6
$\begingroup$

The title is kind of misleading because the task actually to show

Every monotonic increasing and bounded sequence $(x_n)_{n\in\mathbb{N}}$ is Cauchy

without knowing that:

  • Every bounded non-empty set of real numbers has a least upper bound. (Supremum/Completeness Axiom)
  • A sequence converges if and only if it is Cauchy. (Cauchy
    Criterion)
  • Every monotonic increasing/decreasing, bounded and real
    sequence converges to the supremum/infimum of the codomain (not sure if this is the right word).

However, what is allowed to use listed as well:

  • A sequence is called covergent, if for $\forall\varepsilon>0\,\,\exists N\in\mathbb{N}$ so that $|\,a_n - a\,| < \varepsilon$ for $\forall n>N$. (Definition of Convergence)
  • A sequence $(a'_k)_{k≥1}$ is called a subsequence of a sequence $(a_n)_{n≥1}$, if there is a monotonic increasing sequence $(n_k)_{k≥1}\in\mathbb{N}$ so that $a'_{k} = a_{n_{k}}$ for $\forall k≥1$. (Definition of a Subsequence)
  • A sequence $(a_n)_{n≥1}$ is Cauchy, if for $\forall\varepsilon>0\,\,\exists N=N(\varepsilon)\in\mathbb{N}$ so that $|\,a_m - a_n\,| < \varepsilon$ for $\forall m,n>N$. (Definition of a Cauchy Sequence)
  • (Hint) The sequence $(\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}$ is unbounded for $\varepsilon>0$. (Archimedes Principle)

Would appreciate any help.

$\endgroup$
  • 3
    $\begingroup$ Suppose you have a monotonic sequence that is not a Cauchy sequence. Show that it is unbounded. $\endgroup$ – Daniel Fischer Nov 14 '13 at 11:01
4
$\begingroup$

If $x_n$ is not Cauchy then an $\varepsilon>0$ can be chosen (fixed in the rest) for which, given any arbitrarily large $N$ there are $p,q \ge n$ for which $p<q$ and $x_q-x_p>\varepsilon.$

Now start with $N=1$ and choose $x_{n_1},\ x_{n_2}$ for which the difference of these is at least $\varepsilon$. Next use some $N'$ beyond either index $n_1,\ n_2$ and pick $N'<n_3<n_4$ for which $x_{n_4}-x_{n_3}>\varepsilon.$ Continue in this way to construct a subsequence.

That this subsequence diverges to $+\infty$ can be shown using the Archimedes principle, which you say can be used, since all the differences are nonnegative and there are infinitely many differences each greater than $\varepsilon$, a fixed positive number.

$\endgroup$
  • 2
    $\begingroup$ Just to make sure I understand the concept of your proof: You assume that the monotonous increasing and bounded sequence is not Cauchy. Therefore there exists an $\varepsilon>0$ for $\forall N\in\mathbb{N}$ so that $x_q - x_p > \varepsilon$ for $\forall p,q≥N$. Now you start constructing a subsequence which will diverge to infinity => $x_n$ is not bounded => $x_n$ is Cauchy. - Is this already the contradiction we're looking for? Sorry for these questions, but I really want to understand what I'm doing and writing down. Kind regards. $\endgroup$ – Nhat Nov 14 '13 at 17:39
  • 2
    $\begingroup$ Yes. To show the monotone increasing bounded sequence is cauchy, we assume it is not and proceed to select a fixed $\varepsilon$ and so on, eventually deriving that the sequence is not bounded, which goes against one of the hypotheses. Since the construction is from a monotone increasing sequence, the only conclusion is that the "bounded" hypothesis is false, which by contraposition shows the required theorem. $\endgroup$ – coffeemath Nov 14 '13 at 17:41
  • 1
    $\begingroup$ I am sorry, but could you elaborate the part where you are constructing the subsequence out of the not-Cauchy-sequence? I'm still having trouble understanding that part. Thanks. $\endgroup$ – Nhat Nov 14 '13 at 20:53
  • $\begingroup$ @kitkat4.4 We have a fixed $\varepsilon$ for the whole thing, and for any $N$ may find terms beyond that whose difference is at least $\varepsilon$. For the first two terms of the subseq. let $N=1$ and select say $x_5$ and $x_8$ with $x_8-x_5>\varepsilon$. Now we need to reset $N$ to $N=9$ to avoid rechoosing the same terms, and at this next step maybe we select $x_{15}$ and $x_{27}$ with $x_{27}-x_{15}>\varepsilon.$. For step 3 we would now reset $N$ to $N=28$ and so on. Each time we reset $N$ so as not to select the same terms twice. $\endgroup$ – coffeemath Nov 14 '13 at 22:54
  • $\begingroup$ OK, I think I understand now. In this case our subsequences first term would be $x_8$ and the second $x_27$, if I'm not mistaken. Due to our assumption I know that for every arbitrarily large N (which keeps resetting) I can find a difference between two following terms which $> \varepsilon$. Thus the difference between all of the terms of my subsequence have to be $> \varepsilon$ as well. And in order to show that the subsequence is unbounded - Knowing that the sequence $(\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}$ is unbounded for $\varepsilon>0$. $\endgroup$ – Nhat Nov 15 '13 at 0:48
-1
$\begingroup$

1)Every monotonous increasing/decreasing, bounded and real sequence converges to the supremum/infimum of the codomain (not sure if this is the right word).

2)A sequence converges if and only if it is Cauchy. (Cauchy Criterion).

You wrote it yourself.

$\endgroup$
  • 2
    $\begingroup$ The OP said explicitly that the facts you cite cannot be used in the proof. $\endgroup$ – coffeemath Nov 14 '13 at 12:38
  • $\begingroup$ @coffeemath haha,didn't notice that:P $\endgroup$ – Haha Nov 14 '13 at 12:51

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.