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.

 

[CellCoree]

for your question. The Bolzano-Weierstrass Theorem is a fundamental theorem in real analysis that states that any bounded sequence in a metric space has a convergent subsequence. In other words, if a sequence has values that are restricted to a certain range, there will always be a subsequence that converges to a limit within that range.

To prove this theorem, we can use the nested interval method. This method involves dividing the bounded interval into smaller subintervals and then choosing a subinterval that contains infinitely many terms of the sequence. This process is repeated infinitely, creating a nested sequence of subintervals. By the completeness property of real numbers, this nested sequence of subintervals will have a common point, which will be the limit of the subsequence.

Another approach to proving the Bolzano-Weierstrass Theorem is by contradiction. We assume that the sequence does not have a convergent subsequence. This means that for any point within the bounded interval, there exists an open interval containing that point that does not contain any term of the sequence. By using the compactness property of closed and bounded intervals, we can show that this assumption leads to a contradiction, thus proving the theorem.

Overall, the Bolzano-Weierstrass Theorem is a powerful tool in real analysis and has many applications in various fields of mathematics. I hope this explanation helps in understanding the proof of this important theorem.