Sequences in Metric Spaces

Sequences in Metric Spaces

Definition. A sequence is a metric space $M$ is a function from $\mathbb{P}$ into $M$, and we denote such a sequence as $\{ a_n \}_{n=1}^{\infty}$. 

Definition 37.1

Definition 37.1. Let \(\{a_n\}\) be a sequence in a metric space \((M, d)\). We say that \(\{a_n\}\) converges to (or has limit) \(L\), where \(L \in M\), and write \[ \lim_{n \to \infty} a_n = L \] if for every \(\varepsilon > 0\), there exists a positive integer \(N\) such that if \(n \geq N\), then \(d(a_n, L) < \varepsilon\). If \(\{a_n\}\) has a limit, we call \(\{a_n\}\) a convergent sequence. If \(\{a_n\}\) has no limit, we call \(\{a_n\}\) a divergent sequence.

Note

Note. Let $\{ a_n \}$ be a sequence in a metric space $M$. Then
(i) The limit of a sequence, if it exists, is unique.
(ii) If $\{a_n\}$ is a constant sequence, i.e., $a_n = L, \forall n \in \mathbb{P}$, then $\lim_{n \to \infty} a_n = L$.
(iii) If $\lim_{n \to \infty} a_n = L$, then every subsequence of $\{ a_n \}$ converges to $L$. 

Eventually Constant

Definition. A sequence $\{ a_n \}$ in a metric space $M$ is said to be eventually constant if there exists a positive integer $N$ such that $a_n = a_N$ if $n \geq N$. 
Note that if $\{ a_n \}$ converges in $M$, then $\{ a_n \}$ is eventually constant, and vice versa.

Theorem 37.2

Theorem 37.2. Let \(\{a^{(k)}\}_{k=1}^\infty\) be a sequence of points in \(\mathbb{R}^n\), and let \(a = (a_1, a_2, \ldots, a_n) \in \mathbb{R}^n\). Let \[ a^{(k)} = \left(a_1^{(k)}, a_2^{(k)}, \ldots, a_n^{(k)}\right), \quad \text{for } k = 1, 2, \ldots. \] Then \(\{a^{(k)}\}_{k=1}^\infty\) converges to \(a\) $\iff$ \(\lim_{k \to \infty} a_j^{(k)} = a_j\) for \(j = 1, \ldots, n\). 

Proof. ($\Longrightarrow$) Suppose that $\{ a^{(k)} \}_{k=1}^{\infty}$ converges to $a$. Let $\varepsilon > 0$. Then $\exists N$ such that $d(a^{(k)}, a) < \varepsilon, \forall k \geq N$. If we denote $a^{(k)} = (a^{(k)}_1, ..., a^{(k)}_n)$ for each $k$, then for all $n \geq N$, $$d(a^{(k)}, a) < \varepsilon \\ \Longrightarrow |a^{(k)}_i - a_i| \leq \sqrt{\sum_{i=1}^n (a^{(k)}_i - a_i)^2)} < \varepsilon, \forall i \in \{ 1, ..., n \} \\ \Longrightarrow \lim_{k \to \infty} a^{(k)}_i = a_i, \forall i \in \{ 1, ..., n\}.$$ ($\Longleftarrow$) Suppose that \(\lim_{k \to \infty} a_j^{(k)} = a_j\) for \(j = 1, \ldots, n\). Let $\varepsilon > 0$. Then for each $j, 1 \leq j \leq n$, $\exists N_j$ such that $|a^{(k)}_j - a_j| < \frac{\varepsilon}{\sqrt{n}}, \forall k \geq N_j$. Let $N = \max \{ N_1, ..., N_n \}$. Then $$d(a^{(k)}, a) = \sqrt{\sum_{j=1}^n (a^{(k)}_j - a_j)^2} < \sqrt{\frac{\varepsilon^2}{n} + \cdots \frac{\varepsilon^2}{n}} = \sqrt{n \frac{\varepsilon^2}{n}} = \varepsilon, \forall k \geq N \\ \Longrightarrow \lim_{k \to \infty} a^{(k)} = a. \blacksquare$$

Note

Note. We distinguish the statements: (i) "Let $\{ a^{(k)} \}$ be a sequence in $\ell^1$" and (ii) "Let $\{ a^{(k)} \}$ be a sequence of points in $\ell^1$".
(i) When we say that $\{ a^{(k)} \}$ is a sequence in $\ell^1$, we mean that $a^{(k)}$ is a real number for every positive integer $k$ and that $\sum_{k=1}^{\infty} a^{(k)}$ converges absolutely, that is, $\{ a^{(k)} \} \in \ell^1$. 
(ii) When we saythat $\{ a^{(k)} \}$ is a sequence of points in $\ell^1$, we mean that $a^{(k)} \in \ell^1$ for every positive integer $k$. If we let $a^{(k)} = \{ a^{(k)}_n \}_{n=1}^{\infty}$, then $a^{(k)}_n$ is the $n$th term of the $k$th sequence. A listing of the points $\{a^{(k)} \}_{k=1}^{\infty}$ is given below: $$\begin{align*} a^{(1)} &= \left(a_1^{(1)}, a_2^{(1)}, \ldots, a_j^{(1)}, \ldots, a_n^{(1)}\right) \\ a^{(2)} &= \left(a_1^{(2)}, a_2^{(2)}, \ldots, a_j^{(2)}, \ldots, a_n^{(2)}\right) \\ &\vdots \\ a^{(k)} &= \left(a_1^{(k)}, a_2^{(k)}, \ldots, a_j^{(k)}, \ldots, a_n^{(k)}\right) \\ &\vdots \end{align*}$$ Note that Theorem 37.2 does not hold for $\ell^1$ (nor for $\ell^2, c^0, \ell^{\infty}$).
($\because$) 

Theorem 37.3

Theorem 37.3. Let \(\{a^{(k)}\}\) be a sequence of points in \(\ell^1\), and let \(a = \{a_n\} \in \ell^1\). If \(\{a^{(k)}\}\) converges to \(a\), then \[ \lim_{k \to \infty} a_n^{(k)} = a_n \] for every positive integer \(n\).

Proof. Suppose that $\{ a^{(k)} \}$ converges to $a$. Let $\varepsilon > 0$. Then $\exists N$ such that $d(\{ a^{(k)} \}, a ) < \varepsilon, \forall k \geq N$. Thus for all positive integer $n$, we have $$|a^{(k)}_n - a_n| \leq \sum_{j=1}^{\infty} |a^{(k)}_j - a_j| = d( \{ a^{(k)} \}, a ) < \varepsilon, \forall k \geq N \\ \Longrightarrow \lim_{k \to \infty} a^{(k)}_n = a_n. \blacksquare$$