While reading through the lecture notes here (http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/week2.pdf , page 22, last paragraph), I came across the following " Thus there must be some $n_0 \geq N$ for which $|b_{n_0}| > \epsilon$. Since we already know that $|b_{n_0} - b_n| \leq \frac{\epsilon}{2}$ for all $n \geq N$, we thus conclude from the triangle inequality (how?) that $|b_n| \geq \frac{\epsilon}{2}$ for all $n \geq N$."

Trying to answer how? above took me a lot of time. Even after I could answer it, using the following algebraic manipulation $|b_{n_0}| = |b_{n_0} - b_n + b_n| > \epsilon $ and $|b_{n_0} - b_n + b_n| \leq |b_{n_0} - b_n|+|b_n|$, I still feel I have missed the idea.

My question is the following : What is the underlying idea of this algebraic manipulation? What are the clues that forces me to the underlying idea of this algebraic manipulation?

To put my question in perspective, if the author had not hinted at triangle inequality it would have been very difficult for me to figure out how the result was reached at. So,reading the paragraph (and imagine that the author has not hinted at triangle inequality), what do I "see" that can remind me that triangle inequality underlies this statement.

Is it the case that mathematicians "see" something (some idea, some object) which I fail to "see" in this case? Or, is it the case that "algebraic manipulation" is the only underlying idea and I need to remember it?

  • $\begingroup$ Experience as much mathematics as you can. Some techniques will appear over and over (such as adding and subtracting the same value) and start to become a part of your repertoire naturally. $\endgroup$ – Austin Mohr Aug 28 '15 at 15:48
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    $\begingroup$ Try to visualize what is going on using a number line, and for a first attempt disregard the absolute values. Start with the fact that $b_{n_0}$ lies to the right of $\epsilon$ on the line. Since $b_{n_0}$ and $b_n$ are at most $\epsilon/2$ apart, $b_n$ must lie to the right of $\epsilon /2$. $\endgroup$ – Umberto P. Aug 28 '15 at 15:48
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    $\begingroup$ In this simple example it helps a lot to draw a picture. Draw $b_{n_0}$ with distance larger than $\varepsilon$ from the origin, then draw a circle of radius $\varepsilon/2$ around $b_{n_0}$ $\endgroup$ – Thomas Aug 28 '15 at 15:49
  • $\begingroup$ How do mathematicians find the underlying idea? - They don't. The idea finds them. :-$)$ $\endgroup$ – Lucian Aug 28 '15 at 16:02
  • $\begingroup$ Even without the underlying idea, this particular instance seems straightforward: just write down the triangle inequality in full generality, and then figure out where the thing you want to know about has to fit in, then figure out where the things you already know have to go, and check if you get a result you're looking for. $\endgroup$ – user14972 Aug 28 '15 at 18:12

First, as noted, it's not too hard to visualize this particular problem. When dealing with a few real numbers and their absolute values, a picture usually works. To be even more explicit, you might "shade" the region of the number line where $b_{n_0}$ could be (knowing that $|b_{n_0}|>\epsilon$) and then shade the region where $b_n$ could be (knowing that $|b_{n_0}-b_n|\leq \frac{\epsilon}2$). That's probably as close as you'll get to intuition.

As for the algebraic manipulation: it does help to remember tricks like "add and subtract the same term." However, in this case, the combination of "triangle inequality + manipulation" has already been encoded into the reverse triangle inequality. If you look at its proof, you see that it's already done the work of "add and subtract a term" for you. Using this inequality, one could say $\frac{\epsilon}2 \geq |b_{n_0}-b_n|\geq |b_n|-|b_{n_0}| \geq \epsilon - |b_{n_0}|$, which rearranges to $b_n \geq \frac\epsilon2$.

The above paragraph would be utterly trivial... except that I sympathize with you about these types of manipulations, and somehow remembering the reverse triangle inequality helps. (It usually helps whenever you want to use the triangle inequality but the desired inequality is pointing in the other direction.)

Meanwhile, how could we know the triangle inequality is called for (reverse or or otherwise)? Well, the original question is obviously calling for a relationship between quantities $|x|$, $|y|$, and $|x \pm y|$. This might already shout "triangle inequality" -- but if it doesn't, you might go back to the definition of norms or distances and realize that the triangle inequality is pretty much the only fact you have at your disposal. In fact, most easy proofs are easy only because once you've identified the structures you're working with, you realize there are only a few obvious tools.


From your question I surmise that we are given a certain sequence $(b_n)_{n\geq0}$ which is Cauchy, and an $\epsilon>0$. This means that there is an $N$ such that $$|b_n-b_m|\leq{\epsilon\over2}\qquad\forall \>m, \>n>N\ .$$ In addition we are told that there are arbitrarily large $m$ with $$|b_m|\geq\epsilon\ .\tag{1}$$ The claim then is that under these circumstances we in fact have $$|b_n|\geq{\epsilon\over2}\qquad \forall n>N\ .$$ Proof. For any $m$, $n>N$ we have $$|b_n|\geq|b_m|-|b_m-b_n|\ ,$$ using a standard twist of the triangle inequality. Choosing an $m>N$ such that $(1)$ holds we get $$|b_n|\geq \epsilon-{\epsilon\over2}={\epsilon\over2}\qquad\forall n>N\ .$$


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