symmetry, I also choose none. And I think I have stated that > > already. > > Fine. That means there is no actual infinity. Oh. See above. > > > > Wrong. In mathematics the concept of "infinite sum" is not defined, > > > > > > But if there are infinitely many naturals then the infinite sum is > > > necessary. > > > > Not at all. Why do we need it? > > 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. > > 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). > > > Exactly. And in addition, as long as the numbers are finite, ordinal > > > and cardinal is the same size. That is just why the cardinal cannot be > > > infinite as long as the ordinal is finite (as expressed by "the > > > infinite number of finite numbers"). > > > > But indeed. If the ordinality of a set if 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). > > As long as the set of natural numbers includes only finite numbers, its > cardinal number is also finite. What is the cardinality of the set of all finite natural numbers? But see above. You assert that it is finite. But by the definition, in that case there is a largest number (say n) in the set, but n+1 is also a natural number, so we get a contradiction from asserting it to be finite. -- 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 29 Aug 2006 19:27 In article <1156842646.101182.155640(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1156585631.775313.153510(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > I don't understand these questions. 1/3 and 1/5 are infinite paths > > > consisting of infinitely many edges. > > > > You state that you can number the edges. Each edge terminates at a node, > > and if 1/3 and 1/5 are in your tree, there must be edges that terminate at > > those nodes. > > What do you mean with "those nodes"? 1/3 and 1/5 are not nodes but they > are paths. We can state (if each edge terminates) that at each level we assign the number at that level to both the path and the terminating node of that path. If I am wrong here, tell me why I am wrong. So at each level we have paths with terminating nodes, I meant those terminating nodes (that we can give the same name as the path that terminates there). > >In that case you should be able to tell me what the numbers > > of those edges are. Or do you now claim that there is *no* edge that > > terminates at those nodes? If so, where do the nodes come from? Out of > > thin air? > > 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. > > > > As the set of all edges is countable, you > > > > should be able to state that. The problem with your counting method of > > > > edges (count first level 1, next count level 2, etc.) you will already > > > > have exhausted all natural numbers by counting all edges that are at the > > > > end of finite paths. > > > > > > There are no finite paths and there are no ends at all in my tree. > > > > During your construction process all paths are finite, > > There is no construction process. The tree exists like the real numbers > do exist. 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? > > and there is an > > edge that terminates at a node terminating the path. I meant the paths > > during the intermediate process of construction. > > 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. > > > you think that 0.1 would be represented by a finite path, what should > > > then be represented by the path 0.1000...? > > > > The same. > > > > > By your arguing you would never be able to count the digits of 1/3 > > > because "you will already have exhausted all natural numbers by counting > > > all edges that are at the end of finite" sequences. > > > > Wrong. To count digits we need only one number for each level. > > A set of countably many countable sets is countable. Right. But 1/3 is not in that set. > > > > > Do you believe that the set of edges of 1/3 is uncountable? > > > > > > > > That is not the set of all edges in the tree. > > > > > > No. But would you say that the set of all edges of 1/3 and of 1/5 and > > > of 1/7 were uncountable? > > > > No. Assuming of course that 1/3, 1/5 and 1/7 are in your 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. > > > How many paths are required to obtain an > > > uncountable set of edges, according to your opinion? > > > > Depends quite a bit on a host of questions. In your tree with terminated > > edges, also the paths are terminated, so neither becomes uncountable. > > But also 1/3 is not in that tree. > > 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. > 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? > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: David R Tribble on 29 Aug 2006 19:44 Tony Orlow wrote: >> Any finite number of bit positions produces a finite set of strings. >> >> Any countably infinite set of bit positions produces an uncountable set of >> strings. > Tony Orlow wrote: >> Which explains why the reals are uncountable (when represented as >> infinite binary fractions in [0,1)). > Tony Orlow wrote: > Yes, ala Cantor's second proof of uncountability. David R Tribble wrote: >> But it does not explain your claim that the naturals are uncountable >> when represented as finite bitstrings. It's pretty straightfoward to >> show that a countable number of bits produces a countable number >> of bitstrings. There is no "in-between" cardinality. > Tony Orlow wrote: > You just contradicted yourself. > > You agree that [me] "Any countably infinite set of bit positions > produces an uncountable set of strings", right? That's what [you] > "explains why the reals are uncountable (when represented as infinite > binary fractions in [0,1))"?But, [you] "a countable number of bits > produces a countable number of bitstrings"? Is a countably infinite > number of bits countable or not? If so, then does a countably infinite > number of bit positions produce a countable, or an uncountable, set of > strings? A slight misunderstanding, so I'll rephrase it (even though you've heard it all before). If every bitstring contains a finite number of bits, the set of all finite bitstrings (or binary naturals, or finite-length binary tree paths) is countably infinite. If the bitstrings are infinite, containing a countably infinite number of bits, the set of all infinite bitstrings (or nonterminating binary real fractions, or nonterminating infinite binary tree paths) is uncountably infinite. David R Tribble wrote: >> It also flies in the face of your statements about infinite binary >> trees. If each node is numbered with a natural (being the finite >> bitstring of the left/right paths traversed from the root to the node), >> then the nodes, and thus the naturals, are obviously countable. >> >> On the other hand, you don't understand that the infinite paths in >> the tree (the ones that don't have a terminating node) are uncountable >> and correspond directly to your infinite bitstrings and to the real >> binary fractions in [0,1). > Tony Orlow wrote: > I'll wait for your response on the above, since you're still not ready > for the answer you've already heard for this. Please don't snip it. It's > a good question, and a prime example of standard issues.
From: Dik T. Winter on 29 Aug 2006 19:57
In article <1156842732.043353.163480(a)74g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > Cantor states (analogously) that *all stairs exist*, that the width of > > > all ot them is L, but that none of them has hight L. Width [0, 1] and > > > height [0, 1). > > > > A quote please. But indeed, the width of all of them is L, but there is > > none of them with width L. And the heighth of all of them is L, but there > > is none of them with heighth L. So width [0,1] and heighth [0,1]. That > > is, if you define both as the smallest box containing the completed stair. > > If you define both as the largest numbers that can be obtained, you get > > width [0,1) and heighth [0,1). You are trying to make it difficult, using UTF-8. But this one takes the cake. > I agree with you. But Cantor said (Werke, p. 409): "So stellt uns > beispielsweise eine ver?nderliche Gr??e x, die nacheinander die > verschiedenen endlichen ganzen Zahlwerte 1, 2, 3, ..., v, ... > anzunehmen hat, ein potentiales Unendliches vor, wogegen die durch ein > Gesetz begrifflich durchaus bestimmte Menge (=EF=81=AE) aller ganzen If "=EF=81=AE" is UTF-8, the Unicode character is U+F06E, which is in the private area of characters. I have no idea what that symbol stands for, so I will modify it to N. > endlichen Zahlen N das einfachste Beispiel eines aktual-unendlichen > Quantums darbietet. Again translated (why do you post so much German in an English speaking newsgroup while you should know that most readers are not able to read German?): Cantor: So while a changing quantity x that successively takes the various values of finite numbers 1, 2, 3, ..., v, ... , is a potential infinite, on the other hand, a through the axioms completely determined set (N) of all integral finite number is an example of an actually finite quantity. Nice that you found the quote I have alluded to, and that you did deny of existing, but that I could not find back. What Cantor is stating here (and I did already indicate that in an earlier response), is, translated to current set theory: The set N is potentially infinite, the size of N is actually infinite. (In current terminology a set is not a quantity.) > And I am arguing against this actual understanding of infinity. Try to argue with something beyond Cantor. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |