On 20 Apr., 22:12, Virgil <vir...@ligriv.com> wrote:
> > 1 > > 1, 2 > > 1, 2, 3 > > ... > > contains all naturals, i.e., no one is missing, but that in no line > > there are all naturals. That means, in no line all are together, which > > are together in the union of all lines. Since when unioning, nothing > > is added, that not has been previously in at least one of the lines, > > the union must contain at least two numbers, call them m and n, which > > were in the list, but not in the same line. > > Nonsense! > > Either m > n or n > m or m = n, but only one of these. > > So either m > n and both are in every line that m is in, > or n > m and both are in every line that n is in, > or both are in every line the either is in.
Yeah. You got it. But that implies that all numbers of the first column are also in one and the same line. You could prove that even geometrically, turning the lines by pi/2 and writing all into one and the same column.