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:

$\sequence {x_n y_n}$ is convergent to the limit $\ds \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$

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

\(\ds \norm {x_n y_n - l m}\)

\(=\)

\(\ds \norm {x_n y_n - x_n m + x_n m - l m}\)

\(\ds \)

\(\le\)

\(\ds \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\)

Norm Axiom $\text N 3$: Triangle Inequality

\(\ds \)

\(=\)

\(\ds \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m\)

Norm Axiom $\text N 2$: Multiplicativity

\(\ds \)

\(\le\)

\(\ds K \norm {y_n - m} + \norm m \norm {x_n - l}\)

as $\sequence {x_n}$ is bounded by $K$

\(\ds \)

\(=:\)

\(\ds z_n\)

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:

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

$\blacksquare$