Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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?… – Dan Rust Aug 11 '13 at 11:43

I will assume the Least Upper Bound property of the real numbers. I'm not sure if you are allowed to use this, but for some definitions it is built right in to the real numbers.

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 $l=\sup_n\{a_n\}$.

share|cite|improve this answer

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 $\mathbb{Q}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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