Combination Theorem for Sequences/Normed Division Ring/Sum Rule

Theorem
Let $\left({R, \left\Vert{\,\cdot\,}\right\Vert}\right)$ be a normed division ring.

Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be convergent in the norm $\left\Vert{\,\cdot\,}\right\Vert$ to the following limits:


 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l$
 * $\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Then:
 * $\left \langle{x_n + y_n}\right \rangle$ is convergent in the norm and $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$

Proof
Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon 2 > 0$.

Since $\left \langle {x_n} \right \rangle$ is a convergent in the norm to $l$, we can find $N_1$ such that:
 * $\forall n > N_1: \left\Vert{x_n - l}\right\Vert < \dfrac \epsilon 2$

Similarly, for $\left \langle {y_n} \right \rangle$ we can find $N_2$ such that:
 * $\forall n > N_2: \left\Vert{y_n - m}\right\Vert < \dfrac \epsilon 2$

Now let $N = \max \left\{{N_1, N_2}\right\}$.

Then if $n > N$, both the above inequalities will be true.

Thus $\forall n > N$:

Hence:
 * $\left \langle{x_n + y_n}\right \rangle$ is convergent in the norm to $l + m$.