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I don't understand this one part in the proof for convergent sequences are bounded.


Let $s_n$ be a convergent sequence, and let $\lim s_n = s$. Then taking $\epsilon = 1$ we have:

$n > N \implies |s_n - s| < 1$

From the triangle inequality we see that: $ n > N \implies|s_n| - |s| < 1 \iff |s_n| < |s| + 1$.

Define $M= \max\{|s|+1, |s_1|, |s_2|, ..., |s_N|\}$. Then we have $|s_n| \leq M$ for all $n \in N$.

I do not understand the defining $M$ part. Why not just take $|s| + 1$ as the bound, since for $n > N \implies |s_n| < |s| + 1$?

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Okay, this is may be a strange question, but wouldn't the triangle inequality say: $|s_{n}-s| \leq |s_{n}|+|s|$? – Greg.Paul Oct 6 '15 at 1:12
Yes. I had the same thought. I believe one would just choose a different (larger) $N$. This is from Ross's Elementary Analysis (2ed), where the identical "from triangle inequality" remark is made. Kind of nonsense. – SXibolet Feb 12 at 23:42
Actually, no. Using the standard Triangle Inequality here is a tad naïve. I believe what Ross implies is $\vert s_n - s\vert \ge \vert s_n \vert - \vert s \vert < 1 \implies \vert s_n \vert < 1 + \vert s\vert$ – SXibolet Feb 23 at 23:54
We could still use Triangle Inequality here. Begin with $|s_{n}| = |s_{n}-s+s| \leq |s_{n}-s|+|s|$, which implies $|s_{n}| - |s| \leq |s_{n}-s|$. It follows that $|s_{n}| - |s| \leq |s_{n}-s| < 1$. – nexolute Mar 1 at 6:13
up vote 10 down vote accepted

$|s|+1$ is a bound for $a_n$ when $n > N$. We want a bound that applies to all $n \in \mathbb{N}$. To get this bound, we take the supremum of $|s|+1$ and all terms of $|a_n|$ when $n \le N$. Since the set we're taking the supremum of is finite, we're guaranteed to have a finite bound $M$.

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Taking the Max ensures that we take into account elements of $a_n$ for $n \leq N$ that could be bigger than $|s| + 1$? – CodeKingPlusPlus Oct 15 '12 at 3:05
@CodeKingPlusPlus Yes, you got it right! – Ayman Hourieh Oct 15 '12 at 8:36
Helpful even two years later! – mathjacks Oct 15 '14 at 3:45

Because you want to be sure that the bound is large enough to ensure that $|s_n|\le M$ for all $n\in\Bbb N$, not just for all $n>N$. Taking $M\ge|s|+1$ ensures that the only possible exceptions to $|s_n|\le M$ are $s_1,\dots,s_N$, and taking $M\ge\max\{|s_1|,\dots,|s_N|\}$ takes care of these as well.

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Let's prove uniqueness first. Suppose the sequence has two limits, a and a'. Take any > 0. Then there is an integer N such that:

| aj - a | < 

if j > N. Also, there is another integer N' such that

| aj - a' | < 

if j > N'. Then, by the triangle inequality:

| a - a' | < | a - aj + aj - a' |
      |aj - a | + | aj - a' |
      < + = 2 

if j > max{N,N'}. Hence | a - a' | < 2 for any > 0. But that implies that a = a', so that the limit is indeed unique.

Next, we prove boundedness. Since the sequence converges, we can take, for example, = 1. Then

| aj - a | < 1 

if j > N. Fix that number N. We have that

| aj | | aj - a | + | a | < 1 + |a| 

for all j > N. Define

M = max{|a1|, |a2|, ...., |aN|, (1 + |a|)} 

Then | aj | < M for all j, i.e. the sequence is bounded as requi

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Here's a MathJax tutorial :) – Shaun Oct 24 '14 at 9:23

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