In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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.