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Show that if $(b_n) \to b$, then the sequence of absolute values $\left| b_n \right|$ converges to $\left| b \right|$.

Solution (partial):

By the triangle inequality, $\left| b_n \right| = \left| b_n - b +b \right| \leq \left| b_n - b \right| +\left| b \right|$.

Thus $\left| b_n \right| - \left| b \right| \leq \left| b_n -b \right|$,

and in fact $\left| \left|b_n \right| - \left| b\right| \right| \leq \left|b_n - b \right|$...


Why exactly is the "and in fact" line justified? That is, why are you allowed to just take the absolute value of $\left| b_n \right| - \left| b \right|$ and assume the inequality still holds?

As someone has already kindly pointed out to me in another post, this is the reverse triangle inequality. My understanding is that instead of applying that theorem here, the author sort of "proves" it as part of the solution. So I guess I am looking for an explanation of that step in the proof for the (in a limited sense) general case of the reverse triangle inequality... Please let me know if I need to be more clear or expand somewhere, otherwise, thank you for your help!

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The "and in fact" can be justified switching $b_n$ and $b$ in the first line to get $|b|-|b_n|\leq |b_n-b|$. – Davide Giraudo Sep 22 '11 at 10:37
@Davide: that looks like an answer. Why not make it one? – robjohn Sep 22 '11 at 10:52
May I just say that this question makes me really dislike the standard notation for absolute values. Quick, what's $\left|\left|a\right|+c\left|d-\left|e\right|\right|\right|$? – Rahul Sep 22 '11 at 17:41
up vote 4 down vote accepted

The other answer is wrong. You can't go from $|a|-|b|\leq |a-b|$ to $||a|-|b|| \leq |a-b|$ via the inequality $|a|-|b| \leq ||a|-|b||$.

However, you can argue by symmetry $|a|-|b| \leq |a-b|$ and $|b|-|a| \leq |b-a|=|a-b|$, so $||a|-|b||=\max{\{|a|-|b|,|b|-|a|\}} \leq |a-b|$.

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The idea to prove $||a| - |b|| \le |a - b|$ is very simple:

$|a - b|$ denotes the distance between numbers $a$ and $b$. The inequality says that by doing absolute value on $a$ and $b$ you will never increase the distance - because if $a$ and $b$ have the opposite sign, the distance only can get shortened, and if they had the same sign, the distance will be the same.

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I don't think your answer is really wrong so much as begging the question. You show how to prove $|a|-|b|\le|a-b|$ from $||a|-|b||\le|a-b|$, when the original question (as I read it) was how to prove $||a|-|b||\le|a-b|$ from the triangle inequality. The argument you give for that is quite handwavey, even if one probably could expand it into a real proof if one tried. (The key lemma that's missing is something like $a>0>b\implies|a-b|>|a+b|$, for which you can use the triangle inequality.) – Ilmari Karonen Sep 22 '11 at 16:25
Did you link to the wrong post? In that post you prove $|a|-|b|\le||a|-|b||$, which is indeed trivial, but I don't see how $||a|-|b||\le|a-b|$ follows from that (except for the trivial sense in which any two tautologies follow from each other). – Ilmari Karonen Sep 22 '11 at 16:51
@Ilmari, I admit confused it a litte bit so I deleted the first paragraph of my post - and I think now it is OK - exactly what the OP asked. – Tomas Sep 22 '11 at 17:32

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