An uncountable countable set
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From: mueckenh on 5 Sep 2006 10:06 Dik T. Winter schrieb: > In article <1157367467.816725.158560(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > > > 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. > > > > Cantor's list has as many lines as my tree. The diagonal has as many > > digits as each path of my tree. The only difference is that the paths > > split while the diagonal does not. > > And the difference is that given the list we can immediately state what > the n-th digit of the diagonal is for arbitrary n. You can not state what > the n-th path is for arbitrary n. You intermingle numbers (paths) with digits. I can state the n-th digit of any path you want. > > > The same is true for my tree. > > What is the tenth path in your tree? You intermingle numbers (paths) with digits. I can state the n-th digit of any path you want. (I do not ask what is the tenth entry of Cantor's list.) > What is the tenth path of that tree? What is the tenth entry of the list? > > > > 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? > > > > Give my a tree of infinite paths consisting of 0's and 1's, and I show > > that there are not less edges than paths. > > Indeed. If all edges terminate, also all paths terminate, and both are > countable, and 1/3 is not in the tree. If edges do *not* terminale, What are you talking about? Every edge terminates because it is the connectio between subsquent nodes of a path. > also paths do not terminate, and both are not countable, and 1/3 is > in the tree. And has a countably infinite number of edges as has every path. And the number of paths is half the number of edges. > > > > > 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. > > > > You know that sets of order 2^omega are countable? > > No. Can you prove it? It is precisely Cantor's diagonal proof that > shows that it is not countable. You intermingle 2^aleph_0 and 2^omega. A well known example is the set of the unit fractions 1/n used for the proof of the divergence of the harmonic series: 1 + 1/2 + (1/3 + 1/4) + (1/5 +...+ 1/8) + ... It has omega terms in parentheses and 2^omega fractions which biject to the natural numbers. Even sets of order omega^omega^omega are countable. A representation of omega^omega is the set of natural numbers ordered by the number of their prime factors (according to Hessenberg). Regards, WM
From: mueckenh on 5 Sep 2006 10:07 Dik T. Winter schrieb: > In article <1157367591.092721.79760(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > Because I have the German text available and because I do not want to > > > > be blamed of mistranslating. > > > > > > So you can blame other persons of mistranslating? > > > > Those who cannot yet read German but are interested in the origins of > > set theory should learn to do it. > > Yes, everybody who is not able to read German is not allowed to speak > about set theory and the origins... She needs translations which might be erroneous. (Compare: law, axiom) > > > > > > 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. > > > > > > > > Not through the axioms, but through "a law" (ein Gesetz). > > > > > > What is the difference? > > > > A law is derived from the natural properties of arithmetics. > > Oh. What law derived from the natural properties of arithmetics is he > talking about when he gets the completely determined set of all integral > finite numbers? That is but his conviction. Regards, WM
From: Tony Orlow on 5 Sep 2006 11:16 David R Tribble wrote: > 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? Why did you attribute the sentence, "a countable number of bits produces a countable number of bitstrings", to me, when you clearly wrote it right above? You contradicted yourself. > > 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. According to your theory, if the set contains EVERY finite bit string, yes. > > 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. > Right. So, why did you say that "It's pretty straightfoward to show that a countable number of bits produces a countable number of bitstrings"???? > > 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: mueckenh on 5 Sep 2006 11:47 Virgil schrieb: > In article <1157367209.318653.75760(a)p79g2000cwp.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Counting is the most primitive version of addition. > > Counting is not addition at all. > So you cannot even count, in your infinite unphysical mind? Regards, WM
From: Tony Orlow on 5 Sep 2006 11:45
