Combination Theorem for Sequences/Normed Division Ring/Inverse Rule/Lemma

Theorem

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

Let $\sequence {x_n}$ be a sequence in $R$.

Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$

Let $\sequence {y_n}$ be a subsequence of $\sequence {x_n}$.

Suppose $l \ne 0$, and for all $n: y_n \ne 0$.

Then:

$\displaystyle \lim_{n \mathop \to \infty} y_n^{-1} = l^{-1}$

Proof

By Limit of Subsequence equals Limit of Sequence then $\sequence {y_n}$ is convergent with:

$\displaystyle \lim_{n \mathop \to \infty} y_n = l$

Let $\epsilon > 0$ be given.

Let $\epsilon' = \dfrac {\epsilon {\norm l}^2 } {2}$.

Then:

$\epsilon' > 0$

As $\sequence {y_n} \to l$, as $n \to \infty$, we can find $N_1$ such that:

$\forall n > N_1: \norm {y_n - l} < \epsilon'$

As $\sequence {y_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit:

$\exists N_2 \in \N: \forall n > \N_2: \dfrac {\norm l} 2 < \norm {y_n}$

or equivalently:

$\exists N_2 \in \N: \forall n > \N_2: 1 < \dfrac {2 \norm {y_n} } {\norm l}$

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

Then $\forall n > N$:

$(1): \quad \norm {y_n - l} < \epsilon'$
$(2): \quad 1 < \dfrac {2 \norm {y_n} } {\norm l}$

Hence:

 $\displaystyle \norm { {y_n }^{-1} - l^{-1} }$ $<$ $\displaystyle \dfrac {2 \norm {y_n} } {\norm l} \norm { {y_n}^{-1} - l^{-1} }$ from $(1)$ $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 {\norm l^2} \paren {\norm { {y_n } } \norm { {y_n}^{-1} - l^{-1} } \norm l}$ multiplying and dividing by $\norm l$ $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 {\norm l^2} \norm {y_n \paren { {y_n}^{-1} - l^{-1} } l}$ Axiom (N2) of Norm: Multiplicativity $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 {\norm l^2} \norm {y_n y_n^{-1} l - y_n l^{-1} l}$ Axiom (D) of Ring: Distributivity $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 {\norm l^2} \norm {l - y_n}$ Inverse Property of Division Ring $\displaystyle$  $\displaystyle$ $\displaystyle$ $<$ $\displaystyle \dfrac 2 {\norm l^2} \epsilon'$ from $(2)$ $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac {2} {\norm l^2} \paren { \dfrac {\epsilon \norm l^2 } 2}$ Definition of $\epsilon'$ $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \epsilon$ cancelling terms

Hence:

$\sequence { {y_n}^{-1} }$ is convergent with $\displaystyle \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$.

$\blacksquare$