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Chapter 8: Problem 19

Only the special case of the Bolzano-Weierstrass Theorem (Theorem 8.6) was proved (where \([b, c]\) is the closed unit interval, [0,1]\()\). Generalize the proof to arbitrary \(b, c \in \mathbb{R}\) where \(b \leq c\).

Short Answer

Expert verified
The Bolzano-Weierstrass Theorem applies to any closed interval [b, c] since any bounded sequence in [b, c] has a convergent subsequence.

Step by step solution

01

Identify Requirements

Understand that we need to generalize the proof from the closed interval \( [0, 1] \) to any closed interval \( [b, c] \) where \( b \leq c \). The Bolzano-Weierstrass Theorem in the unit interval states that any bounded sequence has a convergent subsequence.
02

Set Up the General Interval

Consider the closed interval \( [b, c] \). Like \( [0, 1] \), this is a closed and bounded interval, essential for the Bolzano-Weierstrass Theorem, which applies to bounded sequences in any closed interval.
03

Map Sequence to Unit Interval

Given a sequence \( (a_n) \) within \( [b, c] \), apply a transformation to map it into the unit interval using linear transformation. Use \( x_n = \frac{a_n - b}{c - b} \) to transform \( a_n \). Now \( x_n \) is within \( [0, 1] \).
04

Apply Bolzano-Weierstrass to Transformed Sequence

Since \( (x_n) \) is within the closed unit interval \( [0, 1] \), use the Bolzano-Weierstrass Theorem to assert that there exists a convergent subsequence \( (x_{n_k}) \) converging to some limit \( L \) in \( [0, 1] \).
05

Transform Back to Original Interval

Convert the convergent subsequence \( x_{n_k} \) back to the \( [b, c] \) interval with \( a_{n_k} = x_{n_k}(c - b) + b \). This \( (a_{n_k}) \) is a convergent subsequence in \( [b, c] \), converging to \( L(c - b) + b \).
06

Conclude with General Theorem

Thus, we have shown that any bounded sequence in a closed interval \( [b, c] \) has a convergent subsequence. This generalizes the Bolzano-Weierstrass Theorem to arbitrary closed intervals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Sequences
A sequence is simply a list of numbers in a particular order. But when we say 'bounded sequence,' we mean something more specific. A sequence is bounded if there is a real number that the size of each number in the sequence doesn't exceed. Let's make this clearer:
  • Imagine having a "barrier" on both sides of the number line.
  • The numbers of your sequence must stay between these two barriers.
  • If they do, your sequence is bounded.
For the Bolzano-Weierstrass Theorem, it's important that the sequence is bounded. This means there are real numbers, like upper and lower bounds, that contain the entire sequence. If your sequence deviates even slightly outside these bounds, the theorem won't apply.
In mathematical terms, a sequence \(a_n\) is bounded if there exists numbers \M\ and \m\ such that \m \leq a_n \leq M\ for all terms in the sequence.
Convergent Subsequence
The term 'convergent subsequence' might sound complicated, but it's not as tricky as it seems. To break it down, a 'subsequence' is basically selecting specific numbers from a sequence, while still keeping the order intact. Now, what does it mean for this subsequence to be 'convergent'?
A convergent sequence is one where the numbers start to "hone in" on a single value. So, even if you have a wild and unpredictable sequence, there might be a subsequence that behaves nicely and starts zeroing in on a specific number.
  • This important behavior of honing into a value is what we call convergence.
  • The subsequence will get closer and closer to some number, called the limit, as it progresses.
In the context of the Bolzano-Weierstrass Theorem, the presence of a convergent subsequence within a bounded sequence is what allows us to assert something more predictable about the sequence, no matter how complex or random it might be.
Closed Intervals
A closed interval is a fundamental concept in mathematics, often visualized as a continuous part of the number line that includes its endpoints. Understanding closed intervals is key to grasping the Bolzano-Weierstrass Theorem:
  • A closed interval \[ b, c \] means every number between \b\ and \c\ is included, along with the numbers \b\ and \c\ themselves.
  • This inclusion of endpoints sets closed intervals apart from open intervals, which do not include their endpoints.
  • Think of a closed interval as a "closed and secured" range where sequences can reside.
For applying the Bolzano-Weierstrass Theorem effectively, it's crucial that sequences are analyzed within these closed intervals. These intervals provide a structured space where the behavior of sequences, specifically those that are bounded, can be understood and predicted properly.

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Most popular questions from this chapter

Let \(X \subseteq \mathbb{R}\) be bounded above. Prove that the least upper bound of \(X\) is unique.

Let \(X \subseteq \mathbb{R}\) be bounded below. Prove that \(X\) has a greatest lower bound.

Let \(X \subseteq \mathbb{R}, Y \subseteq \mathbb{R}\) and let every element of \(X\) be less than every element of \(Y\). Prove that there is \(a \in \mathbb{R}\) satisfying $$(\forall x \in X)(\forall y \in Y) x \leq a \leq y.$$

Show that there are exactly 4 order-complete extensions of \(\mathbb{Q}\) in which \(\mathbb{Q}\) is dense.

Let \(s\) be an infinite decimal expansion, and for any \(n \in \mathbb{N}^{+},\) let \(s_{n}\) be the truncation of \(s\) to the \(n^{t h}\) decimal place. Prove that the sequence \(\left\langle s_{n}\right\rangle\) is a Cauchy sequence.
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