An uncountable countable set
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From: Virgil on 13 Sep 2006 17:24 In article <1158159038.615731.156100(a)i3g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > Why don't you accept the axiom of > > > infinity? There is an infinite set of natural numbers. All they are in > > > the list (in unary representatition). By the axiom this is accomlished. > > > And there is nothing else. > > > > Indeed, and 0.111... is *not* a natural number, so why should it be in > > the list? Why do you think 0.111... is a natural number? > > There is a set which contains { }, and if it contains A then it > contains A U {A}. This gives an inductive set which we can interpreted > as the set of all natural numbers or the set of all indexes. But the process of indexing involves functions between sets. > > All indexes are in the list. All numbers which can be indexed are in > the list. False. I have, and many others also have, indexed the set of rational numbers, many of which are not in the "list" of all naturals. > > > > Because it is not in the list which, by definition, contains all > > > numbers which can be indexed because there are all numbers which can > > > index. That may be a part of M 's definition but is no part of ZF or NBG or any sane system. > > > > This is nonsense. By the definition of your list the list contains all > > finite natural numbers. That has nothing to do with indexing. > > Index = natural number. There is no infinite index, because indexing is > identifying. Where does "Mueckenh" come up with this definition? It is entirely his own, not a part oaf any sane mathematics. > > > > If not all digits can be indexed, there should be a digit that can not be > > indexed. But *by the definition* I gave above, all digits can be indexed. > > Why is that wrong? Because, among other things, being an index is not necessary in order to be indexed. One can index the words in a dictonary, and that is a part of what dictionaries do, without any word being an index. > > Because all digits which can be indexed are in my infinite list. Only in your dreams. Not in anyone else's reality. Further garbage deleted.
From: Virgil on 13 Sep 2006 17:37 In article <1158159208.048232.260050(a)e3g2000cwe.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <45070bcb(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > writes: > > ... > > > What about an unboundedly large but finite number? > > > > Let me ask you to, before I can answer such a question. What is your > > definition of "number"? (I asked the same from Wolfgang Mueckenheim, > > but his answer was not satisfactory, also not to himself, I think, > > because he never answered to questions about it.) > > Didn't you read my paper on the physical constraints of numbers? > Here I give briefly my definition which is not obligatory but useful to > avoid the weaknesses of set theory. It is derived from the first > principle of the notion "number", i.e., a means to count. In German > this becomes even more evident by the wording "Zahl" and "zaehlen". > A number is an entity which: > 1) either is realized by a fundamental set like the following > fundamental set of 3: > {III, Dik, {a,b,c}, {father, mother, child}, {sun, moon, earth}, > ...} > 2) or is completely determined by a series of digits. Then "Mueckenh"'s so called numbers are not the same as anyone else's. > > Each of these two requirements is sufficient to guarantee that the > numbers satisfy the trichotomy. That is the most important property of > a number. > > In this sense irrational numbers, complex numbers, hyperreals, > transfinite numbers, vectors, etc. are not numbers but ideas which > consist of numbers, like 2 + 3i, or not, like pi. > > > > Once you have answered that question we can probe further into the > > definition of "unboundedly large but finite number". > > > > As far as I understand, for any kind of numbers, they are fixed, so > > by definition not unbounded. > > O course. But the sequence or the set or the variable may be unbounded > though always finite (because it consists of or takes on finite > entities - in contrast to 0.111...). So whatever these non-numbers are, they are simultaneoulsy bounded and not bounded. The rest of us have a set of "natural numbers", being the minimal limit ordinal, but we also have other numbers, such as the rationals and reals. But we are careful, when necessary, to distinguish between a number of any sort and a numeral representing it. Thus '8' and 'VIII' are just different numerals for the same (natural) number. When no confusion is likely to arise, we tend to conflate the numeral with its number, but should never, like "Mueckenh" habitually does, confuse the two notions.
