> On 24 Jan., 13:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> "Jesse F. Hughes" <je...@phiwumbda.org> writes: >> >> > Yes, but let's be perfectly clear here. What we have now is this: >> >> > a function t: N -> R such that, for all i in N, t(i) is a real >> > number with terminating decimal representation. We'll write this as >> > t_i and we'll write t_i(j) for the j'th digit of t_i (in its >> > terminating representation, which is, of course, unique). >> >> > Let d in R be defined by >> >> > d(j) = 7 if t_j(j) != 7 >> > d(j) = 6 if t_j(j) = 6. >> >> ^ Should be 7, of course. >> >> >> >> > Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in >> > N. >> >> > Is this what you mean up 'til now? >> >> So, for instance, take the obvious enumeration of finite decimal >> representations (in [0,1), of course) >> >> 0.0 >> 0.1 >> 0.2 >> 0.3 >> 0.4 >> 0.5 >> 0.6 >> 0.7 >> 0.8 >> 0.9 >> 0.11 >> 0.12 >> ... >> >> Clearly, t_i(i) = 0 for every i, and hence d_i = 7 for every i. Thus, >> d = 0.777..., which is a number without any finite decimal >> representation. > > As long as the line has a finite enumeration, also the column has. In > t_j(j) the first j is identical with the second j. And both belong to > a FIS of lines and columns, respectively. This holds for every j.
Once again, I've no idea what point you're trying to make here.
Let's start simply. Take the above list and the previously stated definition of d. Do you agree or deny that d = 0.777... per that definition?
Second, can you please state what precise claim you're trying to make when you write "As long as the line has a finite enumeration, also the column has?" I haven't a clue.
It would be swell if you could write it in more or less set-theoretic terms, since, after all, you are allegedly providing a proof in ZF.
Thanks much. -- "Do you know some logic? Please apply it. If all a in A also are b in B, then it is not excluded that also a c in C that is not in A nevertheless is a b in B." -- Wolfgang Meuckenheim, an actual professor