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.