Robert Shaw wrote: > >For that matter the calculators we used could calculate trigometric >functions in radians, degrees, or grades, effectively allowing >multiplication >and division by pi/180 and 0.9. All these sets have a G(S) which includes >the rationals. > Oh, please, do not mention grades! This is the only stupid think in SI O:-) >>> We don't know much about G(S) or G(Si) for the particular set >>> Ca of calculator functions, and less for arbitary sets of functions. >>> Even for Ca we don't know if G(Si) includes Q. >>> >> Uh? Wasn't this proven before? > >I was implicitly assuming Ca includes the identity function. >If it does G(S) is a subset of G(Si). > >If it doesn't things are less clear. There are arbitarily long >finite sequences that correspond to any rational but it's >not entirely clear that there are infinitely long sequences >that do. > >It may be simpler to include the identity function in Ca. > But since there are inverse functions in Ca, then all rational numbers are accumulation points of some sequences. I agree that in the general case including the Identity is a good idea. >To get further with this we need to consider when different >elements of Si map to the same element of G(Si). >That happens if some element of S is equivalent to the identity >under G. E.g we can insert ln exp at any point in a sequence >without changing its image under G. >This gives a countably infinite subset of Si mapping to each element >of G(Si) but there may be other conditions under which different >elements have the same image that leave us with uncountably many >elements of Si for each element of G(Si) > I still don't see what we gain by proving that G(Si) is uncountable. Alberto Monteiro
