"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message news:email@example.com... > >>>>> Perhaps if you could point out to me why you believe >>>>> Cantor's proof that not all Reals can be listed (as it >>>>> appears you do) but you don't believe my proof that >>>>> not all computable Reals can be listed. They appear >>>>> identical to me. >>>> >>>> All computable reals can be listed, but there is no >>>> finite algorithm for doing so. An "infinite algorithm" >>>> could list every computable real. An anti-diagonal, >>>> then, could be generated from this list but the >>>> algorithm creating the anti-diagonal is implicitly >>>> relying on the "infinite algorithm" underlying the >>>> list. In that sense the anti-diagonal is not >>>> computable. >>> >>> Agreed. >>> >>>> The set of all reals are a different story. Even with >>>> an "infinite algorithm" generating a list of reals, >>>> there is no way such a list could contain every real. >>>> For a proof, do a google search on Cantor's theorem. >>>> >>> >>> No, Cantor's diagonal construction does not prove this, >>> and nor is any of this machinery part of his proof. >>> >>> Cantor's proof applied to computable Reals proves that >>> the computable Reals cannot be listed. >> >> You have contradicted yourself. I wrote "All computable >> reals can be listed,..." and your reply was "Agreed". > > Typo, sorry. > > My whole argument is that they cannot be listed in their > entirety, or we could use a Cantor construction to produce > a computable Real not on the list.
You are wrongly assuming that only computable sets exist. That leaves out many sets. For example, if A is a subset of N, with |A|>1, most of the functions from N to A don't exist by your thinking. Because of your wrong assumption you falsely conclude that your anti-diagonal produces a computable real from the non-computable list containing all computable reals.