Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 2

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

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

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


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

Then:

Proof
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded.

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

Then for $n = 1, 2, 3, \ldots$:

We note that $\sequence {z_n}$ is a real sequence.

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

So from Definition:Convergent Sequence in Normed Division Ring:
 * $\norm {x_n - l} \to 0$ as $n \to \infty$

Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.

From the Combined Sum Rule for Real Sequences:
 * $\ds \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$

By applying the Squeeze Theorem for Complex Sequences (which applies as well to real as to complex sequences):
 * $\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.

By definition of a convergent sequence in a normed division ring:
 * $\sequence{x_n y_n}$ is convergent in $R$

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