Combination Theorem for Cauchy Sequences/Product Rule

Theorem
Let $\left({R, \left\Vert{\,\cdot\,}\right\Vert_R}\right)$ be a normed division ring.

Let $\left({V, \left\Vert{\,\cdot\,}\right\Vert}\right)$ be a normed vector space over $R$.

Let $\left \langle {x_n} \right \rangle$ be a Cauchy sequence in $V$.

Let $\left \langle {\lambda_n} \right \rangle$ be a Cauchy sequence in $R$.

Then:
 * $\left \langle{\lambda_n x_n}\right \rangle$ is a Cauchy sequence in $V$.

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

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

Because $\left \langle {\lambda_n} \right \rangle$ is a is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.

Suppose $\left\Vert{\lambda_n}\right\Vert_R \le K_2$ for $n = 1, 2, 3, \ldots$.

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

Then both sequences are bounded by $K$.

Let $\epsilon > 0$ be given.

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

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

Similarly, $\left \langle {\lambda_n} \right \rangle$ is a Cauchy sequence, we can find $N_2$ such that:
 * $\forall n, m > N_2: \left\Vert{\lambda_n - \lambda_m}\right\Vert < \dfrac \epsilon {2K}$

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

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

Thus $\forall n, m > N$:

Hence:
 * $\left \langle{\lambda_n x_n}\right \rangle$ is a Cauchy sequence in $V$.