Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

As the interval is closed every sequence in the interval converges to some point $x$ in the interval, and every convergent sequence is a Cauchy sequence, hence $[x, y]$ is complete.

Is that correct? I thought so earlier but now I am not sure where I saw it mentioned that a closed interval/set implies that any sequence in that set converges in that interval/set.

share|cite|improve this question
the notion of being complete is that "every Cauchy sequence in that space is a convergent one" – TTY Nov 1 '12 at 20:20
so what you need to show is given a Cauchy sequence in $[x,y]$, it will converges in $[x,y]$, a Cauchy sequence in $[x,y]$ is a Cauchy sequence in $\mathbb{R}$, which is a complete space, so it converges in $\mathbb{R}$, but as $[x,y]$ is a closed set, this limit must also lies in $[x,y]$. – TTY Nov 1 '12 at 20:24
That is 'idea' I am looking for...that as $[x,y]$ is a closed set the limit of the sequence must also lie in $[x,y]$. Any idea what I should google for to learn more about this theorem? – sonicboom Nov 1 '12 at 20:37
You might want to think about turning to a classic text, e.g., Rudin's Principles of Mathematical Analysis (aka "Baby Rudin", aka PMA) for clear and widely accepted definitions and theorems fundamental to analysis (and hence, to point-set topology, etc.) rather than relying solely on Google! :-) – amWhy Nov 1 '12 at 20:43
Hehe, I will have to do that at some stage but the library is about to shut here so Google will have to do for the moment! – sonicboom Nov 1 '12 at 20:46
up vote 2 down vote accepted

It’s not true that every sequence in $[a,b]$ converges to a point of $[a,b]$: some sequences in $[a,b]$ don’t converges at all! For instance, take $[a,b]=[-1,1]$, and let $x_k=(-1)^k$. What is true is that if $\langle x_k:k\in\Bbb N\rangle$ is a convergent sequence in $[a,b]$, then its limit is also in $[a,b]$; proving this is a good elementary exercise.

There are many ways to show that $[a,b]$ is complete if $a<b$, depending on what tools you have available. For instance, it’s a theorem that every compact metric space is complete, and $[a,b]$ is certainly a compact metric space. Another theorem says that a closed subspace of a complete metric space is complete; if you know that theorem and know that $\Bbb R$ is complete in the usual metrix, it follows immediately that $[a,b]$ is complete.

For a direct proof, let $\langle x_k:k\in\Bbb N\rangle$ be a Cauchy sequence in $[a,b]$. Since $[a,b]$ is compact, $\langle x_k:k\in\Bbb N\rangle$ has a convergent subsequence, say with limit $y$. By the exercise that I mentioned above, $y\in[a,b]$. Now use the fact that $\langle x_k:k\in\Bbb N\rangle$ is Cauchy to prove that it must also converge to $y$, and you’ll have shown that every Cauchy sequence in $[a,b]$ converges in $[a,b]$.

share|cite|improve this answer
Cheers, I haven't done compactness yet so I don't understand that part but the ideas you mentioned in the second paragraph, that a closed subspace of a complete metric space is complete...that is the goal I am after, but I don't understand the last part of the theorem behind it -… – sonicboom Nov 1 '12 at 20:45
@sonicboom: The part that you’re having trouble with is the exercise that I mentioned. I’ve given a quick proof in an answer to the other question. – Brian M. Scott Nov 1 '12 at 20:48

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.