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I am trying to show the completeness of R, using the LUB property.

Problem is that I don't know, given a Cauchy sequence,where the limit would come from; I can check if a sequence {an} converges to a specific value, but I don't know how to come up with the limit value that the sequence would converge to.

I imagine as the intervals containing the terms am (m>N) become smaller, as |am-ak|N, maybe an converges to the limiting point of intervals (am,ak).

I used limsup and iminf to show that an converges , but this does not tell me what an should converge to.

Any Suggestions?

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Step 1: Show that any Cauchy sequence is bounded.

(This is true in any metric space and has nothing to do with completeness.) It follows that the limsup and liminf of (the underlying set of terms of) your sequence are finite.

Step 2: Show that any sequence with limsup $L < \infty$ has a subsequence converging to $L$.

Step 3: Now you have something specific to try to show the Cauchy sequence converges to: namely, $L$. Show in fact that if a subsequence of a Cauchy sequence converges to some $L$, then the sequence itself converges to $L$.

Step 4 (optional): Think about what would happen if in Steps 2 and 3 you chose to use the liminf instead of the limsup.

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If one wants to avoid reference to limsups and liminfs, there is a slight variant of Pete's argument: Show that some subsequence of the original sequence must be monotone (always increasing, or always decreasing; this is just combinatorics), so its sup (or inf) exists as long as the subsequence is bounded. But it must be, by Pete's Step 1. – Andrés E. Caicedo Feb 4 '11 at 6:20
@Andres: yes, that's a nice variant, maybe even nicer from an elementary perspective. But it is really based on the monotone sequence lemma rather than LUB directly. I went the way I did because the OP mentioned limsups/liminfs. (FWIW, I am currently teaching a course on sequences and series which treats the monotone sequence lemma as basic, so my students will see your argument rather than mine!) – Pete L. Clark Feb 4 '11 at 6:25
Now I remember that in my first university math class, there was one week in which the teacher (actually a substitute!) started from LUB and proved all the big theorems (Bolzano-Weierstrass and so forth). On Monday he stated and proved your combinatorial result -- he called it the Rising Sun Lemma -- then the Monotone Sequence Lemma, then everything else came quickly and easily. I was struck by how beautiful this argument was. Upon reflection, you should probably leave it as an answer! – Pete L. Clark Feb 4 '11 at 6:29
1 that my memory is really being jogged, I find that I (re)wrote the wikipedia article on Bolzano-Weierstrass so as to use this lemma in the proof. (The name "Rising Sun Lemma" is not used here; wikipedia means something else by that -- although possibly it is related; I didn't look too carefully.) – Pete L. Clark Feb 4 '11 at 6:36
@Andres: okay, but no one is going to give you any points for that! :) – Pete L. Clark Feb 4 '11 at 6:36

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