Combination Theorem for Cauchy Sequences/Inverse Rule

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

Let $\sequence {x_n}$ be a Cauchy sequences in $R$. Suppose $\sequence {x_n}$ does not converge to $0$.

Then:
 * $\exists K \in \N : \forall n > K : x_n \ne 0$

and the sequence
 * $\sequence { \paren {x_{K+n}}^{-1} }_{n \in \N}$ is well-defined and a Cauchy sequence.

Proof
Since $\sequence {x_n}$ does not converge to $0$, by Cauchy Sequence Is Eventually Bounded Away From Zero then:
 * $\exists K \in \N$ and $C \in \R_{\gt 0}: \forall n \gt K: C \lt \norm {x_n}$

or equivalently:
 * $\exists K \in \N$ and $C \in \R_{\gt 0}: \forall n \gt K: 1 \lt \dfrac {\norm {x_n}} C$

By Axiom (N1) of norm (Positive definiteness) $\forall n > K : x_n \ne 0$.

Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ where $y_n = x_{K+n}$.

By Subsequence of a Cauchy Sequence is a Cauchy Sequence then $\sequence {y_n} $ is a Cauchy sequence.

So $\sequence { {y_n}^{-1} } $ is well-defined and $\sequence { {y_n}^{-1} } = \sequence { \paren {x_{K+n}}^{-1} }_{n \in \N}$.

Let $\epsilon > 0$ be given.

Let $\epsilon' = \epsilon C^2$, then $ \epsilon' > 0$.

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

Thus $\forall n, m > N$:
 * $ 1 \lt \dfrac {\norm {y_n}} C,\dfrac {\norm {y_m}} C$
 * $ \norm {y_n - y_m} < \epsilon'$

Hence:

So:
 * $\sequence { { {y_n}^{-1} } }$ is a Cauchy sequence in $R$.