Definition 11.1. Let $\{a_n\}$ be a sequence. Let $f$ be a strictly increasing function from $\mathbb{P}$ into $\mathbb{P}$. The sequence $a_{f(n)}$ is called a subsequence of the sequence $\{a_n\}$. [The function $f$ is strictly increasing if $f(m) < f(n)$ whenever $m < n$.]
Lemma
Lemma. Let $f$ be a strictly increasing function from $\mathbb{P}$ into $\mathbb{P}$. Then $n \leq f(n), \forall n \in \mathbb{P}$.
Proof. Let $S(n)$ be the statement that is $n \leq f(n)$. We will use mathematical induction on $S(n)$. Note that $f(n) \geq 1, \forall n \in \mathbb{P}$ because the range is $\mathbb{P}$. Thus $f(1) \geq 1$ and $S(1)$ is true. Suppose that $S(n)$ is true. Note that $n \leq f(n) < f(n+1)$. Since $f(n+1), n \in \mathbb{P}$, $f(n+1) - n \in \mathbb{P}$ by Lemma 6.8. Then $$f(n+1) - n - 1 = 0 \text{ or } f(n+1) - n - 1 \in \mathbb{P} \\ \Longrightarrow f(n+1) = n+1 \text{ or } f(n+1) > n+1 \\ \Longrightarrow n+1 \leq f(n+1).$$ Thus $S(n+1)$ is true. $\blacksquare$
Theorem 11.2
Theorem 11.2. Let $\{a_n\}$ be a sequence with limit $L$. Then any subsequence of $\{a_n\}$ has limit $L$.
Proof. Let $\{a_{f(n)}\}$ be a subsequence of $\{a_n\}$. Let $\varepsilon > 0$. Then $\exists N \in \mathbb{P}$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$. If $n \geq N$, $N \leq f(N) \leq f(n)$ by above Lemma, and therefore $|a_{f(n)} - L| < \varepsilon$ for all $f(n) \geq N$. Thus $$\lim_{n \to \infty} a_{f(n)} = L. \blacksquare$$
Exercise 11.8
Exercise 11.8. Let $\{ a_n \}$ be a sequence. If $$\lim_{n \to \infty} a_{2n} = L = \lim_{n \to \infty} a_{2n-1},$$ then $\lim_{n \to \infty} a_n = L$.
Solution. For a given $\varepsilon > 0$, there exists positive integers $N_1, N_2$ such that $|a_{2n} - L| < \varepsilon, \forall n \geq N_1$ and $|a_{2n-1} - L| < \varepsilon, \forall n \geq N_2$. Let $N = \max \{ 2N_1, 2N_2 - 1 \}$ and let $n \geq N$. If $n$ is even, i.e., $n = 2k$ for some positive integer $k$, then $$n \geq N \geq 2N_1 \Longrightarrow k \geq N_1 \\ \Longrightarrow |a_{2k} - L| = |a_n - L| < \varepsilon.$$ If $n$ is odd, i.e., $n = 2k-1$ for some positive integer $k$, then $$n \geq N \geq 2N_2-1 \Longrightarrow k \geq N_2 \\ \Longrightarrow |a_{2k-1} - L| = |a_n - L| < \varepsilon.$$ Thus $\forall n \geq N, |a_n - L| < \varepsilon$, which means that $\lim_{n \to \infty} a_n = L$. $\blacksquare$