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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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  • $\begingroup$ You have to be more explicit. How do you construct the sequence $(a_n)$? It is possible to give a proof of Bolzano-Weierstrass along similar lines in which $(a_n)$ does not have to be a subsequence. $\endgroup$ – Michael Greinecker Oct 24 '12 at 7:42
  • $\begingroup$ @MichaelGreinecker I could not get. $\endgroup$ – danny gotze Oct 24 '12 at 7:45
  • $\begingroup$ @MichaelGreinecker I meant I could not get what you meant. $\endgroup$ – danny gotze Oct 24 '12 at 7:46
  • $\begingroup$ You start with some bounded sequence $(x_n)$ of real numbers. How do you construct a convergent subsequence? How do you construct your nested sequence of closed intervals? Your question cannot be answered before you have clarified that. $\endgroup$ – Michael Greinecker Oct 24 '12 at 7:50
  • $\begingroup$ @MichaelGreinecker I started with Xn,....Let its least upperbound be K. So, we can keep on shrinking the interval [-k,k] to [-k/2,k/2]...[-k/2^n,K/2^n] and then apply Cauchy principle to assert the convergence, as it is done while proving the bolzanno weirstrass. Now the sequence of intervals converge to a single value, say ,L. Then consider the sequence of closed intervals [-K,l],[-k/2,l]..... $\endgroup$ – danny gotze Oct 24 '12 at 7:57

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