Kreyszig 1.4, Metric Spaces - Convergence, Cauchy Sequence, Completness

Problem 1. Convergence of Subsequences in a Metric Space

Given: A sequence \((x_n)\) in a metric space \(X\) is convergent and has limit \(x\).

To Prove: Every subsequence \((x_{n_k})\) of \((x_n)\) is convergent and has the same limit \(x\).

Proof:

Let \((x_{n_k})\) be an arbitrary subsequence of \((x_n)\).

  1. Given that \((x_n)\) is convergent with limit \(x\): This means that for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), the distance between \(x_n\) and \(x\) is less than \(\epsilon\). Mathematically, this is: \(d(x_n, x) < \epsilon \quad \text{for all} \quad n \geq N\)

  2. Consider the subsequence \((x_{n_k})\): Since \(n_k\) represents the indices of the subsequence and \(n_k\) is increasing (because it's a subsequence), for every \(k \geq K\) (for some \(K\)), we have \(n_k \geq N\).

  3. Using the convergence of \((x_n)\): For the same \(\epsilon > 0\) as before, for all \(k \geq K\), we have: \(d(x_{n_k}, x) < \epsilon\) This is because \(n_k \geq N\) for all \(k \geq K\), and we know from the convergence of \((x_n)\) that the distance between any term beyond \(N\) and the limit \(x\) is less than \(\epsilon\).

  4. Conclusion: The above expression shows that the subsequence \((x_{n_k})\) also converges to the same limit \(x\).

Hence, every subsequence \((x_{n_k})\) of \((x_n)\) is convergent and has the same limit \(x\).

This completes the proof.

Crux of the Proof for Convergence of Subsequences

  1. Definition of Convergence: A sequence \((x_n)\) in a metric space converges to a limit \(x\) if, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), the distance between \(x_n\) and \(x\) is less than \(\epsilon\).

  2. Nature of Subsequences: A subsequence \((x_{n_k})\) retains the order of the original sequence \((x_n)\). This means that if \(n_k\) is the index of a term in the subsequence, then for every \(k' > k\), \(n_{k'} > n_k\).

Given these two points:

  1. If \((x_n)\) converges to \(x\), then beyond a certain index \(N\), all terms of \((x_n)\) are close to \(x\).

  2. Since a subsequence \((x_{n_k})\) retains the order of \((x_n)\), beyond some index \(K\), all terms of \((x_{n_k})\) will also be terms of \((x_n)\) that are close to \(x\).

Thus, the subsequence \((x_{n_k})\) will also be close to \(x\) beyond this index \(K\), meaning it converges to the same limit \(x\).

In essence, the convergence of the original sequence ensures that its terms get arbitrarily close to the limit, and the nature of subsequences ensures that their terms, being a subset of the original sequence's terms, will also get arbitrarily close to the same limit.

A Practical Example

Consider the sequence \((x_n)\) defined by \(x_n = \frac{1}{n}\) for all natural numbers \(n\). This sequence represents the reciprocals of natural numbers and is defined in the metric space of real numbers with the usual metric (absolute value of the difference).

Observation: As \(n\) grows larger, \(x_n\) gets closer and closer to 0. Hence, the sequence \((x_n)\) converges to 0 in the real numbers.

Subsequence: Let's consider a subsequence \((x_{n_k})\) where \(n_k = 2^k\). This subsequence consists of the terms of \((x_n)\) at the positions which are powers of 2. So, the subsequence is: \(x_{n_1} = \frac{1}{2}\), \(x_{n_2} = \frac{1}{4}\), \(x_{n_3} = \frac{1}{8}\), and so on.

Observation for the Subsequence: Just like the original sequence, as \(k\) grows larger, \(x_{n_k}\) gets closer and closer to 0. Hence, the subsequence \((x_{n_k})\) also converges to 0 in the real numbers.

Conclusion: Both the sequence \((x_n)\) and its subsequence \((x_{n_k})\) converge to the same limit, 0, in the metric space of real numbers. This is consistent with our earlier proof that if a sequence in a metric space converges to a limit, then every subsequence of it also converges to the same limit.

Problem 2. Convergence of Cauchy Sequences with Convergent Subsequences

Given: - A sequence \((x_n)\) is Cauchy. - There exists a convergent subsequence \((x_{n_k})\) such that \(x_{n_k} \rightarrow x\).

To Prove: - \((x_n)\) is convergent and its limit is \(x\).

Proof:

  1. Cauchy Sequence: By definition, for every \(\epsilon > 0\), there exists an \(N_1\) such that for all \(m, n \geq N_1\), we have \(d(x_m, x_n) < \frac{\epsilon}{2}\).

