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.