Cantor Confusion
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From: Andy Smith on 23 Jan 2007 14:42 Dave Seaman <dseaman(a)no.such.host> writes > >You can't have an actually infinite integer for exactly the same reason >that you can't have an actually 200-cm. meter or an actually 4-sided >triangle. It's not that we can't imagine an actually infinite cardinal, >an actually 200-cm. object or an actually 4-sided polygon, it's just that >such things are by definition not actually examples of the previously >defined word. > > Yes, I understand that! You might have said is that an "actually infinite integer" is an oxymoron. -- Andy Smith
From: Virgil on 23 Jan 2007 14:52 In article <1169548200.625843.205790(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > hagman schrieb: > > > mueckenh(a)rz.fh-augsburg.de schrieb: > > > > > Virgil schrieb: > > > > > > > Cantor, using his original diagonal argument, and anyone else who wants > > > > to emulate him, can show that any /list/ of paths in a complete infinite > > > > binary tree is incomplete. > > > > > > So you do not believe in complete trees? Somewhere in the mist of > > > infinity the paths cease to split, or become unobservable? WM demonstrates again his inability to read what is written. Complete infinite binary trees are possible, and indeed necessary in ZFC or NBG, it is just that their sets of paths are not countable in any system. > > > > No, there are, at the edge to infinty, > > > Where can we find this edge? Ijn my opinion you are at the edge either > if you begin to count 1,2,3,... --- ore nowhere. > > > an awful lot of new paths > > that are brought into light, namely non-terminating paths. > > Finite trees have absolutely zero (non-repeating) non-terminating > > paths, > > The union of all finite trees has no terminating path because the union > of all finite segments {1,2,3..., n} (with in N) does nowhere > terminate. By any mathematical definition of "union" a union of two or more distinct trees is not a tree at all. What WM is proposing is a concatenation of the union of node sets with the union of edge sets to form a new tree. Such a concatenation of unions is not itself a union, but might be called a "conjunction"of trees. (While "conjunction" has a standard meaning in logic, I am not aware of it having any definition in this context, so it should be usable here without conflict.)
From: Virgil on 23 Jan 2007 15:12 In article <1169548709.607738.69810(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > >> Even if we now use > > >> > > >> T(m) U T(n) := T(sup(m, n)). (fin-u') > > >> > > >> and and try to define (inf-u) by > > >> > > >> T(1) U T(2) ... := T(omega) > > >> > > >> it is still left open what T(omega) shall mean. One has to take > > >> special care for the notation: The union symbol "U" does not mean the > > >> usual set theoretical union. > > > > > > It does. > > > > No, it does not. The usual set theoretical union is defined as > > > > A U B := { x | x e A v x e B } > > > > If A and B are trees {M_A, E_A} and {M_B, E_B} then the union is > > > > A U B = { x | x e {M_A, E_A} v x e {M_B, E_B} } > > = { M_A, E_A, M_B, E_B } > > > > This union is not a tree. > > unless M_A, E_A and M_B, E_B have same elements at all levels they have > in common. Actually, FN is not quite right about what a tree is. It is an /ordered pair/ of sets, (M,E), with the first set being a set of edges and the second set, of edges, being a subset of the cartesian product, MxM. And a union of two or more ordered pairs is not, itself, an ordered pair of the same type, so is not a tree at all. One realization of an ordered pair (X,Y) is as the set {X,{X,Y}}. Then let A = (M_A, E_A) = {M_A,{M_A,E_A}}, where E_A \ss M_A x M_A, and let B = (M_B, E_B) = {M_B,{M_B,E_B}}, where E_B \ss M_B x M_B, then A U B = {M_A,{M_A,E_A},M_B,{M_B,E_B}}. And is not itself an ordered pair unless A = B. > > Then forget about the graph theoretic union for the moment. Then we are > doing soemthing new (but not really). Then don't call it union. > > > It is an equivocation. > > No. The set theoretic union of the trees (edges are implied by the > general type of tree considered) The set theoretical union , see above, is not anything like a tree. Call your conjoining of trees by some other name.
