Is the converse of the absolute value limit theorem true?

[nitro]

Hi all,
I'm wondering about this question

I can prove that if lim_{n->inf} (a) = L then lim_{n->inf}abs(a) = abs(L)
however.. is the converse true?
thx

 

[quasar987]

Consider the map f:R-->R defined by

f(x)= 1 if x is rational and =-1 if x is irrational

Then |f| is identically 1 so it is everywhere continuous, while f is nowhere continuous by the density property of the rational and irrational numbers.

In particular, for instance, let x_n be a sequence converging to 0 that contains an infinity of rational numbers and an infinity of irrational numbers and set a_n :=f(x_n). Then |a_n|-->1 but a_n does not converge because it has a subsequence converging to 1 and another converging to -1.

However, if |a_n|-->L and a_n converges, then it must be that a_n-->±L because if a_n-->P, and if

|a_n-P|<epsilon,

as soon as n>N, then because of the triangle inequality

||a_n|-|P||<=|a_n-P|,

it follows that

||a_n|-|P||<epsilon

as soon as n>N also.

But this is the same as saying that |a_n|-->|P|. So L=|P|. So P=±L.

 

[Pere Callahan]

Considering the constant sequnce -1 would also do the job. Its absolute value convreges to 1, so you can choose L=1 (of course you could also choose l=-1, but you don't have to.)
Then the sequnece does not converge to L...