Problem 2 If $\left(x_{n}\right)$ is Cau... [FREE SOLUTION]

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Introductory Functional Analysis with Applications

Erwin Kreyszig

$Math Studyset 91影视 Explanations$ Math

1978 Edition

Chapter 1: Problem 2

If $\left(x_{n}\right)$ is Cauchy and has a convergent subsequence, say, $x_{n_{4}} \longrightarrow x$, show that $\left(x_{n}\right)$ is convergent with the limit $x$.

Short Answer

Expert verified

The sequence $(x_n)$ is convergent with the limit $x$.

Step by step solution

Definition of a Cauchy sequence

Recall that a sequence $ (x_n) $ is Cauchy if for every $ \epsilon > 0 $, there exists $ N $ such that for all $ m, n > N $, $ |x_m - x_n| < \epsilon $. This implies that the terms of the sequence get arbitrarily close to each other as the sequence progresses.

Definition of a convergent sequence

A sequence converges to a limit $ x $ if for every $ \epsilon > 0 $, there exists an $ N $ such that for all $ n > N $, $ |x_n - x| < \epsilon $.

Use the property of the convergent subsequence

We are given that the subsequence $ x_{n_4} \to x $. This means that for every $ \epsilon > 0 $, there exists a $ K $ such that for all $ k > K $, $ |x_{n_k} - x| < \epsilon/2 $.

Apply the Cauchy condition to the original sequence

Since $ (x_n) $ is Cauchy, for the given $ \epsilon/2 $, there exists an $ N $ such that for all $ m, n > N $, $ |x_m - x_n| < \epsilon/2 $.

Show $(x_n)$ converges to $x$

By the convergence of the subsequence, there is an index $ n_k > N $ such that $ |x_{n_k} - x| < \epsilon/2 $. For any $ n > N $, making $ m = n_k $, we have $ |x_n - x| \leq |x_n - x_{n_k}| + |x_{n_k} - x| < \epsilon/2 + \epsilon/2 = \epsilon $. This satisfies the definition of convergence, hence $ (x_n) \to x $.

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Key Concepts

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

Convergent Sequence

In mathematics, a convergent sequence is a sequence of numbers that approaches a specific value, known as the limit, as the sequence progresses. Think of it like walking towards a destination: each step you take gets you closer to your final stop. A sequence

Converges to a limit $ x $ if, for every tiny positive number ($ \epsilon $), you can find a point in the sequence (after which all terms are close to this limit) so that the distance between each term and $ x $ is smaller than $ \epsilon $.
Formally, this means for every $ \epsilon > 0 $, there exists a number $ N $ such that for all $ n > N $, $ |x_n - x| < \epsilon $.

This means the terms of the sequence get arbitrarily close to the limit $ x $ as $ n $ becomes larger. Understanding this concept is fundamental, as many results in analysis depend on whether sequences converge to certain values.

Subsequence

A subsequence is a sequence derived from another sequence by omitting some of its elements, without changing the order of the remaining elements. If you imagine cutting a string of pearls but only slipping off some pearls, the pearls you leave make up the subsequence.

A subsequence of a sequence $ (x_n) $ is denoted by $ (x_{n_k}) $, where $ (n_k) $ is a strictly increasing sequence of indices.
Subsequences maintain the order of appearance as in the original sequence, they just might "skip" some elements.
Importantly, if a subsequence converges to some limit $ x $, it can provide key information about the convergence of the full sequence.

Subsequences are particularly useful in convergence proofs, where demonstrating that a given subsequence converges can help prove that the whole sequence does as well.

Convergence Proof

Proving that a sequence converges requires demonstrating that the terms of the sequence get arbitrarily close to a certain number, called the limit. In the case of a Cauchy sequence $ (x_n) $ with a known convergent subsequence $ (x_{n_k}) \rightarrow x $, convergence can be shown as follows:

First, note that a subsequence converges to a limit means for every $ \epsilon > 0 $, there exists $ K $ such that for all $ k > K $, $ |x_{n_k} - x| < \epsilon/2 $.
The original sequence being Cauchy implies that for the $ \epsilon/2 $ chosen, there exists an $ N $ so that for all $ m, n > N $, we have $ |x_m - x_n| < \epsilon/2 $.
Pick $ m = n_k $, with $ n_k > N $. Due to the convergence of $ (x_{n_k}) $, we know $ |x_{n_k} - x| < \epsilon/2 $.
Therefore, for any $ n > N $, the triangle inequality gives $ |x_n - x| \leq |x_n - x_{n_k}| + |x_{n_k} - x| < \epsilon/2 + \epsilon/2 = \epsilon $.

This chain of reasoning shows that $ (x_n) \rightarrow x $, and hence, it converges to the same limit as its subsequence, reconfirming the sequence's convergence to $ x $.

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91影视

If \(\left(x_{n}\right)\) is Cauchy and has a convergent subsequence, say, \(x_{n_{4}} \longrightarrow x\), show that \(\left(x_{n}\right)\) is convergent with the limit \(x\).

Short Answer

Step by step solution

Definition of a Cauchy sequence

Definition of a convergent sequence

Use the property of the convergent subsequence

Apply the Cauchy condition to the original sequence

Show \((x_n)\) converges to \(x\)

Key Concepts

Convergent Sequence

Subsequence

Convergence Proof

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