Cantor Confusion
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From: Andy Smith on 26 Jan 2007 10:35 Dave Seaman writes (snip) > >Showing that one particular mapping fails to be a bijection is not >sufficient. It doesn't rule out the possibility that some other mapping >might be a bijection. The point of the diagonal argument is that >addresses all possible mappings f: N->R at once, by showing that each one >fails to be a bijection. > 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? >> even if one buys the argument above, there is still an issue as to >> whether the failure of the counting scheme might not be circumvented by >> some other means i.e. does the failure of the counting scheme show that >> the reals taken in that order are truly uncountable. > >Uncountability has nothing to do with order. > Just so, but not necessarily obvious? >> 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? -- Andy Smith
From: Andy Smith on 26 Jan 2007 10:41 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. -- Andy Smith
From: Randy Poe on 26 Jan 2007 11:16 On Jan 26, 10:35 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > Dave Seaman writes > (snip) > > >Showing that one particular mapping fails to be a bijection is not > >sufficient. It doesn't rule out the possibility that some other mapping > >might be a bijection. The point of the diagonal argument is that > >addresses all possible mappings f: N->R at once, by showing that each one > >fails to be a bijection. > 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? We like to be precise in our statements. In this case, I would define "arrangement of the reals" as a bijection R->R. Under that definition, I agree with your statement. > >> even if one buys the argument above, there is still an issue as to > >> whether the failure of the counting scheme might not be circumvented by > >> some other means i.e. does the failure of the counting scheme show that > >> the reals taken in that order are truly uncountable. > > >Uncountability has nothing to do with order. > Just so, but not necessarily obvious? If it starts by saying "consider any mapping N->R" then where does the concept of order ever enter into it? It starts by considering all maps, so it applies to all maps. > >> 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? That's not an argument. It's a statement about why several of your intuitions combine in a certain way to lead you to an intuitive conclusion. I don't know what "uncountable" means to you, or why "necessarily infinite binary expansion" implies whatever that means to you. I do know that the set of infinite binary strings is uncountable, but I know that because it is easily proven. - Randy
From: Dave Seaman on 26 Jan 2007 11:25 On Fri, 26 Jan 2007 15:35:49 GMT, Andy Smith wrote: > Dave Seaman writes > (snip) >>Showing that one particular mapping fails to be a bijection is not >>sufficient. It doesn't rule out the possibility that some other mapping >>might be a bijection. The point of the diagonal argument is that >>addresses all possible mappings f: N->R at once, by showing that each one >>fails to be a bijection. > 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? The argument I was referring to, which you snipped, is the one where you said: >>> 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 ? So the point of "this argument" was *not* to show that "if <something> is countable, then ...". That assumption does not appear anywhere in your argument. Quite the contrary, you were attempting to show that if a particular mapping from N->R is *not* a bijection, then no other mapping from N->R can be a bijection. That claim is false. Consider the natural injection i:Z->N. This map is not a a bijection, but yet bijections between Z and N do exist. >>> even if one buys the argument above, there is still an issue as to >>> whether the failure of the counting scheme might not be circumvented by >>> some other means i.e. does the failure of the counting scheme show that >>> the reals taken in that order are truly uncountable. >>Uncountability has nothing to do with order. > Just so, but not necessarily obvious? The existence of a bijection has nothing to do with order. >>> 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 is a "necessarily infinite binary expansion"? Doesn't 1/3 have a "necessarily infinite binary expansion"? Aren't there infinitely many rationals having such expansions? Does this mean the rationals are uncountable? The diagonal argument has the advantage of being valid. Your argument does not. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: davidmarcus on 26 Jan 2007 12:10
On Jan 26, 10:35 am, Andy Smith <A...(a)phoenixsystems.co.uk> wrote: > Dave Seaman writes > (snip) > > >Showing that one particular mapping fails to be a bijection is not > >sufficient. It doesn't rule out the possibility that some other mapping > >might be a bijection. The point of the diagonal argument is that > >addresses all possible mappings f: N->R at once, by showing that each one > >fails to be a bijection. > 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? You are misusing the word "order". And, you seem to be using "arrangement" as a synonym for "order". To you, an order or arrangement seems to be a listing. To us, it means that pi is less than 4 (for example). With our meaning, it is obvious that order has nothing to do with countability. With your meaning, it is false. If you have a listing of all the reals, then the reals are countable. If no listing of all the reals is possible, then the reals are uncountable. But, if you have a listing that includes some reals, but not all, then this tells you nothing about whether the reals are uncountable. For example, here is a listing of some reals: 0.1 0.01 0.001 .... Do you agree? > >> even if one buys the argument above, there is still an issue as to > >> whether the failure of the counting scheme might not be circumvented by > >> some other means i.e. does the failure of the counting scheme show that > >> the reals taken in that order are truly uncountable. > > >Uncountability has nothing to do with order. > Just so, but not necessarily obvious? With our meaning of the words, obvious. |