  2. Convergent Subsequence: Since \((x_{n_k}) \rightarrow x\), for the same \(\epsilon > 0\), there exists a \(K\) such that for all \(k \geq K\), we have \(d(x_{n_k}, x) < \frac{\epsilon}{2}\).

  3. Combining the Two: Let's choose \(N = \max(N_1, n_K)\) where \(n_K\) is the \(K\)-th term of the sequence \((n_k)\). Then, for all \(n \geq N\), we have \(d(x_n, x) \leq d(x_n, x_{n_K}) + d(x_{n_K}, x)\) \(< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\).

Therefore, \((x_n)\) converges to \(x\).

If a sequence is "tightening up" (Cauchy) and a part of it (subsequence) is getting close to a specific value (converging to x), then the entire sequence must also be getting close to that value (converging to x).

Practical Example:

Consider the sequence \((x_n)\) defined as follows: \(x_n = (-1)^n + \frac{1}{n}\).

  1. Cauchy Sequence: \((x_n)\) is not a Cauchy sequence as it oscillates between positive and negative values without settling down as \(n\) increases.

2. Convergent Subsequence: However, we can find a convergent subsequence. Consider \(x_{n_k}\) where \(n_k = 2k\). This subsequence is: \(x_{n_1} = \frac{3}{2}\), \(x_{n_2} = \frac{5}{4}\), \(x_{n_3} = \frac{7}{6}\), and so on, which converges to 1.

According to the proof, if \((x_n)\) was Cauchy, it would converge to the same limit as its subsequence, which is 1. However, since \((x_n)\) is not Cauchy, we cannot conclude that \((x_n)\) converges to 1, which aligns with our observation of the sequence.

Note: This example demonstrates that the condition of being Cauchy is crucial for the sequence to converge to the same limit as its convergent subsequence.

Problem 3. Convergence and Neighborhoods in Metric Spaces

Given:

  • A sequence \((x_n)\) in a metric space.

  • A point \(x\) in the same metric space.

To Prove: \(x_n \rightarrow x\) if and only if for every neighborhood \(B\) of \(x\), there exists an integer \(n_0\) such that \(x_n \in B\) for all \(n > n_0\).

Proof:

(⇒) Forward Direction: Assume \(x_n \rightarrow x\).

By the definition of convergence, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), the distance between \(x_n\) and \(x\) is less than \(\epsilon\). This means that \(x_n\) lies in the \(\epsilon\)-neighborhood of \(x\) for all \(n \geq N\).

Given any neighborhood \(B\) of \(x\), there exists some \(\epsilon > 0\) such that the \(\epsilon\)-neighborhood of \(x\) is contained in \(B\). By the convergence of \(x_n\), there exists an \(n_0\) such that \(x_n\) lies in this \(\epsilon\)-neighborhood (and hence in \(B\)) for all \(n > n_0\).

(⇐) Reverse Direction: Assume that for every neighborhood \(B\) of \(x\), there exists an integer \(n_0\) such that \(x_n \in B\) for all \(n > n_0\).

Given any \(\epsilon > 0\), consider the \(\epsilon\)-neighborhood of \(x\). By assumption, there exists an \(n_0\) such that \(x_n\) lies in this \(\epsilon\)-neighborhood for all \(n > n_0\). This means that the distance between \(x_n\) and \(x\) is less than \(\epsilon\) for all \(n > n_0\).

Therefore, \(x_n \rightarrow x\).

Conclusion: The sequence \(x_n\) converges to \(x\) if and only if for every neighborhood \(B\) of \(x\), there exists an integer \(n_0\) such that \(x_n \in B\) for all \(n > n_0\).

This completes the proof.

Problem 4. Boundedness of Cauchy Sequences

Given: - A sequence \((x_n)\) is Cauchy.

To Prove: - The sequence \((x_n)\) is bounded.

Proof:

  1. Definition of Cauchy Sequence: By definition, a sequence is Cauchy if, for any given \(\epsilon > 0\) (let's choose \(\epsilon = 1\) for simplicity), there exists an \(N\) such that for all \(m, n \geq N\), we have \(|x_m - x_n| < 1\).

  2. Boundedness of Terms Beyond \(N\): For any \(n \geq N\), using the triangle inequality, we get:

    \(|x_n| = |x_n - x_N + x_N|\) \(\leq |x_n - x_N| + |x_N|\) \(< 1 + |x_N|\)

    Let's denote \(M = 1 + |x_N|\). So, for all \(n \geq N\), \(|x_n| < M\).

