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As $\{x_n\}$ is a Cauchy sequence, there exists a positive integer $N$, such that for any $n \geq N$ and $m \geq N$, $d(x_n,x_m) \lt 1$; that is, $|x_n-x_m| \lt 1$. Put $M = |x_1| + |x_2| + |x_3| + ... + |x_N |+ 1$. Then $|x_n| \lt M, $ for all n in the set of Natural numbers.

I don't understand how this can be a bound, won't this sum approach infinity since it's an infinite sequence? I know another way to prove this is to let $M=max\{|x_1|, |x_2|, |x_3|, ..., |x_N| + 1 \}$, but I can't figure out how to derive this from $|x_n| \lt 1 + |x_N|$.

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  • $\begingroup$ $N$ is a fixed number. $\endgroup$ – user42761 May 19 '16 at 15:19
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$N$ is fixed and $$M=|x_1|+|x_2|+|x_3|+\cdots+|x_N|+1$$ is a finite sum.

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