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:
 * $\sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ 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 Non-Limit then:
 * $\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C < \norm {x_n}$

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

By :
 * $\forall n > K : x_n \ne 0$

Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ defined as:
 * $y_n = x_{K + n}$

By Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
 * $\sequence {y_n}$ is a Cauchy sequence.

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

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 $\forall n, m > N$:
 * $(1): \quad 1 < \dfrac {\norm {y_n} } C, \dfrac {\norm {y_m} } C$
 * $(2): \quad \norm {y_n - y_m} < \epsilon'$

Hence:

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