# Real Analysis

Sequences

# Sequences

A *sequence* \(\{p_n\}\) in a metric space \(X\) is a function \(f : \mathbb{N} \rightarrow X\) which maps \(n\) to \(p_n\), a point in \(X\).

A sequence \(\{p_n\}\) *converges* if there exists a point \(p \in X\) such that for any \(\epsilon > 0\), there exists an index \(\mathcal{N} \in \mathbb{N}\) such that \(n \geq \mathcal{N}\) implies \(d(p_n, p) < \epsilon\). A convergent sequence is denoted as \(p_n \rightarrow p\), read as \(p_n\) converges to \(p\), or \(\lim\limits_{n \rightarrow \infty} p_n = p\), read as \(p\) is the limit of a sequence \(p_n\).

For example,

In \(\mathbb{R}\), the sequence \(p_n = \frac{n + 1}{n}\) converges to \(1\).

*Proof.*Consider any \(\epsilon > 0\). Choose \(\mathcal{N} = \left\lceil\frac{1}{\epsilon}\right\rceil + 1\). For any integer \(n \geq \mathcal{N}\), \(n > \frac{1}{\epsilon}\). Hence, \(\frac{1}{n} < \epsilon\). So, \(d(p_n, 1) = \left|\frac{n + 1}{n} - 1\right| = \left|\frac{1}{n}\right| < \epsilon\). QED.If \(p_n \rightarrow p\) and \(p_n \rightarrow p'\) then \(p = p'\).

*Proof (by contradiction).*Assume \(p_n \rightarrow p\) and \(p_n \rightarrow q\). Let \(\epsilon = d(p, q)\). Then, due to convergence, there exists \(\mathcal{N}_p\) such that \(n \geq \mathcal{N}_p\) implies \(d(p_n, p) < \frac{\epsilon}{2}\). Also, there exists \(\mathcal{N}_q\) such that \(n \geq \mathcal{N}_q\) implies \(d(p_n, q) < \frac{\epsilon}{2}\). Let \(\mathcal{N} = \max \{\mathcal{N}_p, \mathcal{N}_q\}\). Then \(n \geq \mathcal{N}\) implies \(\epsilon = d(p, q) \leq d(p_n, p) + d(p_n, q) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\), which is a contradiction. QED.If \(\{p_n\}\) is bounded (the range of \(p_n\) is bounded) then \(p_n\) converges.

False.

If \(p_n\) converges then \(p_n\) is bounded.

*Proof.*Use \(\epsilon = 1\). Then there exists \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(d(p_n, p) < 1\). Let \(R = \max\{1, d(p, p_1), \dots, d(p, p_{\mathcal{N} - 1})\}\). All \(\{p_n\}\) are in the open ball \(B_{R + 1}(p_n)\). QED.If \(p_n \rightarrow p\) then \(p\) is a limit point of the range of \(\{p_n\}\).

False.

If \(p\) is a limit point of \(E\) in \(X\), there exists a sequence \(\{p_n\}\) in \(E\) such that \(p_n \rightarrow p\).

*Proof.*Choose a sequence \(\{p_n\}\), where each \(p_n\) is in an open ball \(B_\frac{1}{n}(p)\). Then \(p_n \rightarrow p\). QED.\(p_n \rightarrow p\) if and only if every neighborhood of \(p\) contains all but finitely many \(p_n\).

TRUE.

IMPORTANT IDEA: to show convergence, you must fine an \(\mathcal{N}\) for every \(\epsilon\).

Consider sequences \(\{s_n\}\), \(\{t_n\}\) in \(\mathbb{C}\) where \(s_n \rightarrow s\) and \(t_n \rightarrow t\).

**Theorem.** \(\lim\limits_{n \rightarrow \infty} (c + s_n) = c + s\).

*Proof.* Fix \(\epsilon > 0\). Then, there exists \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(|s_n - s| < \epsilon\). Hence, for this \(\mathcal{N}\), \(n \geq \mathcal{N}\) implies \(|(c + s_n) - (c - s)| = |s_n - s| < \epsilon\), as desired. QED.

**Theorem.** \(\lim\limits_{n \rightarrow \infty} (s_n + t_n) = s + t\).

*Proof.* Given \(\epsilon > 0\), there exist \(\mathcal{N}_s\) and \(\mathcal{N}_t\) such that if \(n \geq \mathcal{N}_s\) then \(|s_n - s| < \frac{\epsilon}{2}\) and if \(n \geq \mathcal{N}_t\) then \(|t_n - t| < \frac{\epsilon}{2}\). Let \(\mathcal{N} = \max\{\mathcal{N}_s, \mathcal{N}_t\}\). Then, for \(n \geq \mathcal{N}\), \(|(s_n + t_n) - (s + t)| \leq |s_n - s| + |t_n - t| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\), as desired. QED.

