Clarification on Proof. If a sequence is convergent, then the modulus of the sequence is bounded below. There is a lemma in my introduction to analysis book for which the proof eludes me. I was hoping to get some clarification. Here goes:
Lemma: If $\lbrace b_n \rbrace_{n=1}^\infty$ converges to B and B $\not=$ 0, then there is a positive real number $M$ and a positive integer $N$ such that if $n \geq N$, then $\mid b_n \mid \geq M$.
Proof: Since $B \not= 0, \frac{\mid B \mid}{2} = \epsilon > 0.$ There is $N$ such that if $n \geq N, \mid b_n - B \mid < \epsilon.$ Let $M = \frac{\mid B \mid}{2}.$ Thus for $n \geq N$:
$\mid b_n \mid = \mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid \geq \mid B \mid - \frac{\mid B \mid}{2} = \frac{\mid B \mid}{2} = M$
The lemma makes intuitive sense to me because of course if a sequence converges to a number other than 0, its absolute value will always be above some number $M$ for $n \geq N$. However, I get lost in the inequalities of the proof. I don't understand how the author came up with $\mid b_n - B + B \mid \geq \mid B \mid - \mid b_n - B \mid$. I understand that he added and subtracted $B$ to $b_n$ for the left side, but I'm not so clear on the how he came up with $\mid B \mid - \mid b_n - B \mid$. It looks like an iteration of the triangle inequality, but I don't quite see it.
As an aside, the book had an exercise for a variation on this lemma that involved proving $b_n \geq M$ for all $n$. The proof for that involved setting $M = min \lbrace b_1, b_2, ..., b_n \rbrace$, which made a lot more sense to me than the inequalities for this proof.
 A: You are right, it's just the triangle inequality. $B=b_n+(B-b_n)$. So $|B|\leq |b_n| + |B-b_n|$.
By the way, shouldn't it be $n \geq N$ in your second para?
A: Just a remark (that is too long to put in a comment box): The idea of the proof is just this, that if $x_n$ approaches $B$,
then eventually (i.e. for large enough $n$) it has to be within distance $|B/2|$ of $B$, 
and once this happens we can be sure that $x_n \geq B/2$ (if $B$ is positive) or that
$x_n \leq B/2$ (if $B$ is negative).
If this isn't clear just from how I've written it, draw  picture: label $B$,
draw the interval of points within distance $|B/2|$ of $B$, which is the interval
$[B/2, 3B/2]$ (or switch the endpoints if $B$ is negative), and then just randomly
draw in a sequence of points getting closer and closer to $B$.  You'll see that eventually
you have to put them all inside this interval, so they will all be at least $B/2$
(or at most $B/2$ if $B$ is negative).
The manipulations with absolute values are just a slightly tortuous verification that
if $x$ is within distance $|B/2|$ of $B$, then $x$ is at least $B/2$ (or at most $B/2$
is $B$ is negative).  It is easy to write down statements that are obvious when you say
them out like this, or when you draw a picture, but which become a bit of a battle when
you try to derive them formally from the triangle inequality.  The fact that the author
is treating the case when $B$ is negative at the same time adds another small but distracting layer of complexity, because you have to write $|B/2|$ rather than $B/2$ for the distance under consideration.  
As you become more familiar with these sorts of manipulations, when you come to a point
like this in a text where you recognize some kind of triangle inequality manipulations, it is often easier (and good training!) just to draw a picture and figure out the manipulations for yourself, rather than working through somebody else's slightly painful manipulations on the page.  
