Lemma: Cauchy sequences are bounded.

Proof: Given Cauchy sequence $(s_n)$, for $\varepsilon=1$ we obtain $N\in\Bbb N$ such that $m, n > N$ implies $|s_n − s_m| < 1$.

(2) implies $|s_n − s_{N+1}| < 1$ for all $n\geq N$, (from this step (2) to the next step (3) how does one get there? is it through the reverse triangle inequality ? do we simply say that $||s_n|-|s_{N+1}||\leq|s_n-s_{N+1}|<1 $

then $||s_n|-|s_{N+1}|| < 1$ for all $n\geq N$

then add $|s_{N+1}|$ to both sides of the inequality? is it that simple?)

(3) implies $|s_n| < |s_{N+1}| + 1$ for all $n>N$. If $M = \max\{|s_{N+1}| + 1, |s_1|, |s_2|,\ldots, |s_N |\}$, then $|s_n| \leq M$ for all $n \in\Bbb N$ hence the sequence $(s_n)$ is bounded. QDE

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    $\begingroup$ $|s_n|=|s_n-s_N+s_N|\leq |s_n-s_N|+... $\endgroup$ Nov 4 '16 at 19:46

If $n>N$, then $$|s_n| = |s_n-s_{N+1} + s_{N+1}| \leq |s_n-s_{N+1}| + |s_{N+1}|< |s_{N+1}| + 1$$

This ensures that $|s_n|\leq M$ for any $n$: $|s_1|,|s_2|,\ldots, |s_N|\leq M$ and $|s_n| < |s_{N+1}| + 1\leq M$ if $n> N$.

It is not hard to visualize this geometrically (by visualize I mean you can easily draw a picture):

1) Pick any positive radius $\varepsilon$ (you picked $\varepsilon = 1$).

2) By sequence being Cauchy, you can find $N$ such that almost all (all but finitely many) elements of the sequence are within interval $I =\langle s_{N+1}-\varepsilon,s_{N+1}+\varepsilon\rangle$.

3) Now you can find $M$ such that $[-M,M]$ contains both the elements not contained in $I$ (there are finitely many of them) and interval $I$.

4) QED


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