# 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:

$\forall n, m > N_1: \norm {x_n - x_m} < \dfrac \epsilon {2 K}$

Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N_2$ such that:

$\forall n, m > N_2: \norm {y_n - y_m} < \dfrac \epsilon {2 K}$

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

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

Thus $\forall n, m > N$:

 $\ds \norm {x_n y_n - x_m y_m}$ $=$ $\ds \norm {x_n y_n - x_n y_m + x_n y_m - x_m y_m}$ $\ds$ $\le$ $\ds \norm {x_n y_n - x_n y_m} + \norm {x_n y_m - x_m y_m}$ Norm Axiom $\text N 3$: Triangle Inequality $\ds$ $=$ $\ds \norm {x_n \paren {y_n - y_m } } + \norm {\paren {x_n - x_m} y_m}$ $\ds$ $=$ $\ds \norm {x_n} \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot \norm {y_m}$ Norm Axiom $\text N 2$: Multiplicativity $\ds$ $\le$ $\ds K \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot K$ as both sequences are bounded by $K$ $\ds$ $\le$ $\ds K \cdot \dfrac \epsilon {2K} + \dfrac \epsilon {2K} \cdot K$ $\ds$ $=$ $\ds \dfrac \epsilon 2 + \dfrac \epsilon 2$ $\ds$ $=$ $\ds \epsilon$

Hence:

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

$\blacksquare$