Question

a real sequence (a_n) converges to "c", iff for each e>0 there exists a natural number N such that
|a_n-c|=N. Can we define this the other way round? like, (a_n) converges to "c" iff for any natural number N there exists a positive real e, such that for all n>=N |a_n-c| 14 years ago

Answer 1

can u elaborate please

Answer 2

hmm...actually ignore all that, thats wrong. Thats not the reason why you cant switch the order. Its because you can have an epsilon that satisfies the statement for every N, but the sequence could still be divergent.
For example:
0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ....
For every N, i can claim my limit is .5, and make epsilon = 1. Then the statement:
\[(\forall N)(\exists \epsilon) (n\ge N \Rightarrow |a_n-c|<\epsilon)\]
is true (each a_n is in between (.5-1, .5+1) = (-.5, 1.5)), but the sequence doesnt converge.