(2.6.3) Cauchy sequences are bounded.

For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Proof:

Let $(a_n)$ be a cauchy sequence and let $\epsilon = 1$. By definition, this means that there exists a number $N \in \mathbf{N}$ such that

$$ \begin{align*} |a_n - a_m| < 1 \quad \text{whenever $m, n > N$}. \end{align*} $$

Fix $a_m$ such that $m = N+1$. Since $N+1 > N$, then the following holds

$$ \begin{align*} |a_n - a_{N+1}| \leq 1 \quad \text{whenever $n > N$}. \end{align*} $$

Now we can expand the above inequality to

$$ \begin{align*} -1 \leq a_n - a_{N+1} &\leq 1 \\ a_{N+1} - 1 \leq a_n &\leq 1 + a_{N+1}. \end{align*} $$

From this, we can see that we have established a bound on all the terms of the sequence starting after the $N$th term. For earlier terms, we can establish a bound by defining the lower and upper bounds as

$$ \begin{align*} L = \min\{a_1, a_2, a_3, ..., a_N, a_{N+1}-1\}, \\ U = \min\{a_1, a_2, a_3, ..., a_N, a_{N+1}+1\}. \end{align*} $$

$L$ is a lower bound on any term that comes on or before the $N$th term and similarly $U$ is an upper bound on all the terms that comes on or before the $N$th. Therefore $(a_n)$ is bounded. $\blacksquare$

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