From: Mike Kelly on 13 Sep 2006 18:35 Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Tony Orlow wrote: > >>>> Mike Kelly wrote: > >>>>> Tony Orlow wrote: > >>>>>> Virgil wrote: > >>>>>>> In article <44fe2642(a)news2.lightlink.com>, > >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>> > >>>>>>>> Virgil wrote: > >>>>>>>>> In article <44fd9eba(a)news2.lightlink.com>, > >>>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>> > >>>>>>>>>> Dik T. Winter wrote: > >>>>>>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>>>>>>>>>> writes: > >>>>>>>>>>> Your axiom uses things that are not defined. What is the *meaning* of > >>>>>>>>>>> "x<z"? > >>>>>>>>>> Geometrically it means that x is left of z on the number line. > >>>>>>>>> And for someone standing on the other side of the number line would x be > >>>>>>>>> on the right of z? > >>>>>>>>> > >>>>>>>>> And does the line stay horizontal as one moves around earth? Which way > >>>>>>>>> is larger if the line ever goes vertical. And how does the "larger" work > >>>>>>>>> at antipodes? > >>>>>>>>> > >>>>>>>> Silly questions. > >>>>>>> In response to a silly definition. > >>>>>>>>>> It means > >>>>>>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >>>>>>>>>> needs to, wouldn't you say? > >>>>>>>>> Not hardly. > >>>>>>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > >>>>>>>>> is a bit better but still insufficient. > >>>>>>>> True, I should have specified y<>x and y<>z. I guess it's usually done > >>>>>>>> using <= for this reason, eh? > >>>>>>>> > >>>>>>>>>>> > > That is not a definition, because it makes no sense. "The set of > >>>>>>>>>>> > > naturals > >>>>>>>>>>> > > is as large as every natural"? > >>>>>>>>>>> > > >>>>>>>>>>> > It is not larger than all naturals > >>>>>>>>>>> > >>>>>>>>>>> That is something completely different again. > >>>>>>>>>> It's not LARGER than every finite. > >>>>>>>>> Which natural(s) is it "not larger" than", in the sense of not being a > >>>>>>>>> proper superset of that natural or having that natural as a member? > >>>>>>>> ....11111 binary (all bit positions finite) > >>>>>>> Unless that string has only finitely many bit positions as well as only > >>>>>>> finite bit positions, it is not a natural at all, as it is then neither > >>>>>>> the first natural nor the successor of any natural, and every natural > >>>>>>> has to be one or the other. > >>>>>> It is the successor to ....11110. Duh. I've already proven that this is > >>>>>> a finite value, given that all bit positions are finite, and that > >>>>>> therefore no place in that string can achieve an infinite value, and > >>>>>> that any such number has predecessor and successor. The cute thing is > >>>>>> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) > >>>>> Does it not bother you that nobody else agrees with, or even > >>>>> understands, your proof? > >>>>> > >>>> I find it disappointing, but not surprising, that you don't understand > >>>> such a simple proof, since it's contradictory to your education. I do > >>>> find it annoying that you feel the right to disagree with it without > >>>> understanding it. If you feel there is a problem with the proof, please > >>>> state the logical error I made. If the string is all finite bits, and > >>>> none of them ever can possibly achieve an infinite value, then how can > >>>> the string have an infinite value? There's nowhere in the string where > >>>> that can occur. It's that simple. Grok it. > >>> 1) A finite string of 1s represents a (finite) natural number. > >>> 2) An infinite string of 1s represents a (finite) natural number. > >>> > >>> 1) doesn't imply 2). > >>> > >> If the string is unbounded but finite, then 2) follows. > > > > What's a finite but unbounded infinite string? > > > > One with all finite bit positions but no greatest. So... why does 2) follow from 1? 1) A finite string of 1s represents a finite natural number. For example, 101 represents 1* (2^0) + 0* (2^1) + 1*(2^2) = 1 + 0 + 4 = 5 this representation bijects finite bit strings of 1s and 0s and natural number. 2) doesn't work however... Take ...11111. Then the natural this represents would be 1 + 2 + 4 + 8 + 16 + 32 + .... but there is no such natural. This sum is divergent. -- mike.
