On 25 Jan., 09:47, Virgil <vir...@ligriv.com> wrote:
> > > > 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. > > > Is there a first square where my observation would fail? > > Since you claim every line is necessarily finite, but the number of > lines is not, there will be a number of lines greater than the number of > digits in your finite first line.
In an ordered set like the sequence of squares above, we have for every subset a first element. If you claim to know a square that is not a square, then there must be a first square that is not a square. That could be proved by AC if it was necessary, but it is not, since the set of squares is countable. However, the axiom of choice cannot be circumvented in ZFC. Every set can be well-ordered such that every non-emty subset has a first element.
Or should the set of squares that are not squares be empty? Then, of course, it need not, in fact, it can not have a first element.
Or would you like to work in ZFC without AC? I think, you have no choice.