An uncountable countable set
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From: Tony Orlow on 5 Sep 2006 23:27 Dik T. Winter wrote: > In article <1157465168.349273.136370(a)i42g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1157367467.816725.158560(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > 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. > > I do not. The digits are countable because I can immediate state the > n-th digit when n is given. The paths are not countable because I can > not state the n-th path when n is given. The edges within a given path > are countable because you can state which is the n-th edge when n is > given. If you mirror the digits to the other side of the digital point and order like the digital naturals, you have a linear ordering of the paths. Since binary ten is 1010, the tenth infinite path is ..010100000.... Of course, that would make 1/3 the ....01010101010th path, which might rub you the wrong way. :) > > > > > 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.) > > No, of course you do not ask that. You should ask that to the one who > provided the list. Which is *not* Cantor, because he does start with > an arbitrary list, not a specific list. And shows that given an > *aribitrary* list, the anti-diagonal produced is not in the list. > As this proof holds without any reference to the particulars of the > list, it holds for *all* lists. So, if someone gives me a list and > tells me that it contains all infinite sequences of two symbols, I > can show him such a sequence not in the list. That's because the list is square, but no list of all strings beyond unary is as wide as it is long. N=S^L. There are 2^L binary strings of length L. 2^L>L for all L. The list is infinitely longer than it is wide. That doesn't mean the list doesn't exist. That's a consequence of mistakenly assuming no list can be longer than the set of naturals. > > > > What is the tenth path of that tree? > > > > What is the tenth entry of the list? > > That is completely irrelevant. See above. But also, the list is a > *given* thing with some assumptions. And a list is *by definition* > a function from N to a (sub)set something. So in the case of Cantor, > the list is the function > f: N -> S > where S is the set the set of infinite sequences of two symbols. You > ask what the tenth element is, and I answer: f(10). I need not go > further because the list is a given entity. > > On the other hand, you *claim* that your set of paths is countable, > so you should be able to state what the tenth path is. But you refuse > to do so. What does "countable" mean, except that all representations in the set are finite? > > > > > 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. > > And so all paths terminate, and 1/3 is not in your tree. Dik, an edge is the connection from one node to one of its child nodes. A path is a sequence of edges from the root. Every edge terminates. In an infinite tree, no path terminates. > > > > > 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. > > No, I did not know the strange exponentiation used in ordinals. Apparently > it can be defined as the set of functions from a set with ordinal number > omega can be mapped to a set with ordinal number 2, with the proviso that > only finitely elements are mapped to the second element. No comment. > > > 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. > > How do you come at 2^omega fractions? (Yes, I sort of understand.) > Exponentiation can also defined as: > (1) a^0 = 1 > (2) a^(b+1) = (a^b).a (note right handed multiplication) > (3) if d is a limit ordinal, a^d = lim{b -> d} a^b. > So this implies that 2^omega = lim{n -> omega} 2^n, which is omega. 2^omega=omega, but 2^aleph_0=c>aleph_0? No comment. ToeKnee
From: Virgil on 6 Sep 2006 00:18 In article <44fe1466(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > A better way to view the universe of sets is as a relation between the > objects within it and 1-place predicates concerning objects, where each > object has a truth value associated with each predicate, between 0 and 1 > inclusive. Thus, these predicates define subsets of the universe, eh? TO appears to be suggesting that there are truth values strictly between 0 and 1, as in "fuzzy logic". That doesn't work well in mathematics, TO. And allowing unrestricted predicates leads to all sorts of anomalies, like sets which are not members of themselves. Besides which mathematics has no need to include the entire universe, and those why try to make it do so get themselves in over their heads.