**Theorem.** \(\lim\limits_{n \rightarrow \infty} (cs_n) = cs\).

*Proof.* Fix \(\epsilon > 0\). Then, there exists \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(|s_n - s| < \frac{\epsilon}{|c|}\). Hence, for this \(\mathcal{N}\), \(n \geq \mathcal{N}\) implies \(|cs_n - cs| = |c||s_n - s| < \epsilon\), as desired. QED.

**Theorem.** \(\lim\limits_{n \rightarrow \infty} (s_nt_n) = st\).

*Proof.* Fix \(\epsilon > 0\). Let \(k = \max\{1, \epsilon, s, t\}\). Then, there exist \(\mathcal{N}_s\) and \(\mathcal{N}_t\) such that if \(n \geq \mathcal{N}_s\) then \(|s_n - s| < \frac{\epsilon}{3k}\) and if \(n \geq \mathcal{N}_t\) then \(|t_n - t| < \frac{\epsilon}{3k}\). Let \(\mathcal{N} = \max\{\mathcal{N}_s, \mathcal{N}_t\}\). Then, for \(n \geq \mathcal{N}\), \(|s_nt_n - st| = |(s_n - s)(t_n - t) + s(t_n - t) + t(s_n - s)| < \frac{\epsilon^2}{9k^2} + \frac{\epsilon}{3} + \frac{\epsilon}{3} < \frac{\epsilon}{9k} + \frac{\epsilon}{3} + \frac{\epsilon}{3} < \epsilon\), as desired. QED.

# Subsequences

Suppose \(\{p_n\}\) is a sequence. Let \(n_1 < n_2 < n_3 < \dots\), where \(n_i \in \mathbb{N}\), be an increasing sequence. Then, \(\{p_{n_i}\}\) is a subsequence.

If \(p_n \rightarrow p\) then every subsequence converges to \(p\).

For example,

- The sequence \(\{1, \pi, \frac{1}{2}, \pi, \frac{1}{3}, \pi, \dots\}\) does not converge. But it has a subsequence \(\{1, \frac{1}{2}, \frac{1}{3}, \dots\}\) that converges to \(0\) and another subsequence \(\{\pi, \pi, \pi, \dots\}\) that converges to \(\pi\), which is called a
*subsequential limit*.

A metric space is *sequentially compact* if every sequence has a convergent subsequence.

**Theorem.** If a metric space \(X\) is compact then \(X\) is sequentially compact. Or, equivalently, in a compact metric space, every sequence has a subsequence converging to a point of \(X\).

(In fact, compactness is equivalent to sequential compactness.)

*Proof.* Let \(R = \text{range}\{p_n\}\). If \(R\) is finite, then some \(p \in p_n\) is achieved infinitely many times. Use that subsequence. If \(R\) is infinite, then by previous theorem, since \(X\) is compact, \(R\) has a limit point \(p\). Use that point \(p\) and its nested neighborhoods to construct a subsequence convergent on \(p\). QED.

**Corollary.** Every bounded sequence in \(\mathbb{R}^k\) contains a convergent subsequence.

# Cauchy Sequence

A sequence \(\{p_n\}\) is a *Cauchy sequence* if for every \(\epsilon > 0\), there exists \(\mathcal{N}\) such that \(m, n > \mathcal{N}\) implies \(d(p_m, p_n) < \epsilon\).

**Theorem.** If \(\{p_n\}\) converges, then \(\{p_n\}\) is a Cauchy sequence.

*Proof.* Given \(\epsilon > 0\), there exists \(\mathcal{N}\) such that if \(n \geq \mathcal{N}\) then \(|p_n - p| < \frac{\epsilon}{2}\). Then, for this \(\mathcal{N}\), \(m, n > \mathcal{N}\) implies \(d(p_m, p_n) \leq d(p_m, p) + d(p_n, p) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\), as desired. QED.

# Complete Spaces

A metric space \(X\) is *complete* if every Cauchy sequence converges to a point of \(X\).

For example,

- \(\mathbb{Q}\) is not complete.

**Theorem.** Compact metric spaces are complete.

*Proof.* Let \(\{x_n\}\) be a Cauchy sequence in \(X\). Since \(X\) is compact, it is sequentially compact. So, there exists a subsequence \(\{x_{n_k}\}\) converging to a point \(x \in X\). Fix \(\epsilon > 0\). The fact that sequence \(\{x_n\}\) is a Cauchy sequence implies there exists \(\mathcal{N}_1\) such that if \(i, j > \mathcal{N}\) then \(d(x_i, x_j) < \frac{\epsilon}{2}\). The fact that \(\{x_{n_k}\}\) converges to \(x\) implies there exists \(\mathcal{N}_2\) such that if \(n_k \geq \mathcal{N}_2\) then \(d(x_{n_k}, x) < \frac{\epsilon}{2}\). Let \(\mathcal{N} = \max\{\mathcal{N}_1, \mathcal{N}_2\}\). Fix any \(n_k, n > \mathcal{N}\). By triangle inequality, \(d(x_n, x) \leq d(x_n, x_{n_k}) + d(x_{n_k}, x) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). Since \(\{x_n\}\) is arbitary, \(X\) is complete. QED.

