Convergent Sequence in Metric Space is Bounded

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

Let $\left \langle {x_n} \right \rangle$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.

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

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

Let $\left \langle {x_n} \right \rangle$ be a sequence in $M$ which is convergent, and so $x_n \to l$ as $n \to \infty$.

From the definition, in order to prove boundedness, all we need to do is find $K \in \R$ such that $\forall n \in \N: d \left({x_n, l}\right) \le K$.

Since $\left \langle {x_n} \right \rangle$ converges, it is true that:
 * $\forall \epsilon > 0: \exists N: n > N \implies d \left({x_n, l}\right) < \epsilon$

In particular, this is true when $\epsilon = 1$, for example.

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

So now we set:
 * $K = \max \left\{{d \left({x_1, l}\right), d \left({x_2, l}\right), \ldots, d \left({x_{N_1}, l}\right), 1}\right\}$

The result follows.