How Do You Prove the Bolzano-Weierstrass Theorem?

[maximus101]

Hello, please could someone explain how to prove the Bolzano-Weierstrass Theorem?

thank you

 

[HallsofIvy]

Starting from what axioms? It is possible to take "Bolzano-Weierstrasse" as an axiom for the real numbers and prove other properties (such as the Cauchy Criterion or least upper bound property from that. But it is more common to start with the least upper bound property or monotone convergence as an axiom.

You can, for example, prove that every infinite sequence contains a monotone subsequence:

Let $\{a_n\}$ be a sequence of real numbers. Then Define the sequence $\{a_{i}\}$ for i in some subset S of the positive integers by: i is in S if and only if $a_i\ge a_m$ for all m> i. That is, $a_i$ is in the subsequence if and only if $a_i$ is greater than or equal to all subsequent numbers in the sequence. Of course, it is quite possible that this subsequence is "empty"- for example, if the sequence is increasing, this is never true. It is also possible that the subsequence is non-empty but finite. Or it is possible that the subsequence is infinite. For a decreasing sequence this "subsequence" is, in fact, the original sequence.

Now there are two cases:
1) This subsequence is infinite.
Then we are done! This subsequence is itself a decreasing sequence.

2) The subsequence is either empty or finite.
Then the set, S, of indices, is empty of finite. If finite, then there exist an index, $i_1$ that is larger than any number in S (If empty, $i_1= 1$ will do). Since $i_1$ is not in S, there must exist $i_2> i_1$ such that $a_{i_2}< a_{i_1}$. Since $i_2> i_1$, and $i_1$ was larger than any number in S, $i_2$ is not in S and so there must exist $i_3> i_2$ such that $a_{i_3}< a_{i_2}$. Since $i_3> i_2> i_1$ it also is not in S and so there exist $i_4> i_3$ such that $a_{i_4}< a_{i_3}$. Continuing in that way we get a decreasing subsequence.

Now Bolzano-Weierstrasse follows easily from monotone convergence- If $\{a_n\}$ is a bounded sequence then it has both upper and lower bounds. If that monotone subsequence is increasing, it has an upper bound and so converges. If that monotone subsequence is decreasing, it has a lower bound and so converges. In either case, a bounded sequence contains a convergent subsequence.

 

[sachinism]

http://cool-rr.com/bw.htm

 

[mathwonk]

subdivide.