Convergent Sequence in Metric Space is Bounded

Theorem
All convergent sequences are 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 $$\forall \epsilon > 0: \exists N: n > N \Longrightarrow d \left({x_n, l}\right) < \epsilon$$.

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.