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The question from my textbook is as follows:

Prove any sequence $\{a_n\}$ that $\lim_{n\to\infty} a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$.

My attempt:

If $\lim_{n\to\infty} a_n = 0$, then for every $\epsilon > 0$, there exists an $n\in N$ such that $|an -0| = |a_n| = ||a_n|-0| < \epsilon$. Thus $\lim_{n\to\infty} |a_n| = 0$.

If $\lim_{n\to\infty} |a_n| = 0$, then for every $\epsilon > 0$, there exists an $n\in N$ such that $||a_n| - 0| = a_n = |a_n - 0| < \epsilon$. Thus $\lim_{n\to\infty} a_n = 0$.

I thought that my proof is a bit sketchy. Is my proof a valid proof? If not, how would I go about solving this?

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    I'd say you did fine on the first part. For the second part, use the fact that $-|a_n| \le a_n \le |a_n|$. – 2017-02-16
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    I think this proof is correct. You should specify that the same $n$ works both the sequences. – 2017-02-16
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    You reasoning is correct but your definition of limits is mangled. It isn't there exist an n so that ... blah.... that is just one value. The def is there exist and M so that for all n > M then .... blah. Do you see the difference and why it is important? But your reasoning $|a_n| = |a_n - 0| = ||a_n|-0|$ is just fine and that is indeed all you need to demonstrate to prove this. (Just use the *correct* definition of limits though.) – 2017-02-16

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Your only problem is that you don't have the correct definition of limit.

$\lim_{n\rightarrow \infty} a_n = c$ means that for any $\epsilon$ there exists and $M \in \mathbb R$ so that $n > M; n \in \mathbb N$ will imply $|a_n - c| < \epsilon$.

You stated something entirely different.

Then as $|a_n - 0| = |a_n| = ||a_n| - 0|$ we have $|an - 0| < \epsilon \iff ||a_n| - 0|<\epsilon$ and so $\lim_{n\rightarrow \infty} a_n =0 \iff \lim_{n\rightarrow \infty} |a_n| =0$.

The part of your proof that felt sketchy was just fine. But the part that felt just fine (the definition of limit) was wrong.

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It doesn't look fine to me.

For example for the first part, you argued that $$ \forall \epsilon >0, \exists n\in\mathbb{N} : ||a_n|-0|<\epsilon\tag{1} $$ and concluded that $$ \displaystyle\lim_{n\to\infty}|a_n|=0\tag{2} $$ While $(1)$ is true, it is not equivalent to $(2)$. The meaning of $(2)$ is $$ \forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies ||a_n|-0|<\epsilon\tag{3} $$ Of course the same argument you gave will lead you to $(3)$.

Note: Define $$ a_n := \begin{cases}\frac{1}{n} & n \text{ even}\\1 & n \text{ odd}\end{cases} $$ Then the sequence $(a_n)$ satisfies $(1)$ but not $(2)$.

As for the second implication you tried to prove, you ran into the same problem and also the equality $||a_n| - 0| = a_n$ isn't true if $a_n<0$ (which can happen).

All you need in order to prove the result is to note (as you did) that $$ |a_n-0| = ||a_n|-0| $$ Then $$ \lim_{n\to\infty} a_n=0\\ \Updownarrow \\ \forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies |a_n-0|<\epsilon \\ \Updownarrow \\ \forall \epsilon >0, \exists N\in\mathbb{N},\forall n\in\mathbb{N} : n>N \implies ||a_n|-0|<\epsilon \\ \Updownarrow \\ \lim_{n\to\infty} |a_n|=0 $$

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It's not correct. You're emphasizing that there exists (at least) one $n$ that fits. But what you should say is those n large enough will fit.

$\epsilon$-$\delta$$(\epsilon$-$K)$ method is like a machine: For any given $\epsilon$, you put the $\epsilon$ into it, and get a corresponding $\delta$$(K)$.

In your proof, you don't take the advantage of the correspnding $K^\dagger$ from the definition of $\lim_{n\to\infty}(\bullet)$.

$^\dagger K\in \Bbb{Z}^+$.

Assume LHS is true: $$\forall \epsilon\gt 0, \exists K \in \Bbb{N}, \forall n\in \Bbb{N}: n\ge K\implies |a_n - 0|\lt\epsilon,$$

which implies the RHS is true, by the reasoning you given, if you choose the just corresponding K to limit your $n$. The reverse is similar.

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    No need for triangle inequality, it's a bit redudant – 2017-02-16
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    @YoTengoUnLCD Ok I've fixed it, thanks for the good catch. – 2017-02-16