**Corollary.** \([0, 1]\) is complete.

**Corollary.** \(k\)-cells in \(\mathbb{R}^k\) are complete.

**Corollary.** Any closed subset of a compact set is complete.

**Corollary.** \(\mathbb{R}^k\) is complete.

*Proof.* If \(\{x_n\}\) is Cauchy, it’s bounded. So it’s in some \(k\)-cell in \(\mathbb{R}^k\). Hence, \(\{x_n\}\) converges because \(k\)-cells are complete. QED.

For example,

\(x_n = 1 + \frac{1}{2} + \dots + \frac{1}{n}\) does not converge.

*Proof.*Given \(n > m\), consider \(|x_n - x_m| = \frac{1}{m + 1} + \frac{1}{m + 2} + \dots + \frac{1}{n} \geq \frac{n - m}{n} = 1 - \frac{m}{n}\). Let \(n = 2m\). We have \(|x_{2m} - x_m| \geq \frac{1}{2}\). So the sequence is not Cauchy. Thus, it does not converge. QED.Let \(x_1 = 1\), \(x_2 = 2\), and \(x_n = \frac{x_{n - 1} + x_{n - 2}}{2}\). Every subsequence of it is Cauchy so it is convergent.

**Theorem.** Every metric space \((X, d)\) has a *completion* \((X^\star, d)\).

Idea: Given \(X\), let \(X^\star\) be the set of all Cauchy sequences in \(X\) under an equivalent relation \(\sim\), where \(\{p_n\} \sim \{q_n\}\) if \(\lim\limits_{n \rightarrow \infty} d(p_n, q_n) = 0\). For \(P, Q \in X^\star\), let \(\Delta(P, Q) := \lim\limits_{n \rightarrow \infty} d(p_n, q_n)\), where \(\{p_n\}\) and \(\{q_n\}\) are representatives of \(P\) and \(Q\), respectively.

# Bounded Sequences

*Monotonically increasing* sequences: \(\dots \leq s_{n - 1} \leq s_n \leq \dots\)

*Monotonically decreasing* sequences: \(\dots \geq s_{n - 1} \geq s_n \geq \dots\)

**Theorem.** Bounded monotonic sequences converge to their suprema (if increasing) or to their infima (if decreasing).

*Proof.* Fix a bounded monotonically increasing sequence ${x_n}. Let \(s = \sup(\text{range}\{x_n\})\). So, for all \(\epsilon > 0\), there exists \(\mathcal{N}\) such that \(s - \epsilon \leq x_\mathcal{N} \leq s\). Then, for all \(n \geq \mathcal{N}\), \(s - \epsilon \leq x_\mathcal{N} \leq x_n \leq s\), as desired. QED.

# Divergent Sequences

Write \(x_n \rightarrow +\infty\) if for all \(\mathcal{M} \in \mathbb{R}\), there exist \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(x_n > \mathcal{M}\).

Write \(x_n \rightarrow -\infty\) if for all \(\mathcal{M} \in \mathbb{R}\), there exist \(\mathcal{N}\) such that \(n \geq \mathcal{N}\) implies \(x_n < \mathcal{M}\).

Given sequence \(\{x_n\}\), let \(E\) be the set of all possible subsequential limits of \(\{x_n\}\) (allowing \(+\infty\) and \(-\infty\), i.e. \(E\) is in extended \(\mathbb{R}\)). Then \(E\) is closed. Let \(s^\star = \sup E\) and \(s_\star = \inf E\). Then \(\limsup\limits_{n \rightarrow \infty} s_n = s^\star\), called the *upper limit* of \(s_n\), and \(\liminf\limits_{n \rightarrow \infty} s_n = s_\star\), called the *lower limit* of \(s_n\).

Alternatively, \(\limsup\limits_{n \rightarrow \infty} s_n = \lim\limits_{n \rightarrow \infty}\left(\sup\limits_{k \geq n} s_k\right)\). Similarly, \(\liminf\limits_{n \rightarrow \infty} s_n = \lim\limits_{n \rightarrow \infty}\left(\inf\limits_{k \geq n} s_k\right)\).

For example,

- If \(s_n \rightarrow s\) then \(\liminf s_n = \limsup s_n = s\).
- If \(\{s_n\} = \{0.1, \frac{3}{2}, 0.11, \frac{4}{3}, 0.111, \frac{5}{4}, 0.1111, \frac{6}{5}, \dots\}\) then \(\limsup s_n = 1\) and \(\liminf s_n = \frac{1}{9}\).