Cauchy Sequence is Bounded

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Then every Cauchy sequence in $M$ is bounded.

Proof
Let $\left \langle {x_n} \right \rangle$ be a Cauchy sequence in $M$.

By definition:


 * $\forall \epsilon > 0: \exists N \in \N: \forall m, n > N: d \left({x_n, x_m}\right) < \epsilon$

Particularly, setting $\epsilon = 1$:


 * $\exists N_1: \forall m, n > N_1: d \left({x_n, x_m}\right) < 1$

Note that since $N_1 \geq N_1$, this means that:
 * $\forall n \geq N_1: d \left({x_n, x_{N_1}}\right) < 1$

To show $\left \langle {x_n} \right \rangle$ is bounded, we need to show that there exists $a \in A$ and $K \in \R$ such that $d \left({x_n, a}\right) \le K$ for all $x_n \in \left \langle {x_n} \right \rangle$.

Let $K'$ be the maximum distance from $x_{N_1}$ to any of the earlier terms in the sequence.

That is, $K' = \max \left\{d \left({x_{N_1}, x_1}\right), d \left({x_{N_1}, x_2}\right), \ldots, d \left({x_{N_1}, x_{N_1 - 1}}\right)\right\}$

Then:
 * Each $x_n$ for $n \ge N_1$ satisfies $d \left({x_{N_1}, x_n}\right) \le 1$ by choice of $N_1$ as mentioned above
 * Each $x_n$ for $n < N_1$ satisfies $d \left({x_{N_1}, x_n}\right) \le K'$ by choice of $K'$.

Thus, taking $a = x_{N_1}$ and $K = \max \left\{{K', 1}\right\}$, we have shown that each $x_n$ satisfies $d \left({a, x_n}\right) \le K$.

So, $\left \langle {x_n} \right \rangle$ is bounded.