> [!definition] > > Let $(x_n)$ be a [[Sequence|sequence]]. A [[Real Numbers|real number]] $x \in \real$ is called a [[Limit|limit]] (accumulation) point if there exists a subsequence $(x_{n_k}): \lim_{k \to \infty}x_{n_{k}} = x$. $\mathcal L$ denotes the [[Set|set]] of all limit points of $(x_n)$. If $(x_n)$ is bounded, then $\mathcal L$ must be a bounded set. > [!theorem] > > If $(x_n)$ is a bounded sequence, then by the [[Bolzano-Weierstrass Property|Bolzano-Weierstrass Theorem]], it has at least one subsequence ($\mathcal L \ne \emptyset$). Since the sequence is bounded, $\mathcal L$ is bounded as well. > [!theorem] > > Let $(x_n)$ be a sequence. Then > $ > \forall x \in \mathcal{L}, \varepsilon > 0, \abs{\bracs{n: |x_n - x|<\varepsilon}} = \aleph_0 > $ > *Proof*. Suppose that $x \in \mathcal L$ and let $(x_{n_k})$ be a subsequence where $\lim_{k \to \infty}x_{n_k} = x$. Meaning that $\exists K \in \nat: \forall k \ge K, |x_{n_{k}} - x| < \varepsilon$. > $ > \bracs{n_k: k \ge K} \subseteq \bracs{n: |x_n - x| < \varepsilon} > $ > Meaning that > $ > \aleph_0 = \abs{\bracs{n_k: k \ge K}} \le \abs{\bracs{n: |x_n - x| < \varepsilon}} \le \aleph_0 > $ > > Take $\varepsilon = 1$ and suppose that $\abs{\bracs{n: |x_n - x| < \varepsilon}} = \aleph_0$. Take $n_1 = \min{\bracs{n: |x_n - x| < \varepsilon}}$. Define $n_k$ recursively, suppose that $n_k$ is chosen, then take $\varepsilon = \frac{1}{k + 1}$ and consider the set $\bracs{n: n > n_k, |x_n - x| < \frac{1}{k + 1}}$. This set is infinite because we are excluding a finite number of elements from an infinite set. > $ > n_{k + 1} = \min \bracs{n: n > n_k, |x_n - x| < \frac{1}{k + 1}} > $ > This creates a strictly increasing sequence of indices, where for any $k$, > $ > |x_{n_k} - x| < \frac{1}{k} > $ > which leads to the subsequence converging to $x$. > [!theorem] > > Let $(x_n)$ be a bounded sequence, then the [[Limit Superior and Inferior|limit superior and inferior]] > $ > \overline\lim_{n \to \infty}x_n = \sup \mathcal L \quad > \underline{\lim}_{n \to \infty} = \inf \mathcal L > $ > and > $ > \sup \mathcal L, \inf \mathcal L \in \mathcal L > $ > *Proof*. Let $x \in \mathcal L$, and let $(x_{n_a})$ be a subsequence such that $\lim_{a \to \infty}x_{n_{a}} = x$. Let $y_n = \sup \bracs{x_m: m \ge n}$, by construction $y_n \ge x_n \forall n \in \nat$ and $y_{n_k} \ge x_{n_k} \forall k \in \nat$. As $(y_n)$ represents the sequence converging to the limit superior, its subsequence $(y_{n_k})$ converges to the same value. Then by the [[Order Limit Theorem|order limit theorem]], $\limsup x_n \ge x$, making $\limsup x_n$ an upper bound of $\mathcal L$. > > Set $n_1 = 1$ and define > $ > y_{n_k + 1} = \sup\bracs{x_m: m \ge n_k + 1} > $ > Then $y_{n_k + 1} - \frac{1}{k + 1} < y_{n_k + 1}$, and is not an upper bound of $\bracs{x_m: m \ge n_k + 1}$. There has to exist some $n_{k + 1} \in \bracs{x_m: m \ge n_k + 1}$, $x_{n_{k + 1}} \ge n_k + 1 \gt n_k$ where $x_{n_{k + 1}}$ is between $y_{n_k + 1} - \frac{1}{k + 1}$ and $y_{n_k + 1}$. $n_k$ then is a strictly increasing sequence of indices where for every $k \ge 1$, $|x_{n_{k + 1}} - y_{n_{k + 1}}| < \frac{1}{k + 1}$. As $y_n$ converges, $x_{n_k}$ must converge to the same value as $y_n$, creating a subsequence with its limit being the limit superior. > [!theorem] > > Let $(x_n)$ be an *unbounded* sequence, then the limit superior and inferior belong to the set of limit points in the context of an extended set.