On 26 Jun., 23:14, "Mike Terry" <news.dead.person.sto...@darjeeling.plus.com> wrote: > "WM" <mueck...@rz.fh-augsburg.de> wrote in message > > news:firstname.lastname@example.org... > > > On 26 Jun., 00:32, "Mike Terry" > > > > > Please let me know: Does the list consisting of A0, A1, A2, A3, ... > > > > contain its antidiagonal or not? > > > > No, of course not. > > > The set of all these antidiagonals A0, A1, A2, A3, ... constructed > > according to my construction process and including the absent one is a > > countable set. > > Yes, I've agreed that several times. > > > > > > > That is not a problem. It shows however, that there are countable sets > > > > that cannot be listed. > > > > How exactly does it show that there are countable sets that cannot be > > > listed? > > > See the one above: The results of my construction process. > > No. The above shows that you've constructed a countable sequence of numbers > (A0, A1, A2, A3,...) which CAN be listed, namely consider the list (A0, A1, > A2, A3,...). > > Read carefully!
In order to do so, I posed the question: Does the list consisting of A0, A1, A2, A3, ... contain its antidiagonal or not?
You said no. Therefore your assertion "which CAN be listed" is plainly wrong. >
Of course it is not possible to list a list and its antidiagonal. > > > > > > > OK, but you've still not given any explanation why you think the > > > antidiagonals of your process cannot be listed! (I'm genuinely > baffled.) > > > You said so yourself, few lines above. > > Rubbish. I have never said anything other than that (A0, A1, A2, ...) is a > countable sequence of reals that CAN be listed
but not with all its antidiagonals. > > Go on - have one more go, then I'll give up...
You'd better give up than make contradictory claims.