Basic Analysis - Proof Bolzano Wierestrass by Least Upper Bound

[rakalakalili]

Homework Statement

Let (a_n) be a boundedd sequence, and define the set
S= {x\in R : x < a_n for infinitely many terms a_n\}
Show that there exists a subsequence (a_n_k)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 \epsilon > 0 there are finitely many terms a_n s.t. a_n> s+\epsilon
I thought of defining the subsequence to be a_n_k = min(a_n | a_n > s+\frac{1}{k}) 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!

 

[FanofAFan]

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.

 

[rakalakalili]

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.