Combination Theorem for Cauchy Sequences/Product Rule

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

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

Then:
 * $\sequence {x_n y_n}$ is a Cauchy sequence.

Proof
Because $\sequence {x_n} $ is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.

Suppose $\norm {x_n} \le K_1$ for $n = 1, 2, 3, \ldots$.

Because $\sequence {y_n} $ is a is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.

Suppose $\norm {y_n} \le K_2$ for $n = 1, 2, 3, \ldots$.

Let $K = \max \set {K_1, K_2}$.

Then both sequences are bounded by $K$.

Let $\epsilon > 0$ be given.

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

Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
 * $\forall n, m > N_1: \norm {x_n - x_m} < \dfrac \epsilon {2K}$

Similarly, $\sequence {y_n} $ is a Cauchy sequence, we can find $N_2$ such that:
 * $\forall n, m > N_2: \norm {y_n - y_m} < \dfrac \epsilon {2K}$

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

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

Thus $\forall n, m > N$:

Hence:
 * $\sequence {x_n y_n}$ is a Cauchy sequence in $R$.