Real Analysis

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 $\liminf s_n = \frac{1}{9}$.