2008/1/12, Cristian Baboi
Suppose lim a_n = a , lim b_n = b, c_2n = a_n, c_2n+1 = b_n.
What is lim c_n ?
This reminds me of these good old days in "Classe prépa" (something typically french) doing these silly things with sequences and Epsilons... So let's try to do it : First, if we assume that a = b, that's ok : lim c_n = a = b Second, we consider that a \neq b. We will show that c_n do not have a limit. By contraction, we assume that c_n has a limit c. Then, by definition of limit, we have : \exists N, \forall n \geq N, | c_n - c | \leq \epsilon Therefore, this result holds for the suits (c_2n) and (c_{2n+1}), ie. : \exists N, \forall n \geq N, | c_2n - c | \leq \epsilon and \exists N, \forall n \geq N, | c_{2n+1} - c | \leq \epsilon By definition of c_2n and c_{2n+1}, we can replace them by a_n and b_n : \exists N, \forall n \geq N, | a_n - c | \leq \epsilon and \exists N, \forall n \geq N, | b_n -c | \leq \epsilon By definition of limit, this means that : lim a_n = c and lim b_n = c By unicity of the limit, this means that : a = c and b = c Leading to a contradiction as we have assumed that a \neq b. Q.E.D. But I'm still wondering about the relation between this and Haskell... -- Pierre-Evariste DAGAND