Robert Shaw wrote: > >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. > Ah, but our initial definition [some messages ago] was that we were considering the set of accumulation points of the sequences, not the limits. So, convergence is not required. The sequence that begins with exp, exp and then goes with 1/x, 1/x, etc has two accumulation points and does not converge. >> I still don't see what we gain by proving that G(Si) is uncountable. > > It's information about the set. > I still think it's too much trouble for too little gain, as "almost all" G(Si) will be uncountable - the exceptions come from very weird combinations of allowed functions. >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. > Using Probability for generic sets is very dangerous :-) Alberto Monteiro
