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Consider a bounded sequence. By Bolzanno Weierstrass theorem we can conclude that it contains a sub sequence which is convergent . The proof of Bolzanno uses Cantor's completeness principle.

Let us consider the sequence of nested closed intervals $[a_1,b_1][a_2,b_2],\ldots,[a_n,b_n]$. Suppose this sequence of intervals satisfies all the conditions required for the cantor's completeness principle (described and discussed in my previous question).

Now both the sequences converge to a value K , which is the common intersection of all the closed intervals.Now let us consider the sequence of closed intervals $[a_1,k], [a_2,k],\ldots$. This can be thought as the outcome of considering the sequences $a_1,a_2,a_3,\ldots$ and $k,k,k.,k,\ldots$. So we can apply the cantor's completeness principle here. Then we can apply the idea used in proving Bolzanno Weierstrass theorem to conclude that every bounded sequence has a monotonic convergent sub sequence.

Is this idea of mine correct?

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    Let $x_n=1/n^2$ for all $n$. Your construction doesn't provide a subsequence.2012-10-24

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