Cantor Confusion
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From: Dave Seaman on 26 Jan 2007 12:28 On Fri, 26 Jan 2007 15:41:28 GMT, Andy Smith wrote: > Andy Smith <Andy(a)phoenixsystems.co.uk> writes >>> >>The point of that argument was intended precisely to show that if a >>particular arrangement or the reals is countable, then any arrangement >>is countable, and vice-versa, if a specific ordering of the reals is >>uncountable, then any arrangement is. I think that is probably not very >>controversial? >> > By "that argument" I meant that advanced in the previous post, not > Cantor. Sorry for the ambiguity. I understood that, but it would have been better if you had preserved the argument you were referring to instead of snipping it. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Franziska Neugebauer on 26 Jan 2007 12:43 mueckenh(a)rz.fh-augsburg.de wrote: > > > On 26 Jan., 11:11, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 25 Jan., 15:57, Franziska Neugebauer >> > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> >> > If T(oo) as defined by me is T as defined by me, concerning the >> >> > set of nodes and edges, >> >> >>There is no room in equality for "concerning" this or that. Either >> >>two >> >> symbols refer to the same entity or they do not. >> >> > That is not so obvious. [indentation corrected:] >> That *is* obvious. > > Indeed? Indeed. Two sets a and b are equal if A c (c e a <-> c e b) Of course only in orthodox set theory possibly not in Augsburg. F. N. -- xyz
From: Franziska Neugebauer on 26 Jan 2007 12:51 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > On 25 Jan., 15:34, Franziska Neugebauer >> > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> context: >> >> ,----[ <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> >> >> ] >> >> | >> Again: Your notations >> >> | >> >> >> | >> T(1) U T(2) U ... >> >> | >> >> >> | >> and >> >> | >> >> >> | >> U {T(i) | i e N } >> >> | >> >> >> | >> are undefined. >> >> | > >> >> | > You are in error. The union of the trees T(n) and T(n+1) is >> >> | > defined. n is a natural number. Therefore the union of all >> >> | > finite trees is defined. >> >> | >> >> | You have misunderstood the induction principle. It is not made >> >> | for "counting over to the infinite". >> >> `---- > > Look: Above, there is written "i in N", not "i = N". The latter would > be missed by induction. ??? >> You have obviously been interested in the "union of all finite trees" >> which still waits for a definition. > > I defined level n and defined the bijection of levels and natural > numbers. That is enough. You did not *define* the "union of _all_ finite trees". >> > It is irrelevant for the consideraton of trees. >> >> It is highly relevant to show where your reasoning is malfunctioning. > > No. You only try to huddle around, avoiding any concrete discussion. I have a different understanding of what concrete means. Do you mean the concrete in your head? > You cannot make me believe that your intelligence is insufficient to > understand this discussion. This may be due to the concrete. >> > In particular we can prove by induction that the tree, formally >> > called T(oo), >> >> 1. Define T(oo). > > Has been done. > >> 2. Prove that the "formally called" entity T(oo) is really a tree. > > T(oo) is a tree by definition. > >> >> > contains all paths if it contains all levels enumerated by >> > natural numbers. >> >> 1. Define "path". > > Look at my tree. Perhaps I need the Third Eye for that. >> 2. Define "contains". > > Look at Wiki: subset. "Subsets" is a formal notion which applies to sets not to graphical sketches on your paper seen by the Third Eye. >> > We have no use for your set N in the tree (in case it >> > should differ significantly from the ensmble of all natural >> > numbers.) >> >> Define "ens[e]mble of all natural numbers". >> > With pleasure (though it is not simple). > > The "ensemble of all natural numbers" is a somehing which cannot be > named without raising wrong impressions. I got every nesecessary impression. F. N. -- xyz
From: imaginatorium on 26 Jan 2007 13:44 Andy Smith wrote: > Dave Seaman writes > (snip) > >> I would say yes, > >> because there is no scope for any more compact non-infinite > >> representation of the reals than their infinite binary expansions. > > > >Can you prove that? > > > And there is the nub of the matter. For me, at this moment, I would say > that the reals are uncountable because they are infinite, have a > necessarily infinite binary expansion. That is more than you get from > Cantor's diagonalisation argument? What exactly do you mean by "the reals are infinite"? (If just that the set of reals is infinite - that is, you cannot count all of them and stop on a number - then so are the natural numbers) What do you mean by a "necessarily infinite binary expansion"? In what sense do the algebraic numbers not have a "necessarily infinite binary expansion"? (Trouble is that Cantor's diagonalisation argument gives a proof of something, whereas your somewhat uncollected thoughts are merely suggestive to someone who thinks they already know the answer.) Brian Chandler http://imaginatorium.org
