Here is the classic proof that Cauchy sequences are bounded:

Let $\{x_n\}$ be a Cauchy sequence. For $\epsilon > 0$, choose $N$ such that if $n, m \geq N$, then $|a_n - a_m| < \epsilon$. Then $$|x_n| = |(x_n - x_N) + x_N| \leq |x_n - x_N| + |x_N| \leq \epsilon + |x_N|$$ for all $n \geq N$.

Let $M = \max\{|x_1|, |x_2|, ..., |x_N\|\} + \epsilon$. Then $|a_n| \leq M$ for all $n$.

I see how the sequence is bounded for every $\epsilon > 0$, but I am having confusion because the particular bound $M$ depends on the value of $\epsilon$. From more elementary proofs in calculus, I did not think that the chosen bound could be dependent on the $\epsilon$ given. What I am trying to ask is, why is this dependency allowed?

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    $\begingroup$ Take $\varepsilon=1$ $\endgroup$ – Pedro Feb 3 '14 at 22:54
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    $\begingroup$ You pick one $\epsilon > 0$. Usually, one picks $\epsilon = 1$, then the confusion doesn't arise. $\endgroup$ – Daniel Fischer Feb 3 '14 at 22:54

Forget about $\varepsilon$.

Set $\varepsilon=1$. Then, as $\{x_n\}$ is Cauchy, there exists an $N$, such that $$ m,n\ge N\quad\Longrightarrow\quad |x_m-x_n|<1. $$ Thus $$ |x_n|\le 1+\max_{1\le k\le N}\{|x_k|\}. $$


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