Prove if $x_n$ is a Cauchy sequence, it is bounded above and below.

Curious as to whether this proof also works, the proof I know is that instead of having $|x_m + 1|, |x_m - 1|$ in the max you have $|x_N| + 1$. I'm not sure if the method below holds, as it is only true for $m \geq N$.

Let $\epsilon = 1$, then we know that $\exists N \ s.t. \forall n,m \geq N$:

$|x_n - x_m| < 1$.

Let $b = \max\{|x_1|, |x_2|, ..., |x_{N-1}|, |x_m + 1|,|x_m - 1|\}$. Then we know that for $1 \leq n \leq N - 1$ that $|x_n| \leq b$, and for $n,m \geq N$:

$ x_m - 1 < x_n < x_m + 1$ and thus $|x_n| \leq b$. Therefore, for all $n$:

$-b \leq x_n \leq b$, is bounded above and below.


When you define $b$ you need to specify $m$, but that isn't hard to do.

Note that in particular $n \ne N$ implies $|x_n - x_N| < 1$, so that $|x_n| < |x_N| + 1$. Thus you can just take $b = \max \{ |x_1|,\ldots,|x_{N-1}|,|x_N|+1\}$.

  • $\begingroup$ Sorry, could you explain a bit more what you mean by I have to specify $m$? $\endgroup$ – student_t Jan 30 '17 at 18:39
  • $\begingroup$ Your definition of $b$ contains the symbol $m$ which has not been defined. $\endgroup$ – Umberto P. Jan 30 '17 at 18:53
  • $\begingroup$ Yes, where $m\geqN$ $\endgroup$ – student_t Jan 30 '17 at 19:00
  • $\begingroup$ Exactly. That does not define $m$. $\endgroup$ – Umberto P. Jan 30 '17 at 19:01

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