Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 1

Theorem
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.

Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the following limits:


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

Then:

Proof
Because $\sequence {x_n}$ is convergent, it is bounded by Convergent Sequence in Normed Division Ring is Bounded.

Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.

Let $M = \max \set {K, \norm m}$.

Then $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$.

Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon {2 M} > 0$.

Since $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that:
 * $\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$

Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
 * $\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$

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

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

Thus $\forall n > N$:

Hence:
 * $\sequence {x_n y_n}$ is convergent

It follows that:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$