Basic Analysis - Proof Bolzano Wierestrass by Least Upper Bound

  • #1

Homework Statement


Let (an) be a boundedd sequence, and define the set
[itex]S= {x\in R : x < a_n [/itex] for infinitely many terms [itex]a_n\}[/itex]
Show that there exists a subsequence [itex](a_n_k)[/itex]converging to s = sup S


Homework Equations


This is supposed to be a direct proof of BW using the LUB property, so no monotonic convergence, Cauchy criterion, nested interval property etc...


The Attempt at a Solution


I am having trouble thinking of a way to define the subsequence. What I can show is that, if [itex]\epsilon > 0[/itex] there are finitely many terms [itex] a_n [/itex] s.t. [itex]a_n> s+\epsilon[/itex]
I thought of defining the subsequence to be [itex]a_n_k = min(a_n | a_n > s+\frac{1}{k})[/itex] But I was having trouble proving that this subsequence converges to s. I would greatly appreciate a tip or prod in the right direction of defining a subsequence that will work. Thank you!
 

Answers and Replies

  • #2
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Note that the set S is bounded above, since (a_n) is bounded and any upper bound of
the sequence will be an upper bound of S. Therefore, sup S exists, define s = sup S.
Now, consider an arbitrary ε > 0. Since s + ε cannot be an element of the set S, there must be only a finite number of an such that an > s + ε. But by the definition of the supremum, for any ε > 0 there exists some x ∈ S with s − ε < x. Thus, for any ε > 0 there are an infinite number of terms an with s − ε < an and only a finite number of those terms satisfy an > s + ε. Therefore, we must have an infinite number of an with s−ε<an <s+ε.
Now, to construct a subsequence (a_nk ) → s, consider εk = 1/k. So, start by choosing some an1 so that s−1 < an1 < s+1, from this point on, we choose ank+1 so that nk+1 >nk ands−1/k<ank+1 <s+1/k.We know that we can always do this, since at every step k there are an infinite number of a_n with s−1/k < a_n <s+1/k.
 
  • #3
That's perfect thank you! I had known that there was an infinite number of terms less than s+ \epsilon, but I did not connect that with making an interval, and simply choosing a term from each interval to form the subsequence. Thanks again.
 

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