From: Mike Kelly on 13 Sep 2006 18:36 Tony Orlow wrote: > Mike Kelly wrote: > > Tony Orlow wrote: > >> Mike Kelly wrote: > >>> Tony Orlow wrote: > >>>> Mike Kelly wrote: > >>>>> Tony Orlow wrote: > >>>>>> Mike Kelly wrote: > >>>>>>> Tony Orlow wrote: > >>>>>>>> Virgil wrote: > >>>>>>>>> In article <44fd9eba(a)news2.lightlink.com>, > >>>>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>>>>> > >>>>>>>>>> Dik T. Winter wrote: > >>>>>>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>>>>>>>>>> writes: > >>>>>>>>>>> Your axiom uses things that are not defined. What is the *meaning* of > >>>>>>>>>>> "x<z"? > >>>>>>>>>> Geometrically it means that x is left of z on the number line. > >>>>>>>>> And for someone standing on the other side of the number line would x be > >>>>>>>>> on the right of z? > >>>>>>>>> > >>>>>>>>> And does the line stay horizontal as one moves around earth? Which way > >>>>>>>>> is larger if the line ever goes vertical. And how does the "larger" work > >>>>>>>>> at antipodes? > >>>>>>>>> > >>>>>>>> Silly questions. > >>>>>>>> > >>>>>>>>>> It means > >>>>>>>>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >>>>>>>>>> needs to, wouldn't you say? > >>>>>>>>> Not hardly. > >>>>>>>>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > >>>>>>>>> is a bit better but still insufficient. > >>>>>>>> True, I should have specified y<>x and y<>z. I guess it's usually done > >>>>>>>> using <= for this reason, eh? > >>>>>>>> > >>>>>>>>>>> > > That is not a definition, because it makes no sense. "The set of > >>>>>>>>>>> > > naturals > >>>>>>>>>>> > > is as large as every natural"? > >>>>>>>>>>> > > >>>>>>>>>>> > It is not larger than all naturals > >>>>>>>>>>> > >>>>>>>>>>> That is something completely different again. > >>>>>>>>>> It's not LARGER than every finite. > >>>>>>>>> Which natural(s) is it "not larger" than", in the sense of not being a > >>>>>>>>> proper superset of that natural or having that natural as a member? > >>>>>>>> ....11111 binary (all bit positions finite) > >>>>>>> That isn't a natural number, Tony. > >>>>>>> > >>>>>> Are you sure? Pay close attention. > >>>>>> > >>>>>> For any finite bit position n, it and all predecessors can only sum to a > >>>>>> finite bit string value of 2^(n+1)-1. > >>>>> OK. Any string 111....1111 with a finite number of 1s represents a > >>>>> natural in binary. > >>>>> > >>>>>> Since there are only finite bit positions in the string, it can never achieve any infinite value at any position in the unending string of bits. > >>>>> OK. Any string 111....1111 with a finite number of 1s represents a > >>>>> natural in binary. > >>>>> > >>>>>> Therefore the value must be finite. > >>>>> Why? You're supposed to be *demonstrating* that the string represents a > >>>>> value, but you're *assuming* it instead. > >>>>> > >>>>> You've shown that any finite bit string of all 1s represents a finite > >>>>> natural number. And concluded from this that an infinite string of 1s > >>>>> represents a finite natural number. Why? Total non sequitur. > >>>> At which bit does this string attain an infinite value? > >>> It doesn't. I never said it does so please STOP asking that question. I > >>> say that 111..... doesn't represent a (natural) value at all. > >> If it is a whole number (no fractional component), is finite, and has > >> successor and precedessor, then ....1111 is a finite natural. > > > > It isn't. Now what? > > It's not what? A whole number? Show me the fractional component. Finite? > SHow me the bit position where an infinite sum is possible. With > successor and predecessor? There is no successor to this largest > natural, unless the number line is taken to be circular. 1 + 2 + 4 + 8 + 16 ... doesn't sum to a natural number. It doesn't sum to anything. It's a divergent sum. So, no, ....1111 isn't a finite natural number. > >> Unless you > >> have another kind of finite positive whole numbers. > > > > Nope. ...11111 isn't any kind of finite positive whole number. > > If you say so. You haven't presented any representation where it does. > >>>> All bits are > >>>> finite, and have a finite number of predecessors. There is none where > >>>> the string can ever become infinite in value, or even extent. > >>> Yeah, finite strings are finite and represent naturals, got it. > >>> > >>> Why does that imply that an infinite string represents a natural? > >> Because no bit position is infinite so all points in the string are > >> finite in value. > > > > Yes, all finite bit strings are finite and represent naturals. > > > > Why does that imply that an infinite string represents a natural /at > > all/? > > > > Because it is only potentially infinite, and can never achieve actual > infinity. Stop babbling, please. By what scheme can each infinite (finite bit positions, no maximum length bla bla) bit string represent a natural number? Your suggestion so far doesn't work because 1 + 2 + 4 + 8 + 16... is divergent i.e. it does not sum to a natural number. -- mike.