From: MoeBlee on 26 Jan 2007 15:11
On Jan 26, 4:25 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > As I understand it an injection f:A->B means that for every element in A > , f identifies a unique element of B, but E elements of B not mapped; I guess by 'E' you mean 'there exist'. No, we can't conclude that. For a given f, it might be the case that there are member of B that are not mapped to by f, but just being an injection from A into B does not entail that there are members of B not mapped to by f. > it > is- one-to-one but not onto. It is one-to-one and possibly onto B and possibly not onto B. Here's the definition: f is an injection from A into B <-> f is a 1-1 function with domain A and range subset of B. So the range could be a proper subset of B or it could be B itself. (By the way, EVERY function is onto the range of the function. So here it's a question of whether B is the range of f or whether the range of f is a proper subset of B; and f is still an injection into B either way.) > A surjection f:A->B means that f maps every > element of A onto an element of B such that for every element in B E at > least one corresponding member of A, but this is not unique (onto but > not one-to-one) No, it could also be 1-1, but you got the first part right. Definition: f is a surjection from A onto B <-> f is a function with domain A and range B. So f might also be 1-1 or it might not be 1-1. > For the reals to be countable, there has to be an injection from R->N, Correct. > but since there is a bijection from N to any infinite subset of N, It's correct that there is a bijection from N onto any infinite subset of N, but I don't know why you think this is relevent here. > for > the reals to be countable there also has to be a bijection R->N. I > think? For R be countable, there has to be a bijection between N and R or a bijection between a natural number and R. For R to be denumerable, there has to be a bijection between N and R. Meanwhile, we have a theorem that, for any X and Y, there is a bijection between X and Y iff there is both an injection from X into Y and and injection from Y into X. Now it's trivial to show an injection from N into R. So if we could show an injection from R into N, then, by that theorem I mentioned, we'd know that there is a bijection between R and N. And that's all correct, but it's not usually the route we take witht the diagonal argument. Rather, we do this: The diagonal argument proves, for any f, if f is a function from N into R, then f is NOT ONTO R. So there is no bijection from N onto R. In other words, the "list" of reals is a function f from N into R. And the diagonal argument shows that for any such f, it is NOT ONTO R. So since a bijection between N and R would have to be a 1-1 function from N onto R, there can be no such bijection since there is no function of any kind that is from N onto R. > If there exists a bijection R->N , there will then be a 1:1 mapping > between N and R taken in any order (given a bijection between R->N we > can swap the mappings from N to R as a chain so that we can map R in ANY > order (i.e. any permutation) that we choose). I don't know what YOU'RE definition of 'chain' is, but a permutation is a 1-1 function from a set onto itself. > So if we can show that there is no one to one mapping between N and R > taken in some given order, then that would imply that there is no > bijection R->N ? > > I think. If there is no bijection between X and Y, then no permuation of X nor any permutation of Y will lead to a bijection between X and Y, is the best I can answer without further specification of what you're tyring to say. Or, maybe you mean a permuation of f itself. In other words, maybe you are thinking of taking the function f and mapping itself onto itself. If that's what you have in mind, then, no, that won't work to construct a bijection between X and Y. Yes, CERTAIN functions may fail to be a bijection between X and Y while certain other functions are a bijection between X and Y. But if there does not exist any bijection between X and Y, then no permuting or anything else will make a bijection between X and Y. MoeBee |