Dik T. Winter wrote: > In article <44ef3b88(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > > Dik T. Winter wrote: > ... > > > > > There is no such specific natural number. It is when we have them > > > > > all, but as there is no largest number, this can not be achieved by > > > > > taking them one by one. > > > > > > > > The set of all naturals numbers consists of only natural numbers. There > > > > is NO natural number where the count becomes infinite. So there is no > > > > point in the set, even if you COULD get to the "last" one, where any > > > > infinite set has been achieved. > > > > > > And there is no point in the set where you have the complete set. Yes. > > > Indeed. So what? > > > > So, if there is no point in the set which can even remotely be > > considered infinitely far from the beginning, what makes it actually > > infinite? > > What the set of all finite numbers makes infinite? The axiom of infinitys > states that the set of all finite (natural) numbers does exist. From that > is is easy to prove that that set is not finite, hence infinite. Yes, according to the Dedekind definition of an infinite set, but that definition leads to some conclusions that offend some people's sensibilities. > > > If no element of the set can be an infinite number of steps > > from the start, you may not be able to find an end. > > And indeed, when you go step by step you will not get at the end. Right, you will always have gone a finite number of steps and have arrived at a finite natural. > > > But does that mean > > it's "greater than" every finite, or only "greater than or equal"? > > What is the difference? Assuming you mean "aleph-0" when you write "it", > it is easily proven that: > aleph-0 is greater than or equal to each natural > gives the theorem: > aleph-0 is greater than each natural. > > Because: suppose aleph-0 >= each n in N. Now suppose in addition that it > is equal to some particular n. Well, n + 1 is in N, and so aleph-0 > should also be larger or equal to n + 1. Hence it can not be equal to n. > So it is not equal to any n at all. And so aleph-0 > each n in N. Only assuming that you have identified some largest natural. This is the form of most proofs in this theory: assume a largest finite, find a contradiction, and blame it something else. Of course you cannot find a largest natural such that this is the size of the set. You also cannot take an infinite number of increments without achieving an infinite value. > > > > Indeed. The set of all natural numbers is just sufficient. > > > > No, it is far too great. If you have a countably infinite number of bit > > positions, then you have an uncountably infinite set of strings. Where > > bit positions are indexed by the naturals, the naturals are the power > > set of the number of bit positions, > > Wrong. This is plain nonsense. Suppose there are three bit positions. > The set of naturals {1, 2, 3} is sufficent to index them. In what way > is {1, 2, 3} the power set of the number of bit positions (3)? Let me explain, using your example. We have the first three bit positions 0, 1 and 2, representing 1, 2 and 4, the fist three powers of 2. With those bit positions we can produce eight unique binary strings, as follows: 0 1 2 0 0 0 0 1 1 0 0 2 0 1 0 3 1 1 0 4 0 0 1 5 1 0 1 6 0 1 1 7 1 1 1 The set of binary naturals which can be represented by this set of bit positions is the power set of the bit positions, each being a unique combination, in the form of a sum, of those powers of 2. Given n bit positions, we can produce 2^n unique strings. Given n unique powers of 2, we can produce 2^n unique sums. Thus the set of binary strings is the power set of the number of bit positions. > > > > No one is obligated to accept the theory at all. Whether it is proven > > > to be "correct" or not, as I have no idea what "correct" in this context > > > means. Is Euclidean geometry "correct"? Is hyperbolic geometry "correct"? > > > Is elliptic geometry "correct"? > > > > Ah, now you bring up a prime example. Euclid set down laws for flat 2D > > geometry, and questioning those axioms led to new shapes for space. > > Accrdingly, the axioms of set theory might work together to describe a > > system, but it is not impossible that entirely other systems might arise > > from different starting assumptions. > > And indeed, I never did state the opposite. But if you want to get at a > new system, provide axioms, definitions, and whatever. Tell us what > axioms to retain and what axioms to reject. And if there are axioms to > be rejected, come up with alternatives. Well, that's what I'm trying to do, to provide an alternative perspective and system. I certainly reject the Dedekind definition of an infinite set as being inconsistent with notions of infinite measure when it comes to the finite naturals. By viewing the situation in terms of density and range, the conclusion can be quite different. Since no two naturals are infinitely distant, and since each natural occupies a unit of measure on the real line, there cannot be an infinite number of them, because you cannot fit an infinite number of unit segments in a finite range. > > > > But what is the case is that if you accept the axioms, you also have > > > to accept what follows from the axioms. > > > > Yes, I understand that, and much to the consternation of some, I don't. > > Yes, much consternation, I can understand that. So apparently you are > accepting the axioms, but not what follows from the axioms. What kind > of logic are you using? I am NOT accepting the axioms. The axioms are artificial statements which can be made to work together, but which do not follow from elementary logic the way they should. Every axiom should be justifiable based on fundamental concepts. I don't see that here. > > > Rusin had the gall to tell me that if I don't accept that there are an > > infinite number of finite naturals, then I will join JSH and others on > > his "kill list". I don't claim that my conclusions are derived purely > > from ZFC or NBG, but that there are more fundamental concerns which > > contradict both, and that som |