From: Andy Smith on 23 Jan 2007 15:12 David Marcus writes (snip) >> >You still seem to be missing it. >> >> Maybe. The issue wasn't with a finite list, it was whether you could >> have an infinite list when all the indices of the rows i.e. all natural >> numbers, must be finite ... resolved by considering the list as "an >> infinite set" just as the set of "all natural numbers" can be considered >> as "an infinite set", even though no member of the natural numbers are >> infinite; "an infinite set" is an abstract mental concept. > >Let's ignore sets and start at the beginning. The natural numbers are >the following numbers: > >0 >1 >2 >3 >... > >The dots mean we keep going "forever". All of these numbers are finite. >Probably we should just take that as the definition of the word >"finite". So, saying a "finite natural number" is redundant. > Yes. >Now, I ask you: how many natural numbers are there? One? Two? Five? Ten? >One hundred? Which natural number could be the answer to this question? >Any of them? If your answer is the natural number n, this can't be right >because all of 0,1,...,n are natural numbers and we see that there are >n+1 of them. So, the answer to the question "How many natural numbers >are there?" has to be "not a finite number" or "infinite". > >Note that we haven't said anything about sets. > Yes, this is still Aristotle. >> If you see this as straightforward it is because your mindset has been >> conditioned by your education to see this as normal. I can safely say >> that if your concepts of infinite sets was placed in front of the >> population at large 99 % would think that this is barking mad >> doublethink ... > >Maybe. Maybe not. You might ask some of your friends. You might be >surprised. I think that what I wrote above about natural numbers would >be intuitively clear to a significant fraction of the population. And, I >think they would also find what I write below believable. > >> Cantor provided a perspective to view infinity, and his insight >> underpins, as I understand it, modern set theory. It may be consistent, >> but I don't see that the philosophical rational is trivial - and, as I >> understand it, in Cantor's day there were many eminent and far from >> stupid mathematicians who couldn't get a handle on it. From posts on >> this site I can see that their descendants are still here and active ... > >Well, the number of cranks is actually quite small. > >As for Cantor, his great idea was that it made sense to compare sets by >whether you could biject or inject them. Before him, people thought that >this idea didn't work. > >However, I think we can understand the fact that (in some sense) there >are more reals than naturals without mentioning sets. A list is a bunch >of rows where each row is labeled with a natural number, e.g., > >0. adfna >1. afdkj >2. dfajhadf >3. adfj;df >... > >The question is can we construct a list such that every real number in >[0,1] appears on the list? We argue as follows. Let's agree to write >real numbers in [0,1] as infinite decimals. Some infinite decimals have >two representations, one ending in 0's, the other ending in 9's. Let's >agree to use the one ending in 0's, except let's use 0.999... for 1. >Suppose we have a list of reals: > >0. 0.1234... >1. 0.4893... >2. 0.3839... >... > >For convenience, let's number the digits (after the decimal point) >starting with zero. We construct a real x as follows. Let the n-th digit >of x be 4 if the n-th digit of the n-th number on the list is 5. >Otherwise, let the n-th digit of x be 5. > >Is x on our list? Well, it can't be the 0th number because it has a >different 0th digit. It can't be the 1th number because it has a >different 1th digit. In fact, it can't be any of the numbers. So, it >isn't on our list. (Note that x can't be equal to a number on our list >without the digits matching because the only way it could do that would >be if it ended in 9's.) > >Therefore, given any list of real numbers in [0,1], we see there is a >real number in [0,1] that is not on the list. Hence, no such list can be >complete. > Yes, but, as you have observed to me in the past (and is my natural understanding) that arguments which work with finite quantities can turn into paradoxes or nonsense when things become infinite. The nub of my original question was "why is this valid reasoning". >Notice that I haven't used the word "set" or said anything about the >"sizes" (or "cardinalities") of "infinite sets". > >> If I was asked to sum it up, at present I would say that my >> understanding is that you can't have an actually infinite integer, > >What in the world would an "actually infinite integer" be and where did >you get the idea there could be such a thing? It seems so >counterintuitive. Yes, but on a simplistic view of infinity (e.g. mine as of 2 weeks ago) there is no reason not to think that you can have ...1111 as well as ..1111... - or that the Universe has to have had a beginning (everything has to succeed to something previous, so there must be an origin, so you can't have infinite negative time - but can conceive of an infinite future time, Aristotle again. Hence my (ignorant) initial perspective that there ought to be negative integers, and that in some sense everything ought to wrap-around) > >> but >> reals can be defined as having an actually infinite binary >> representation .. (with apologies for the adjective "actually"). > >Isn't it intuitive that the decimal representation for 1/3 "never ends"? Yes, but ... > >> So no surprise that the reals are "uncountable". > >On the contrary, I think the fact that the reals are "uncountable" is >very surprising. The algebraic numbers are countable. > did you read my comment on transcendentals with an infinite random bit string (a "conceptual" number, as I suggested)? Any one of such numbers, that make up essentially all reals, requires an actually infinite number of bits to represent it - it cannot be represented with a finite set of integers. BTW, my copy of Spivak turned up - it is beautifully presented and written, absolutely at my level, Feynmann would have approved of it. Thanks. -- Andy Smith
From: Virgil on 23 Jan 2007 15:17
In article <1169549271.288532.26240(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > So you acknowledge the complete infinite tree as a tree and also every > finite tree as a tree. Burt the union of all infinite trees is not a > tree? Why. Because "union" has a definition for sets and "rooted binary tree" has a definition as a set and under these definitions, the union of two distinct trees form a set which is not a tree. |