| I had to find time to think about this. > The technical step is where this fails: is there a bijection f that maps integers to their representation (as you have constructed) that also maps the sequence not in the list back to an integer? > We don't have that because f(f\inv(x)) must be on the list, but x is not on the list. Or: the diagonal complement is not an integer, so doesn't enter into any mapping of integers to integers. In contrast, for the presentation I saw for reals, the diagonal complement with infinite digits is of course a real, so doesn't have this problem. BTW: the presentations I've seen used a randomly-ordered list, which avoids the all-1 diagonal. Not sure why they this - but it led to my question. Anyway, all-1 is a real, but not an integer, so that answers my question. Thanks for taking the trouble and time to help unravel my question! Your last para, I think, is a simpler way around the article's problem that a real can have mor one representation in. infinite digits. e.g I think .999... = 1.000... |