  3. Boundedness of Terms Before \(N\): For terms \(x_1, x_2, ... x_{N-1}\), they are finitely many, so they have a maximum absolute value, say \(M'\).

  4. Combining the Two: The sequence \((x_n)\) is bounded by \(\max(M, M')\) for all \(n\).

Conclusion: Every Cauchy sequence is bounded.

This completes the proof.

Crux of the Proof for Boundedness of Cauchy Sequences

The core idea behind proving that a Cauchy sequence is bounded revolves around leveraging the defining property of Cauchy sequences: as the sequence progresses, its terms get arbitrarily close to each other.

  1. Cauchy's Property: A Cauchy sequence ensures that, after a certain point (denoted by \(N\)), the distance between any two terms is less than any given positive value. For the sake of the proof, we chose this value as \(\epsilon = 1\).

  2. Boundedness Beyond a Point (Using \(M\)): Given the Cauchy property, we deduced that all terms of the sequence beyond the point \(N\) are not just close to each other but are also close to a specific term, \(x_N\). This means that the sequence's terms, after \(N\), are bounded by a value \(M\), which is a little more than the absolute value of \(x_N\).

  3. Boundedness Before the Point (Using \(M'\)): The terms before \(N\) are finitely many. Any finite set of numbers is always bounded because there will be a maximum and minimum value among them. We denote the maximum absolute value of these terms as \(M'\).

  4. Why Two Different \(M\) and \(M'\)?: The reason for using two different bounds, \(M\) and \(M'\), is to separately handle the boundedness of two segments of the sequence:

    • \(M\) handles the terms after the point \(N\), ensuring they don't stray too far from \(x_N\).

    • \(M'\) handles the initial terms, up to \(N\), by simply using the maximum absolute value among them.

By combining these two bounds, we ensure that the entire sequence is bounded by the larger of \(M\) and \(M'\).

In essence, the proof uses the "tightening" behavior of Cauchy sequences to ensure that the sequence remains within a certain "boundary" and doesn't diverge to infinity, thus proving it's bounded.

Problem 6. Convergence of Distance Sequence of Cauchy Sequences

Given: Two sequences \((x_n)\) and \((y_n)\) in a metric space \((X,d)\) are Cauchy.

To Prove: The sequence \((a_n)\), where \(a_n = d(x_n, y_n)\), converges.

Proof:

  1. Cauchy Property of \((x_n)\) and \((y_n)\): Since both \((x_n)\) and \((y_n)\) are Cauchy, for any given \(\epsilon > 0\), there exist integers \(N_1\) and \(N_2\) such that for all \(m,n \geq N_1\) and \(p,q \geq N_2\), we have:

    \(d(x_m, x_n) < \frac{\epsilon}{2}\) \(d(y_p, y_q) < \frac{\epsilon}{2}\)

  2. Using the Triangle Inequality: Consider the difference \(|a_m - a_n|\), where \(a_m = d(x_m, y_m)\) and \(a_n = d(x_n, y_n)\). Using the triangle inequality, we get:

    \(|a_m - a_n| = |d(x_m, y_m) - d(x_n, y_n)|\) \(\leq d(x_m, x_n) + d(y_m, y_n)\)

    Now, using the properties of the metric and the Cauchy nature of the sequences, we can further bound this as:

    \(\leq d(x_m, x_n) + d(y_m, y_n) < \epsilon\) for all \(m,n\) greater than \(N = \max(N_1, N_2)\).

  3. Convergence of \((a_n)\): The above inequality shows that the sequence \((a_n)\) is Cauchy. In metric spaces where every Cauchy sequence converges (like in real numbers), \((a_n)\) will converge.

Illustrative Example:

Consider the metric space \((X,d)\) where \(X\) is the set of real numbers and \(d\) is the usual metric (absolute difference). Let:

\(x_n = \frac{1}{n}\) \(y_n = \frac{1}{n+1}\)

Both \((x_n)\) and \((y_n)\) are Cauchy sequences in this metric space. Now, consider:

\(a_n = d(x_n, y_n) = \left| \frac{1}{n} - \frac{1}{n+1} \right| = \frac{1}{n(n+1)}\)

The sequence \((a_n)\) represents the distances between the terms of \((x_n)\) and \((y_n)\). As \(n\) goes to infinity, \(a_n\) goes to 0, showing that \((a_n)\) converges to 0.

This completes the proof and illustrative example.

Problem 8. Equivalence of Cauchy Sequences in Two Metrics

Given: Two metrics \(d_1\) and \(d_2\) on the same set \(X\). There exist positive numbers \(a\) and \(b\) such that for all \(x, y \in X\):

\begin{equation*} a d_1(x,y) \leq d_2(x,y) \leq b d_1(x,y) \end{equation*}

To Prove: The Cauchy sequences in \((X,d_1)\) and \((X,d_2)\) are the same.

