Combination Theorem for Sequences/Normed Division Ring/Product Rule

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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:

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


Proof 1

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:

$\norm m \le M$

and:

$\forall n: \norm{x_n} \le M$


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 $\forall n > N$:

\(\displaystyle \norm {x_n y_n - l m}\) \(=\) \(\displaystyle \norm {x_n y_n - x_n m + x_n m - l m}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) Axiom (N3) of norm (Triangle Inequality).
\(\displaystyle \) \(=\) \(\displaystyle \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m\) Axiom (N2) of norm (Multiplicativity).
\(\displaystyle \) \(\le\) \(\displaystyle M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M\) since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$
\(\displaystyle \) \(\le\) \(\displaystyle M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac \epsilon 2 + \dfrac \epsilon 2\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

Hence:

$\sequence {x_n y_n}$ is convergent

It follows that:

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

$\blacksquare$


Proof 2

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

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

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

\(\displaystyle \norm {x_n y_n - l m}\) \(=\) \(\displaystyle \norm {x_n y_n - x_n m + x_n m - l m}\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) Axiom (N3) of norm (Triangle Inequality)
\(\displaystyle \) \(=\) \(\displaystyle \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m\) Axiom (N2) of norm (Multiplicativity)
\(\displaystyle \) \(\le\) \(\displaystyle K \norm {y_n - m} + \norm m \norm {x_n - l}\) since $\sequence {x_n}$ is bounded by $K$
\(\displaystyle \) \(=:\) \(\displaystyle z_n\)

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

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

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

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

From the Combined Sum Rule for Real Sequences:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$

By applying the Squeeze Theorem for Sequences of Complex Numbers (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:

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

$\blacksquare$


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