"Virgil" <Virgil@home.esc> wrote in message news:Virgil-B10127.13553420062010@bignews.usenetmonster.com... > In article > <7d59f93c-4bc8-43e2-9b40-485733a01305@z8g2000yqz.googlegroups.com>, > WM <mueckenh@rz.fh-augsburg.de> wrote: > >> On 20 Jun., 01:58, Tim Little <t...@little-possums.net> wrote: >> > On 2010-06-19, Virgil <Vir...@home.esc> wrote: >> > >> > > There is, however, some question in my mind about the existence of a >> > > list of all and ONLY computable reals. >> > >> > Why? The computable reals can be proven countable, as you already >> > know, so there is a bijection between N and the set of countable >> > reals. >> >> That is not true. An exclusive list need not exist. >> >> The reals in a certain Cantor-list are countable. And if you form the >> anti-diagonal of that list and add it (for instance at first position) >> to the list, this new list is also countable. Again consttruct the >> anti-diagonal and add it to the list. Continue. The situation remains >> the same for all anti-diagonals you might want to construct. Therefore >> all reals constructed in that way are countable. Nevertheless it is >> impossible to put all of them in one list, because there would be >> another resulting anti-diagonal. >> >> Conclusion: It is impossible to obtain a bijection of all these reals >> with N although all "these" reals are countable with no doubt. > > > Equal cardinality of two sets requires, by definition, a bijection > between them. > > The naturals are, by definition of COUNTABLE cardinality so that any set > of equally countable cardinality must have a bijection with the > naturals. > > Cantor has shown that the set of all reals cannot biject with the > naturals, ergo, the set of all reals is NOT of countable cardinality. >
Actually, he shows that any purported list of all Reals must miss at last one Real.
*not* the same as being uncountable, witness the same argument applied to computable Reals.
> Actually Cantor's original "diagonal" proof was based on the set of all > functions from the naturals to a two element set,
No. It is based upon showing that any purported list of all Reals must neccesarily omit at least one Real, and he does this by providing an explicit construction of the anti-diagonal based upon examining the first n digits of the first n items on the list for all n.
Which is exactly what I am doing.
> but several people > have constructed bijections between that set of functions and the set of > all reals. I have done it myself.
Perhaps if you were to provide an actual list of all computable numbers - and only the computable numbers - your claim that they can be listed would be vindicated.
My bet is that you will have about as much success forming a list of all computable Reals as you would do forming a list of all Reals, but if you think such a list exists, go for it.