From: Virgil on 6 Sep 2006 01:15 In article <44fe1669(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> I find this last sentence vague because of the word "induction". If you > >> refer to inductive proof, then I am familiar with the Peano axioms and > >> the underlying logical construction which makes inductive proof valid. > > > > You have only a faint idea. If you actually read a book on the subject, > > you'd find how much deeper, richer, and rigorous this is. > > > >> If you refer to inductive logic, the formulation of rules from instances > >> of fact, then there are statistical methods coupled with feedback that > >> make it possible. Which were you talking about? > > > > Mathematical induction. Inductive sets, et al. As in mathematical logic > > and set theory, which is deductive. Not inductive logic as in the other > > sense of inductive - empirical based inference (or however you want to > > define it). > > > > MoeBlee > > > > So, what is it you think I DON'T get about inductive proof, sets, and > recursion? As far as I can tell, very few dare to question the > predefined rules as set forth, but I have yet to see any valid > counterexample to my rules regarding inductive proof in the infinite > case. Where an equality between expressions is proven for all n, it is > valid for infinite n. What standard rule of induction or standard definition does TO claim justifies these claims? > Where an inequality is proven for all n greater > than some finite m, and the difference upon which the inequality is > based does not have a limit of 0 as n->oo, it holds also in the infinite > case. What standard rule of induction or standard definition does TO claim justifies these claims? If neither can be justified by any standard rule, then TO must present his system in its entirety to get it in, as it is in conflict with every standard system.
From: Virgil on 6 Sep 2006 01:32 In article <44fe1725(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> And yet, you are saying it's not technically correct. So, 1 is a natural > >> but not a rational or real? If rigorous formulations come to that > >> conclusion, then rigor does not ensure correctness. > > > > You're hopeless. You didn't understand a thing I said, which may be my > > fault for not providing adequate explanation in the context of your > > ignorance of the subject; but it is no my fault that you won't even > > look at a book on mathematics, such as even a introductory text in real > > analysis. > > > > MoeBlee > > > > Your whole point here is ludicrous. You are arguing that the naturals > are not a subset of the rationals, which are not a subset of the reals. In which argument he is perfectly correct. That there is an isomorphic image of the system of naturals in some other systems does not mean that the original naturals are in those systems. The empty set is, at least in the von Neumann model, a natural number but it is not a rational number nor a real number, as the corresponding rational and real numbers are both much more complicated objects. > While each superset may require a more complex construction than the > subset, all elements of the subset are covered by the superconstruction. > As points on the real line, they are all real numbers. What you're > saying amounts to what I said above, which is indeed absurd. So, no, I > don't understand a thing you're saying, when you say naturals aren't > reals. Sorry. Which statement of misunderstanding only underscores the profundity of TO's invincible ignorance of mathematics. There are two ways of constructing the rationals from the naturals (1) by first constructing the integers and from them certain ratios of integers, or (2) By first constructing certain non-negative ratios of naturals and from them the signed ratios. Once one has the rationals, one can construct the reals from them either via Dedekind cuts or as a quotient ring of the ring of Cauchy sequences modulo the ideal of null sequences. For someone of TO's talents, learning the details of this would involve at lest a semester's serious and guided study, but more likely a year or two. After which he just might understand why the set of naturals is not directly a subset of either the set of rationals or the set of reals.
From: Virgil on 6 Sep 2006 01:37
In article <44fe17fb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <1157471262.974912.294510(a)b28g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >> 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? > > > > Counting precedes adding, in that one may count when unable to add. > > > > But how is one able to add without being able to count? > > > > A primitive version of counting, from which the word "calculus" > > allegedly derives, is Greek shepherds in classical times counting their > > flocks by pairing off their sheep with pebbles (calculi). > > Yes, counting does precede addition. Addition is repeated increment, > multiplication is repeated addition, exponentiation is repeated > multiplication, tetration is repeated exponentiation... > > Increment is fundamental. It's not just successor, but successor with > measure, and this is what generates the natural numbers. Except that successorship does not involve any measure of anything, TO is, for a change, nearly right. Set successorship only involves the union of a set with the singleton set which contains it as a member. The successor of any set, x, is (x \/ {x}) |