Proof:

  1. Assume a Cauchy Sequence in :math:`(X,d_1)`:

    Let \((x_n)\) be a Cauchy sequence in \((X,d_1)\).

    This means that for any given \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\):

    \begin{equation*} d_1(x_m, x_n) < \epsilon \end{equation*}

  2. Using the Given Inequality: Using the given inequality, we can deduce:

    \begin{equation*} d_2(x_m, x_n) \leq b d_1(x_m, x_n) < b\epsilon \end{equation*}

    This shows that \((x_n)\) is also a Cauchy sequence in \((X,d_2)\).

  3. Conversely, Assume a Cauchy Sequence in \((X,d_2)\):

    Similarly, if \((x_n)\) is a Cauchy sequence in \((X,d_2)\), then for any given \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\):

    \begin{equation*} d_2(x_m, x_n) < \epsilon \end{equation*}

    Using the given inequality again, we get:

    \begin{equation*} d_1(x_m, x_n) \leq \frac{1}{a} d_2(x_m, x_n) < \frac{\epsilon}{a} \end{equation*}

    This shows that \((x_n)\) is also a Cauchy sequence in \((X,d_1)\).

Conclusion: The Cauchy sequences in \((X,d_1)\) and \((X,d_2)\) are the same.

This completes the proof.

Crux of the Proof for Equivalence of Cauchy Sequences in Two Metrics

The essence of the proof lies in the given relationship between the two metrics \(d_1\) and \(d_2\). The inequalities provided ensure that the "distance" between any two points in \(X\) as measured by \(d_1\) and \(d_2\) are directly proportional. This proportionality ensures that if the terms of a sequence get arbitrarily close to each other in one metric, they must also get arbitrarily close in the other metric.

Crux of the Proof:

  1. Proportional Distances: The given inequalities \(a d_1(x,y) \leq d_2(x,y) \leq b d_1(x,y)\) ensure that distances in \(d_2\) are always bounded by proportional distances in \(d_1\).

  2. Cauchy in \(d_1\) Implies Cauchy in \(d_2\): If a sequence is Cauchy in \((X, d_1)\), then the terms of the sequence are getting closer in the \(d_1\) metric. Due to the proportional relationship, they must also be getting closer in the \(d_2\) metric.

  3. Cauchy in \(d_2\) Implies Cauchy in \(d_1\): Similarly, if a sequence is Cauchy in \((X, d_2)\), the proportional relationship ensures that the sequence is also Cauchy in \((X, d_1)\).

By establishing these two implications, we conclude that the set of Cauchy sequences in both metrics is the same.

Problem 10. Completeness of Complex Numbers Using Completeness of Real Numbers

Given:

  • The real numbers \(\mathbb{R}\) are complete, which means every Cauchy sequence of real numbers converges to a limit in \(\mathbb{R}\).

To Prove:

  • The complex numbers \(\mathbb{C}\) are complete.

Proof:

  1. Representation of Complex Numbers: Every complex number can be represented as:

    \begin{equation*} z = x + yi \end{equation*}

    where \(x\) and \(y\) are real numbers and \(i\) is the imaginary unit.

  2. Assume a Cauchy Sequence in \(\mathbb{C}\): Let \((z_n)\) be a Cauchy sequence in \(\mathbb{C}\). This means that for any given \(\epsilon > 0\), there exists an integer \(N\) such that for all \(m, n \geq N\):

    \begin{equation*} |z_m - z_n| < \epsilon \end{equation*}

  3. Real and Imaginary Parts are Cauchy: The sequences of real parts \((x_n)\) and imaginary parts \((y_n)\) of \((z_n)\) are also Cauchy in \(\mathbb{R}\). This is because:

    \begin{equation*} |x_m - x_n| \leq |z_m - z_n| |y_m - y_n| \leq |z_m - z_n| \end{equation*}

  4. Convergence of Real and Imaginary Parts: Since \(\mathbb{R}\) is complete, the Cauchy sequences \((x_n)\) and \((y_n)\) converge to limits in \(\mathbb{R}\).

  5. Convergence of the Complex Sequence: The sequence \((z_n)\) converges to:

    \begin{equation*} z = x + yi \end{equation*}

    in \(\mathbb{C}\).

Conclusion: Every Cauchy sequence in \(\mathbb{C}\) converges to a limit in \(\mathbb{C}\). Hence, \(\mathbb{C}\) is complete.

This completes the proof.