Alberto Monteiro wrote > > Robert Shaw wrote: > > >>> 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 don't think so. If you consider sequences such as (...,1/x,1/x.,1/x,exp,exp) when acting on 0 they don't converge to anything, That sequence just keeps flipping between e and 1/e. I think the only infinite sequences that converge converge to a fixed point of one of the functions in Ca. > > I agree that in the general case including the Identity is > a good idea. Yes, that guarantees G(S) is a subset of G(Si). If the only sequences in Si that converge are where after some point in the sequence all the functions in the sequence are identical, then that is a countable set of sequences in Si which makes G(Si) countable. > I still don't see what we gain by proving that G(Si) is uncountable. > It's information about the set. If the set is countable it's measure zero and typical numbers aren't in G(Si). Prroving it countable also proves G(Si) isn't R without having to exhibit a number not in G(Si) If the set is uncountable the probability that a randomly chosen number is constructable can be larger than zero. -- Robert
