Convergent Sequence in Metric Space is Bounded

Theorem
All convergent sequences are bounded.

Proof for General Metric Spaces
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.

Proof for Real and Complex Numbers
Although the Real Number Line is Metric Space, and the Complex Plane is Metric Space and therefore the proof for metric spaces is perfectly adequate, it is useful to provide a proof which does not rely on the topological concepts upon which the theory of metric spaces depend.

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$ or $\C$.

Let $x_n \to l$ as $n \to \infty$.

We need to find $K$ such that $\forall n \in \N: \left|{x_n}\right| \le K$ from the definition of boundedness.

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

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

That is:
 * $\exists N_1: \forall n > N_1: \left|{x_n - l}\right| < 1$

By the backwards form of the Triangle Inequality:
 * $\forall n > N_1: \left|{x_n}\right| - \left|{l}\right| \le \left|{x_n - l}\right| < 1$

That is, $\left|{x_n}\right| < \left|{l}\right| + 1$.

So we set:
 * $K = \max \left\{ {\left| {x_1} \right|, \left| {x_2} \right|, \ldots, \left| {x_{N_1}} \right|, \left| {l} \right| + 1} \right\}$

and the result follows.