# Proving completeness of the real number line.

I am trying to prove that the real number line $\Bbb{R}$ is complete. I know that every cauchy sequence in $R$ has a bounded monotone subsequence. Hence, if the subsequence has a limit $l$, the original sequence is also convergent to $l$. However I am having difficulty in proving two things:

1. that there exists such an $l$.
2. that $l$ exists in $\Bbb{R}$, and not outside of it. For example, the bounded cauchy sequence $3,3.14,3.141,\dots$ does not converge in $\Bbb{Q}$

How should I go about proving this? Thanks!

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which topology are you using? under the normal topology $\mathbb{R}$ is not complete. – Dima McGreen Aug 11 '13 at 11:35
$\Bbb{R}$ is a metric space with $d(x,y)=|x-y|$. These are the only facts provided. – hudialala Aug 11 '13 at 11:39
By the Heine-Borel theorem a set in $\mathbb{R}$ is compact iff it is bounded and closed. But $\mathbb{R}$ is not bounded. – Dima McGreen Aug 11 '13 at 11:41
@DimaMcGreen Completeness is a property of metric spaces (in this instance anyway). Complete metric spaces can be homeomorphic to non-complete metric spaces and so it is not a topological invariant. $\mathbb{R}$ is complete but $(0,1)$ is not. – Dan Rust Aug 11 '13 at 11:42
@hudialala Are you allowed to use the Least Upper Bound property of the reals? en.wikipedia.org/wiki/… – Dan Rust Aug 11 '13 at 11:43

You've already shown that every Cauchy sequence in $\mathbb{R}$ has a bounded monotone subsequence, so let's assume wlog that the sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded increasing. We want to show that the sequence converges to $\sup\{a_n\}$. By hypothesis, let $l=\sup\{a_n\}$ which exists by the least upper bound property and the fact that $\{a_n\}$ is bounded above.
Let $\epsilon >0$ be given. There exists an $N\in\mathbb{N}$ such that $a_N>l-\epsilon$ as, if not, $l-\epsilon$ would bound $\{a_n\}$ from above which contradicts the definition of $l$. Now, $\{a_n\}$ is increasing and so for all $n>N$, we get $$|l-a_n|=l-a_n\leq l-a_N<\epsilon,$$ and so by definition of limit we have $\lim_{n\rightarrow\infty} \{a_n\}$ exists and is equal to $l=\sup_n\{a_n\}$.
I like to answer your second question as your first doubt should be clear by Daniel Rust's answer. $Q$ is not complete. So a Cauchy sequence in $\mathbb{Q}$ may not converge in $\mathbb{Q}$, as you have given an example. Such type of sequences will converge at some point of $\mathbb{R}$, a field extension of $\mathbb{Q}$.