In article <4c1c61ae$0$2162$afc38c87@news.optusnet.com.au>, "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
> "Tim Little" <tim@little-possums.net> wrote in message > news:slrni1ol51.jrj.tim@soprano.little-possums.net... > > On 2010-06-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote: > >> Of course there exists a mapping from N to computable numbers. But > >> Cantor's proof requires more than that; it requires the mapping to > >> be recursively enumerable such that we can also explicitly list > >> them. > > > > It does not. If you believe otherwise, feel free to show where in > > Cantor's proof any requirement for recursive enumerability is stated > > or implicitly used. > > > > Its not. > > > The simple fact is that no such requirement exists. Cantor's diagonal > > proof applies to *every* mapping from N to R, recursively enumerable > > or not, showing that no such mapping is surjective. > > No. It requires that any listing be provided in advance, so that the > anti-diagonal can be formed. > > > > > > Your introduction of recursive enumerability into Cantor's proof is > > entirely your own fabrication. > > Well, I did say it was my interpretation of what was going on. > > The simple fact is that if insert the word "computable" in front of every > reference to "Real" in Cantor's proof that there is no list of all Reals, > you produce a proof that thye computable Reals cannot be listed either. yet > they are known to be countable. > > Cantor's proved that the Reals cannot be listed. It is your incorrect > assumption that this means they are also uncountable. Some countable sets > cannot be listed either, such as the set of Computable Reals. This is > because the set of computable Reals is not recursively enumerable, and hence > cannot be listed by a terminating process.
Ah! But that "listed by a terminating process" is not the same as merely "listed"
