Bolzano Weierstrass Theorem

As we know, not all sequences converge. In reality, the only ones that do converge are those that are “very good.” Even “excellent” sequences, however, may fail to converge.

For example, Cauchy sequences are excellent; they’re not much different from convergent sequences in that they converge in “good” spaces (i.e., complete spaces) and fail to converge only when the point that “should be” its limit is not in the space; in other words, it fails to converge because the space is not good (it’s incomplete), not because the sequence is bad. In this article, we will discuss all the Bolzano Weierstrass theorems.

When does a Sequence have a Convergent Subsequence or vice versa?

Consider the alternating real sequences {1, −1, 1, −1, 1, . . .}.  Although the series as a whole does not converge, some subsequences do, such as{1, 1, 1, . . .}.

We now look at an easy-to-prove condition that ensures that a sequence in R or Rn has a convergent subsequence: any limited sequence in Rn has a Cauchy subsequence, which converges in Rn.

The Bolzano-Weierstrass Theorem is about this. (Throughout this section, we’ll assume that  Rn has a norm; we’ve already demonstrated that the norm we employ in Rn has no bearing on convergence — that is, on which sequences converge.)

Bolzano Weierstrass Theorem 

The Bolzano Weierstrass theorem is a theorem that states that a convergent subsequence, or subsequential limit, exists for every bounded sequence of real numbers. Many analytical results are based on the Bolzano Weierstrass theorem. It is, in fact, equivalent to the real-number completeness axiom.

Bolzano Weierstrass Theorem for Sequences

• A convergent subsequence exists for every bounded sequence of real numbers.
• A convergent subsequence exists for any bounded sequence in Rn.
• A convergent subsequence exists for every sequence in a closed and bounded set S in Rn. (which eventually converges on a point in S).

Assume that (an) is a constrained sequence, and each time you choose an interval Ix with an unlimited number of terms in (an), the complement of the interval Ix in Ik includes only finitely many terms in the series. Using the notation from this theorem-proof: You must determine whether or not the bounded sequence converges.

Using the prior problem as an example shows that bounded divergent sequence has two subsequences (don’t and an), which converge to distinct values.

Bolzano Weierstrass Theorem Examples 

As shown, every convergent sequence is bounded, but not every bounded sequence is convergent. (-1) is an example of a non-convergent fixed sequence. As a result, the statement that any convergent sequence is bounded has no complete converse. A partial converse is given by the Bolzano-Weierstrass theorem, which is stated next.

Convergent Proof

Assume that (xn) is constrained. Assume that the sequence’s term is dominant if and only if our proof has now been divided into two cases.

• Case 1: Assume that (xn) has an infinite number of dominant terms 1, 2, 3, 4,… where 1 2 3 is the number. Then we obtain 1 2 3 by the definition of dominant, which is a limited monotonic non-increasing subsequence that converges according to the monotone convergence theorem.

• Case 2: If not, the sequence (xn) only has a finite number of dominant terms. Choose one term that isn’t the last dominant term. (For example, we can define 1=+1 using the last dominating term.) Because one isn’t dominant, there’s 2 1, which equals 1 2, and since two isn’t dominant, there’s 3 2, which equals 2 3, and so on. Continuing in this manner, we have a monotonic bounded growing subsequence 1 2 3 that follows the monotone convergence theorem once more.

Bolzano Weierstrass Theorem Sources

If you are looking for a deeper understanding of the Bolzano Weierstrass theorem, there are several online sources like the Bolzano Weierstrass theorem pdf. The theorem is a significant and powerful finding relating to the so-called compactness of intervals in real numbers. This will likely be addressed more in a metric space or topological space course. It’s worth noting that the proof relies heavily on the completeness of the reals (as expressed by the monotone convergence theorem). We’ll use it in later notes to provide a different perspective on completeness. You can use various sites and Bolzano Weierstrass theorem PDFs to see the long arithmetic proof of this theorem.

Conclusion

Given a bounded sequence, the theorem states that one (or more) convergent subsequences exist. All subsequences of a sequence that converges to LR converge to L. (-1)n is an example of a bounded sequence. Two convergent subsequences can be seen:- 1n and -(1n). The first converges to one, while the second converges to one. The bounded sequence is indeed irregular (it neither converges nor diverges).