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Proposition. Every Cauchy sequence $(a_n)$ is bounded.

Proof. Given any $\varepsilon > 0$ there is an integer $N$ such that $|a_n-a_m| < \varepsilon$ for all $n, m \geq N$. So $|a_n|-|a_m| \leq |a_n-a_m| < \varepsilon$ for all $n,m \geq N$.
Taking $m = N$ and transposing, we have $|a_n| < |a_N| + \varepsilon$ for all $n \geq N$. Thus, for all $n$, $$ |a_n| \leq \max\{|a_1|, \dotsc,|a_{N-1}|, |a_N| + \varepsilon \}. $$ A bound for $\{a_n\}$ is then $M = \max\{|a_1|, \dotsb,|a_{N-1}|,|a_N| + \varepsilon \}$.

My question is why we took $m = N$.

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    $\begingroup$ Because it gives the desired bound. What were you expecting? $\endgroup$ – anomaly Jan 9 '16 at 6:31
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For all $m \ge N$ and all $n \ge N$, we have $|a_n| < |a_m|+\varepsilon$.

Thus, we can pick any integer $m \ge N$, and get the bound $|a_n| < |a_m|+\varepsilon$ for all $n \ge N$.

Therefore, for any choice of $m \ge N$, we can set $M = \max\{|a_1|,|a_2|,\ldots,|a_{N-1}|,|a_m|+\varepsilon\}$ and have the bound $|a_n| \le M$ for all positive integers $n$.

The proof picked $m = N$ because it was a convenient choice of $m$ which satisfies $m \ge N$. We could have also picked $m = N+1$, or $m = N+2$, or $m = N+123456789$, or $m = N+\left\lfloor\pi^{\pi^{\pi}}\right\rfloor$, or any other integer greater than or equal to $N$ and the proof would still work out. Of course, $m = N$ was the most convenient to write.

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  • $\begingroup$ My question would be, why would you even have to specify a "m". Could you not delete the "taking $m=N$ and transposing part and just use $a_m$ whenever needed? $\endgroup$ – Jonathan Nov 14 '16 at 18:07

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