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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.

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the notion of being complete is that "every Cauchy sequence in that space is a convergent one" –  TTY Nov 1 '12 at 20:20
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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

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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 $[a,b]$.

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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 - math.stackexchange.com/questions/227042/… –  sonicboom Nov 1 '12 at 20:45
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@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

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