An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Tony Orlow on 13 Sep 2006 15:32 imaginatorium(a)despammed.com wrote: > Mike Kelly wrote: >> Tony Orlow wrote: >>> David R Tribble wrote: >>>> Mike Kelly wrote: >>>>>> Better in what sense? What mathematics are you hoping to be able to do >>>>>> with your new foundation that cannot be done with ZFC? >>>> Tony Orlow wrote: >>>>> It accounts for changes of a single element between infinite sets and >>>>> therefore provides for a full spectrum of ordered infinite sets, thus >>>>> making the Continuum Hypothesis null and void. It relates the continuum >>>>> to the hypernaturals formulaically, and provides for an integration of >>>>> sets and measure in the infinite case. It rids us of anomalies like >>>>> omega-1=omega. > > <snippo-bippo> > >>> Use IFR. N maps to S using f(n)=n+1. The inverse of that function is >>> g(x)=x-1. So, over the range of 0 to N, |S|=|N|-1. Map N to the Evens E >>> using f(n)=2n. The inverse function is g(x)=x/2, so over the range of N, >>> the evens have |N|/2 elements. Isn't that intuitively satisfying? >> No. Now what? > > Well, isn't this "IFR" thing basically Tony's semicomprehension of > limiting density, which applies to some infinite subsets of the > integers (and to some other sets, but not all). AFAICS, Tony simply > ignores everything known about limiting densities, to assume that his > version gives a "size" of - presumably - any set. > > Brian Chandler > http://imaginatorium.org > IFR is a generalization of limiting density. Limiting density is a special case of IFR where f and g are linear functions.
From: Tony Orlow on 13 Sep 2006 15:34 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.
From: Tony Orlow on 13 Sep 2006 15:37 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. > >> 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. > >>>> 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.
From: Tony Orlow on 13 Sep 2006 15:43 David R Tribble wrote: > Tony Orlow wrote: >>> If you remove an element, the proper subset should ALWAYS be smaller by >>> 1. That is the case for me. For a theory to claim a proper subset is the >>> same "size" as the proper superset is an immediate deal-breaker for me. > > David R Tribble wrote: >>> If by "different size" you mean that you cannot pair up all the >>> elements from both sets, then you're going to have a difficult >>> time proving that for any infinite set. (You have never show this, >>> BTW.) >>> >>> If by "different size" you mean something other than some way of >>> denumerating (counting) the elements of the set (e.g., by assigning >>> them different natural indices), then you should use a different term, >>> because it's confusing. Oh, and you have to prove that it works >>> (you have never shown this, either). >>> >>> Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} >>> and remove one element to get set S = {1,2,3,...}. Now show that >>> the "T-size" of N is exactly one less than the T-size of S. In other >>> words, find a way to show that every counting of S versus every >>> counting of N always leaves one element of N (0) left over. > > Tony Orlow wrote: >> Use IFR. > > A.k.a. a bijection. You see where this is going? > Yes, bijection with measure. > >> N maps to S using f(n)=n+1. The inverse of that function is >> g(x)=x-1. > > Which proves that every n in N has an x in S. Where is that > leftover element that was removed from N? If N has more > elements than S, shouldn't N have a member that can't be > mapped to any member of S? > One can map all sorts of sets. N contains every element of S - map those first, to themselves. Now you have one left over. > >> So, over the range of 0 to N, |S|=|N|-1. > > Funny how you don't define what |X| is. You're using standard > symbols but obviously with a different meaning, since "|X|" means > "cardinality of X" when X is a set. Your IFR bijection proves that > |S| = |N|. > |X| means size of, like the absolute value of a real. > >> Map N to the Evens E using f(n)=2n. The inverse function is >> g(x)=x/2, so over the range of N, >> the evens have |N|/2 elements. Isn't that intuitively satisfying? And >> gee, it works for finite sets accurately too!! > > Same thing as above, it proves that every n has an x. Where are > the leftover unmapped elements of N that make S a smaller proper > subset of N? > All the odds. N contains all elements of E, plus the elements of O. IFR works to the level of accuracy of detecting a change of 0ne element out of an uncountable number.
From: Virgil on 13 Sep 2006 15:50
In article <1158133724.305764.245170(a)d34g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Not in a binary tree in which no node is terminal, which is the case in an "infinite binary tree". > An example: The digits 1 of the number 0.111... do terminate (they are > finite) but the number does not terminate. If "Mueckenh" claims "the digits terminate" , he is claiming that there is a last digit. But then 0.111... must also terminate with that same digit. > > > > > 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. In an "infinite binary tree", (1) each node has two child branches (edges) ending in child nodes, (2) There is a unique "root" node which is not a child node. (3) A path is a either maximal sequence of nodes in which each node has exactly one child node, or, equivalently the sequence of edges connecting those nodes, since either uniquely determines the other. > > This" /" and this "\" are edges. > An edge is the connection between two nodes, which represent the digits > of a number. There is no necessary "digits of a number" corespondence, though many such are possible. > A path is an infinite sequence of digits. No digits need be involved in binary trees. > In my tree every edge is terminating and no path is terminating. OK. > This example says: Countability can be proven without a bijection. Not to the satisfaction of anyone who does not already believe it. > > 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. And there are uncountably many such diagonals possible. > > Why do you accept the axiom of infinity for Cantor's list, but deny it > for my tree? How has anyone denied it for your tree? |