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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.

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2 Answers 2

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?

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Aha, I think the $b_n$ on the far left was confusing me. So I would manipulate $\mid B \mid \leq \mid b_n \mid + \mid B - b_n \mid$ to get $|b_n| \geq |B|-|b_n - B|$? Is it okay to swap the $|B - b_n|$ with $|b_n - B|$ since they are both just a measure of distance between $B$ and $b_n$? I think that was also confusing me. –  Jarrod Sep 30 '10 at 1:51
    
You are right on both points. Regarding the equality of $|b_n-B|$ and $|B-b_n|$ your intuition is good. You can also think in terms of the modulus operator throwing away a number's sign, so that $|a|=|-a|$. –  Jyotirmoy Bhattacharya Sep 30 '10 at 3:47
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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.

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Thanks, that helps. So is the choice of $\epsilon = M = \frac{B}{2}$ sort of arbitrary? I always have a hard time understanding how I would synthesize an $\epsilon$ for these types of proofs. I realize now that the right half of the inequality comes from subtracting $\mid b_n - B \mid \geq \epsilon$ from $B \geq B$, so now I'm confused as to how the triangle inequality comes into play. I don't understand where the minus comes from between $\mid B \mid$ and $\mid b_n - B \mid$ relative to the $\mid b_n + B - B \mid$ portion. –  Jarrod Sep 30 '10 at 1:39
    
Nevermind, I figured out what I was having trouble with. Thanks again! –  Jarrod Sep 30 '10 at 4:01
    
@Jarrod: Dear Jarrod, Yes, the choice of $B/2$ is pretty arbitrary. Any number $M$ that is strictly between $0$ and $B$ would work just as well. One could then take $\epsilon$ to be $|B - M|$, and the same argument would work. The (pretty silly) reason for taking $B/2$ here, and not some other $M$, is just that $1/2$ is an easy fraction to work with (and despite being arbitrary, looks less arbitrary than if you took say $M = (24/25)B$ --- although that would be just as valid). –  Matt E Sep 30 '10 at 7:37
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