Question

The following guides you through the construction of the proof of the fact that a Cauchy sequence of real numbers converges. Let \(\left\\{a_{n}\right\\}\) be a sequence of real numbers. (i) Show that \(\left\\{a_{n}\right\\}\) contains a subsequence which is either nondecreasing or non-increasing. (ii) Use (i) to show that every bounded sequence has a convergent subsequence. Now let \(\left\\{a_{n}\right\\}\) be a Cauchy sequence of real numbers. (iii) Show that \(\left\\{a_{n}\right\\}\) is bounded. (iv) Deduce that \(\left\\{a_{n}\right\\}\) contains a convergent subsequence. (v) Show that \(\left\\{a_{n}\right\\}\) converges.

Short answer

Question: Prove that every Cauchy sequence of real numbers converges. Answer: We can break down the proof into the following steps: 1. Show that every sequence of real numbers contains a subsequence which is either non-decreasing or non-increasing. 2. Show that every bounded sequence has a convergent subsequence. 3. Show that every Cauchy sequence is bounded. 4. Deduce that a Cauchy sequence has a convergent subsequence. 5. Show that a Cauchy sequence converges to the same limit as its convergent subsequence. Following these steps, we first demonstrate that every sequence has a non-decreasing or non-increasing subsequence. Then, using the Bolzano-Weierstrass theorem, we establish that every bounded sequence has a convergent subsequence. Next, we prove that all Cauchy sequences are bounded. Based on the prior steps, we can deduce that a Cauchy sequence has a convergent subsequence. Finally, we show that a Cauchy sequence converges to the same limit as its convergent subsequence, thus proving that every Cauchy sequence of real numbers converges.

Step-by-step solution

Part (i): Find a nondecreasing or non-increasing subsequence

For a sequence of real numbers \(\{a_n\}\), we need to show that it contains a subsequence which is either non-decreasing or non-increasing. To prove this, we will use the following approach: 1. Select \(n_1=1\) 2. For each \(k>1\), if there is an \(n_k > n_{k-1}\) such that \(a_{n_k} \geq a_{n_{k-1}}\), choose the smallest such \(n_k\). 3. If there is no such \(n_k\), choose the smallest \(n_k > n_{k-1}\) such that \(a_{n_k} \leq a_{n_{k-1}}\). Now, we can make two observations: 1. For each \(k>1\), either a non-decreasing subsequence \(\{a_{n_k}\}\) is being constructed, in which case \(a_{n_1} \leq a_{n_2} \leq \dots \leq a_{n_k}\), or, 2. A non-increasing subsequence \(\{a_{n_k}\}\) is being constructed, in which case \(a_{n_1} \geq a_{n_2} \geq \dots \geq a_{n_k}\). Thus, we have shown that every sequence \(\{a_n\}\) has a subsequence which is either non-decreasing or non-increasing.

Part (ii): Show that every bounded sequence has a convergent subsequence

Now, we need to show that every bounded sequence has a convergent subsequence. We will use the result from part (i) and the Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence. Suppose we have a bounded sequence \(\{a_n\}\). Then, by part (i), we know that \(\{a_n\}\) contains a subsequence which is either non-decreasing or non-increasing. Now, using the Bolzano-Weierstrass theorem, we can conclude that this subsequence has a convergent subsequence. As a subsequence of a subsequence is also a subsequence of the original sequence, we have shown that every bounded sequence has a convergent subsequence.

Part (iii): Show that every Cauchy sequence is bounded

Let \(\{a_n\}\) be a Cauchy sequence of real numbers. By definition, for every \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that for all \(m, n > N\), \(|a_m - a_n| < \epsilon\). Let's choose \(\epsilon = 1\). Then, for \(m, n > N\), we have \(|a_m - a_n| < 1\). In particular, we have \(|a_{N+1} - a_n| < 1\) for all \(n > N\). This implies that \(a_n \in (a_{N+1} - 1, a_{N+1} + 1)\) for all \(n > N\). Now, we have that all terms \(a_1, a_2, \dots, a_N\) are finite, and all terms \(a_{N+1}, a_{N+2}, \dots\) are within the interval \((a_{N+1} - 1, a_{N+1} + 1)\). Therefore, the sequence \(\{a_n\}\) is bounded.

