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Let ($a_n)_{n \in \mathbb{N}}$ be a sequence, prove that if $|a_n|$ converges to $0$ then ($a_n)_{n \in \mathbb{N}}$ converges to 0 as well. Now let $|a_n|$ converge to $0$ and let $\epsilon > 0$ that means that $||a_n|-0| < \epsilon$ for an $N \in \mathbb{N}$ such that $n > N$.

But how do I go from here? I should conclude from $||a_n|-0| < \epsilon$ that $|a_n-0| < \epsilon$.

MoebiusCorzer
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JonDoeMaths
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2 Answers2

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You're almost there:

$||a_n|-0| = ||a_n|| = |a_n| = |a_n-0|$

Ovi
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  • I've just seen this topic while I was searching about this fact. Is it true for every absolute value that $||x||=|x|$? How can I see this? Thanks! – robbis Apr 23 '19 at 21:46
  • @robbis By every absolute value, do you mean every norm on any vector space? Or are you only talking about real numbers and the usual absolute value? – Ovi Apr 23 '19 at 22:46
  • I mean every absolute value on a generic field. I have to prove that $|x|<1$ if and only if $|x^n| \to 0$ if and only if $x^n \to 0$, not only for the usual absolute value. Do you have any suggestions? – robbis Apr 24 '19 at 08:22
  • @robbis Hmm I just learned from you that there is a more general notion of absolute value that can be defined for diferent fields... I would like to think about that problem but unfortunately I don't have time. Have you tried following the proof of that fact for the usual absolute value, and adapting it to the general case? – Ovi Apr 24 '19 at 09:49
  • I've read that an absolute value over a field is a function from the field to $\mathbb{R}$ such that $|x|\ge 0$, $|x|=0$ iff $x=0$, $|xy|=|x||y|$ and $|x+y|\le |x|+|y|$ – robbis Apr 24 '19 at 10:19
  • yes, I tried what you say but I don't know how to see that $||a_n||=|a_n|$ – robbis Apr 24 '19 at 10:20
  • @robbis Let $| \cdot|_{R}$ be the regular absolute value on $\mathbb{R}$ and let $| \cdot |_F$ be the absolute value on the field $F$ you're considering. I think $||a_n||$ can only be interpreted as $||a_n|_F|_R$. Now $|a_n|_F$ is a non-negative real number by definition.In general, if $x$ is a non-negative real number, then $|x|_R=x$ (you could see this formally from the formal piecewise definition of $|x|$). Letting $x = |a_n|_F$ and taking absolute value of both siedes, we have that $|x|_R = ||a_n|_F |_R$, or (since $x$ is non-negative) $x = ||a_n|_F |_R$, or $ |a_n|_F = ||a_n|_F |_R$ ... – Ovi Apr 24 '19 at 16:08
  • @robbis (continues) as desired. Does this work? – Ovi Apr 24 '19 at 16:09
  • sorry but I didn't log in until now. Your last answer seems to me correct and right for my purpose, so thank you very much for your help! – robbis May 29 '19 at 14:36
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In fact $||a_n||=|a_n|$ so you're done. If $||a_n|-0|$ is less than $\epsilon$ then so is $|a_n-0|$.

ziggurism
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