2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $b = \max \{ |x_1|,\ldots,|x_{N-1}|,|x_N|+1\}$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.