What's a number greater than ||x|-|y|| but less than |x-y|?

[Eclair_de_XII]

Homework Statement

"Prove that if $\{a_n\}$ converges to $A$, then $\{|a_n|\}$ converges to $|A|$. Is the converse true?"

Homework Equations

Triangle inequality: $|x+y|≤|x|+|y|$

The Attempt at a Solution

I think I solved the first part...

"If $\{a_n\}$ converges to $A$, then there must exist an $N∈ℕ$ such that for each $\epsilon>0$, $|a_n-A|<\epsilon$ for all $n≥N$. In turn, $|a_n|=|a_n-A+A|≤|a_n-A|+|A|$, which implies that $|a_n-A|≥|a_n|-|A|$. Taking the absolute values of both terms gives: $|a_n-A|≥| |a_n|-|A| |$. Thus, $\epsilon>|a_n-A|≥| |a_n|-|A| |$ for all $\epsilon>0$ and for all $n≥N$."

For the second part, I basically came up with a counterexample:

"Define $\{a_n\}$ by $\{a_n:a_n=-1,∀n∈ℕ\}$. We note that $\{a_n\}$ does not converge to $1$, but $\{|a_n|:|a_n|=|-1|,∀n∈ℕ\}$ does converge to $1$."

But I don't think this will cut it... What I want to find now, is an $\epsilon>0$ such that $\epsilon>| |a_n|-|A| |$ but $|a_n-A|>\epsilon$. Hence, I will need to find a number $\epsilon$ such that $|a_n-A|>\epsilon>| |a_n|-|A| |$, to show that the converse is not true. I will need to find an appropriate choice of $\epsilon$ that satisfies these conditions, in other words.

 

[FactChecker]

For the second part, I basically came up with a counterexample:

"Define $\{a_n\}$ by $\{a_n:a_n=-1,∀n∈ℕ\}$. We note that $\{a_n\}$ does not converge to $1$, but $\{|a_n|:|a_n|=|-1|,∀n∈ℕ\}$ does converge to $1$."

But I don't think this will cut it...

Why not? You may want to slightly modify and state the counterexample more carefully so that you can convince yourself if it is right or not.

 

[Eclair_de_XII]

You mean I should rearrange the order in which I state it, to make the implication--or lack thereof--clearer?

 

[FactChecker]

That is one thing that would help. Also, you haven't specifically stated what A is. Is it +1 or -1? Define your {a_n} and A first. Then show that is satisfies the absolute value limits. Finally show that it does not satisfy the non-absolute value limits. I don't think that you need to go into detail of the ε proof for such obvious limits, but I don't know what your class will want.

 

[Eclair_de_XII]

Okay, thanks for the advice.

 

[FactChecker]

Okay, thanks for the advice.

You might also be able to think of a closely related example where the {a_n} do not converge at all.

 

[Eclair_de_XII]

Oh, do you mean, $\{a_n:a_n=(-1)^n,\forall n \in ℕ\}$?