Product of Sequence Converges to Zero with Cauchy Sequence Converges to Zero

Theorem

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero $0$.

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

Let $\sequence {x_n}$ converge to $0$.

Let $\sequence {y_n}$ be a Cauchy sequence.

Then:

$\sequence {x_n y_n}$ and $\sequence {y_n x_n}$ converge to $0$.

$\exists M \in \R_{>0}: \forall n, \norm {x_n} \le M$

Given $\epsilon > 0$.

Since $\sequence {x_n}$ converges to $0$ then:

$\exists N \in \N: \forall n > N, \norm {x_n} < \dfrac \epsilon M$

Hence:

$\ds \norm {x_n y_n - 0}$

$=$

$\ds \norm {x_n y_n}$

$\ds $

$=$

$\ds \norm {x_n} \norm {y_n}$

$\ds $

$<$

$\ds \norm {x_n} M$

$\sequence {y_n}$ is bounded by $M$

$\ds $

$<$

$\ds \paren { \dfrac \epsilon M } M$

$\sequence {x_n}$ converges to $0$

$\ds $

$=$

$\ds \epsilon$

Similarly, $\norm {y_n x_n - 0} < \epsilon$

The result follows by definition of convergence in normed division rings.

$\blacksquare$