> On 24 Jan., 14:16, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> WM <mueck...@rz.fh-augsburg.de> writes:
>> > You will have have recognized that here the diagonal argument is >> > applied. It is obvious that up to every line = column the list is a >> > square. >> >> It is clear that, for all j, d(j) != t_j(j) and hence d != t_j. If >> that's what you mean by the diagonal argument, great! >> >> Once again, however, you say something that has no clear meaning to >> me. Can you clarify "It is obvious that up to every line = column the >> list is a square?" I've no clue what it means. > > Then ponder a while about the following sequence > > d > > d1 > 2d > > d11 > 2d2 > 33d > > and so on. In every square there are as many d's as lines. The same > could be shown for the columns.
Yes, in this sequence of three squares, what you say is true.
But none of this is relevant, because we've explicitly defined the anti-diagonal d and it is a triviality to see that it is an infinite sequence of non-zero and non-nine digits. And this fact really has nothing at all to do with limits of sequences of squares. It is all perfectly explicit.
>> > Neither the set of t_i does have a largest element. Nevertheless there >> > is no t_i of actually infinite length. >> >> Correct, to both claims. So what? The digit d(j) is defined for >> every j, and is neither 0 nor 9, so d is a real number which has no >> terminating decimal representation. >> >> Do you agree with that (obvious) claim or not? > > No. Pleeze try to understand: Presently we work in the system of > terminating decimals, by definition. There is no non-terminating > decimal. If we were so unlucky to met any non-terminating decimal we > would not know what to do. It would not even be defined. If the > diagonal gets non-terminating, we have to stop and cut it before it > becomes non-terminating.
Perhaps we should start more slowly.
Do you agree that (by presumption) t_i is defined for every i in N?
> >> >> Sorry, I could've sworn you were talking about what ZF proves. Can >> you state this in the language of ZF and prove it? Much thanks! > > Can't you imagine to define in ZF the set of all terminating decimals? > Is it too hard? Just take the above set that you couldn't recover in > our arguing, namely the set of all FIS (1, 2, ..., n) with n in n as > indices of the digits.
I don't want to imagine what you are thinking, because I will risk getting it wrong. I'd prefer that you explicitly give an argument in ZF so that we can determine whether it is valid or not.
> >> > In particular, what would be changed in the length of d if we admitted >> > also non-terminating t_i (of infinite length)? >> >> Nothing would change. So what? > > Look, presently we work in the system of terminating decimals - by > definition. If nothing changes when we switch to the system of non- > terminating decimals, do we switch then at all? How could we recognize > that we have switched?
I don't have any idea what these questions mean and will let them pass. Let's instead see a valid proof of a contradiction in ZF.
-- Jesse F. Hughes
"It's not really winning if you don't get to where you want to go." -- An inspirational slogan from James S. Harris