From: Dik T. Winter on 13 Sep 2006 20:39
In article <1158133724.305764.245170(a)d34g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > You keep confusing the situation where edges do terminate (and so all paths > > do terminate) with the situation where edges do not terminate (and so all > > paths do not terminate). > > What do you imagine? Every edge terminates in a node. Yes, so every path made up of a sequence of edges terminates. > An example: The digits 1 of the number 0.111... do terminate (they are > finite) but the number does not terminate. An incomprehensible remark. Digits are fixed entities to which the word terminate does not really apply. Or do you also state that a letter terminates? Or a coin of 1 euro? And the same is true of a "number". Whatever kind of number you allude to, it is not a changing entity, so stating that it does not terminate is nonsense. (About non-terminating paths:) > > > > 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. > > > > > > All the edges are countable, because you can state what is the m-th > > > edge on the n-th level. > > > > Say, which of my statements do you contradict here? > > Impossible to quote a meaningful statement of yours, because you have > not yet grasped what a tree, a path and an edge are. I think I have grasped that sufficiently well. But if you think I did not, please provide *definitions* that are mathematically satisfactory. Not some figure and some handwaving: > This" /" and this "\" are edges. > An edge is the connection between two nodes, which represent the digits > of a number. > A path is an infinite sequence of digits. > In my tree every edge is terminating and no path is terminating. like this. > > > > 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. > > > > > > There are other proofs of 2 + 2 < 10 than executing the addition. > > > If I know that 2 + 2 < 5, then I can conclude that 2 + 2 < 10. > > > Why should some logic conclusion like that be forbidden in set theory? > > > > What is the relevance with respect to my question? > > This example says: Countability can be proven without a bijection. A misapprehention. Countability is *by definition* the presence of a bijection with (a subset) of the set of natural numbers. There is no addition used (or even defined) there. But you still refuse to state what your tenth path is. Now I can imagine what your first path is, but state at least what your second path is. In that way we can tell something about the manner in which you count your paths. And if you can not state what the second path is, how can you even start counting? > > > > > > > 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. > > > > > > But I told you already: The paths do not terminate. The paths are taken > > > from the diagonals in Cantor's list. > > > > Your second sentence makes no sense. > > Oh yes, it does. My paths are as long as the diagonal of Cantor's list. > Every number which can in principle be the diagonal of Cantor's list > can be represented by a path of my tree. > > Why do you accept the axiom of infinity for Cantor's list, but deny it > for my tree? I do not deny it, I only state that in the "completed" tree either there are edges that do not terminate, or all paths do terminate. > > It is well known that the set of *finite* sequences of digits is countable. > > You are telling nothing new here. > > The paths of my tree are not finite paths. If you insist, that no > infinite paths do exist, then you must also deny the existence of an > infinite diagonal of Cantor's list. You should reread what I have written, I have never written that infinite paths do not exist. I have only stated that the paths either terminate at a node (when all edges terminate at a node) or do *not* terminate at a node (when all edges do not terminate at a node). If paths terminate at a node, 1/3 is not in your tree, because that is not a node in your tree. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |