On 15 June, 07:15, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote: > "|-|ercules" <radgray...@yahoo.com> wrote in message > > news:87ocucFrn3U1@mid.individual.net... > > > Consider the list of increasing lengths of finite prefixes of pi > > > 3 > > 31 > > 314 > > 3141 > > .... > > > Everyone agrees that: > > this list contains every digit of pi (1) > > Sloppy terminology, but I agree with what I think you are trying to say. > > > as pi is an infinite digit sequence, this means > > > this list contains every digit of an infinite digit sequence (2) > > Again sloppy, but basically true. > > > similarly, as computable digit sequences contain increasing lengths of ALL > > possible finite prefixes > > Not "similarly", but if you are claiming that all Reals which have finite > decimal expansions can be listed, this is correct. > > > the list of computable reals contain every digit of ALL possible infinite > > sequences (3) > > No. You cannot form a list of all computable Reals. If you could do this, > then you could use a diagonal argument to construct a computable Real not in > the list.
A little more precisely, there does not exist a computable bijection from the natural numbers to the set of all computable reals.
But the set of computable reals is of course countable so there does exist a list of all computable reals - but not a computable list.
> > OK does everyone get (1) (2) and (3). > > No. (3) is not true, as it is based on a false premise (that the computable > Reals can be listed). > > > There's no need for bullying (George), it's just a maths theory. Address > > the statements and questions and add your own. > > > Herc > > -- > > If you ever rob someone, even to get your own stuff back, don't use the > > phrase > > "Nobody leave the room!" ~ OJ Simpson > >