-2
$\begingroup$

Let $\{a_n\}$ be a Cauchy Sequence in $M$. Prove that the set $\{a_n\}$ is bounded.

Proof: let $\{a_n\}$ be a Cauchy sequence in $M$, then there exists a $N \in \Bbb N$ such that $n,m \geq N$ implies $d(a_n,a_m) < \varepsilon$.

Thus, $a_n$ is bounded by $N$.

Is this correct?

$\endgroup$
  • 2
    $\begingroup$ No, it is not. Read what your statements say carefully. $N$ has nothing to do with an upper or lower bound of the values of the $a_n$, but is a lower bound on the index $n$. $\endgroup$ – Simon S Nov 16 '15 at 17:48
  • $\begingroup$ The bound has to do with $\varepsilon$, not with $N$. $\endgroup$ – Alex M. Nov 16 '15 at 18:43
2
$\begingroup$

Hint:

For $\epsilon = 1$ there exists an $N$ such that for all $n, m > N$, $|a_n - a_m| < 1$ and in particular $|a_{N+1} - a_m| < 1$ for all $m > N$.

Can you now find an upper and lower bound on all $a_n$?

$\endgroup$
0
$\begingroup$

But is not $\mathbb{R}$, i.ea Suppose $(\forall \epsilon > 0)(\exists N> 0)(n, m> N \Rightarrow d(x_n,x_m)<\epsilon)$.Prove$ (\exists x\in x) (\epsilon> 0)( \forall n> 1)d(x_n, x)<\epsilon$. Any hint? For your solution.

$\endgroup$
0
$\begingroup$

Cauchy sequences converge. Thus the tail of such sequences are bounded between (L-epsilon,L+epsilon). But then the whole sequence is bounded because any finite set (the first k terms) is also bounded.

$\endgroup$

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.