Part (iv): Deduce that a Cauchy sequence has a convergent subsequence

From part (iii), we know that every Cauchy sequence is bounded. So now, using the result from part (ii), we can deduce that since a Cauchy sequence is bounded, it must contain a convergent subsequence.

Part (v): Show that a Cauchy sequence converges

Let \(\{a_n\}\) be a Cauchy sequence, and let \(\{a_{n_k}\}\) be the convergent subsequence we found in part (iv), which converges to a limit \(L\). We want to show that \(\{a_n\}\) itself also converges to \(L\). Take any \(\epsilon > 0\). Since \(\{a_n\}\) is Cauchy, there exists a positive integer \(N_1\) such that for all \(m, n > N_1\), \(|a_m - a_n| < \frac{\epsilon}{2}\). Simultaneously, since \(\{a_{n_k}\}\) converges to \(L\), there exists a positive integer \(N_2\) such that for all \(k > N_2\), \(|a_{n_k} - L| < \frac{\epsilon}{2}\). Let \(N = \max\{N_1, n_{N_2}\}\). Then, for any \(n > N\), there exists \(k > N_2\) with \(n_k \geq n > N\). Hence, using the triangle inequality, we have $$|a_n - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$ Thus, \(\{a_n\}\) also converges to \(L\), which completes the proof that every Cauchy sequence of real numbers converges.

Key concepts

Real number sequences

In mathematics, a sequence of real numbers is an ordered list of numbers, typically denoted with terms like \(a_n\). Each \(a_n\) represents a real number and \(n\) serves as the position of the number in the sequence, beginning with \(n=1\). Sequences can display various patterns and properties, such as being increasing, decreasing, or bounded.

Understanding the behavior of sequences often involves examining their limits and the conditions under which they converge to certain values. This is fundamental to calculus and analysis because it helps in defining functions and understanding series. The concept of boundedness is important to grasp since it helps set up the stage for discussing properties such as convergence and the applicability of the Bolzano-Weierstrass theorem.

Subsequences

A subsequence is formed by selecting certain terms from a sequence without altering their order, but not necessarily taking every term. In other words, from the original sequence \(a_n\), we create a new sequence \(a_{n_k}\) by choosing terms where \(k\) increases; \(n_k < n_{k+1}\) for every \(k\). It's kind of like picking specific pieces from a trail of breadcrumbs, maintaining the direction but perhaps not taking every crumb.

Subsequences are essential when discussing the convergence of the original sequence because, as we will see with the Bolzano-Weierstrass theorem, if any subsequence of a bounded sequence converges, this has larger implications for the sequence as a whole. Finding a convergent subsequence often provides a vital step towards proving the convergence of the entire sequence.

Bolzano-Weierstrass theorem

The Bolzano-Weierstrass theorem is a cornerstone of real analysis and has critical implications for the study of sequences. It states that every bounded sequence in \(\mathbb{R}\), the set of real numbers, has at least one convergent subsequence. 'Bounded' means that all the terms of the sequence lie within a fixed interval, while 'convergence' means that the terms of the subsequence get arbitrarily close to a real number called the limit.

In practice, this theorem is like a guarantee or insurance policy; it assures us that within any set of numbers where no number is too far away from the point of origin, there's always a pattern (subsequence) that settles down towards a single number. This makes it an invaluable tool for proving the convergence of sequences, as seen in the analysis of Cauchy sequences.

Convergence of sequences

A real number sequence \(\{a_n\}\) is said to converge if its terms approach a specific real number, called the limit, as \(n\) tends to infinity. In simple terms, no matter how small an error margin (\(\epsilon\)) we set, there’s always a point beyond which all terms of the sequence lie within that margin around the limit.

For a sequence to converge, it must be consistent in its approach to the limit, which is precisely what Cauchy sequences exhibit. A Cauchy sequence is one where the terms get arbitrarily close to each other as the sequence progresses. This concept closely ties with the idea of a limit since if the terms of a sequence do not stray from each other, they likely settle towards a common point – indicating convergence. A Cauchy sequence is inherently bounded, and hence, by the Bolzano-Weierstrass theorem, has a convergent subsequence, which intuitively leads to the conclusion that the entire sequence must converge.