# Show that every monotonic increasing and bounded sequence is Cauchy.

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.

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

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.

• 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. – Nhat Nov 14 '13 at 17:39
• 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. – coffeemath Nov 14 '13 at 17:41
• 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. – Nhat Nov 14 '13 at 20:53
• @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. – coffeemath Nov 14 '13 at 22:54
• 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 $\varepsilon>0$. – Nhat Nov 15 '13 at 0:48

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.

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