_Posted on May 08, 2024_ _Revised on September 20, 2024_ ##### Introduction and definitions I use this theorem primarily as a means of introducing the definition of sequences and what it means for one to converge. A _sequence_ in general is a function whos domain is an interval of integers. There are a few ways of denoting sequences, but the sequence $a: \{n \in \mathbb{Z} \mid m \leq n \leq k\} \rightarrow \mathbb{R}$ is typically denoted as $(a_n)_m^k$. For $m = 1$, $k = \infty$, the sequence is denoted as $(a_n)_{n \in \mathbb{Z}^+}$. When the domain is understood from context, or $k = \infty$ and the results do not depend on $m,$ we donote the sequence more simply as $(a_n)$. The value of $(a_n)_m^k$ at $n$ is written as $a_n$. In analysis, sequences with domain $\mathbb{N}$ or $\mathbb{Z}^+$ are typically considered. A sequence $(a_n)$ is said to converge to the value $s$ if and only if for all $\epsilon \in \mathbb{R}$ such that $\epsilon > 0,$ there exists a $N \in \mathbb{N}$ such that, for $n \in \mathbb{N}$ where $n > N$, $|a_n - s| < \epsilon$. If a sequence $(a_n)$ converges to the value $s$, we write this often as $\lim_{n \rightarrow \infty}a_n=s$, where $s$ is called the _limit_ of $(s_n)$. We also may write this as $(a_n) \hookrightarrow s$. ---- ##### Theorem Statement Now, there is a theorem which states that > [!theorem] **Theorem:** _All sequences which converge are bounded. That is, for all sequences $(a_n)$ for which there exists a limit $s$, there exists $m,M \in \mathbb{R}$ such that $m \leq a_n$ and $M \geq a_n$ for all $n$._ --- ##### Proof Let $\epsilon > 0$. We are given $(a_n) \hookrightarrow s$, for some $s$. By the definition of convergence, we have an $N$ such that $n > N \implies |a_n - s|< \epsilon$. Thus, $s - \epsilon < a_n < s + \epsilon$, for $n > N$. Now, let $S = \{a_n \mid n \leq N\} \cup \{s - \epsilon, s + \epsilon\}$. This set is finite and so clearly has both a minimum and a maximum. Take $m = \text{min}(S)$ and $M = \text{max}(S)$. Clearly, $m$ is a lower bound for $(a_n)$ and $M$ is an upper bound for $(a_n)$ $\blacksquare$