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 $\epsilon' > 0$.

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

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

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

$\exists N_2 \in \N: \forall n \gt \N_2: \dfrac {\norm l} 2 \lt \norm {y_n}$.

or equivalently:

$\exists N_2 \in \N: \forall n \gt \N_2: 1 \lt \dfrac {2 \norm {y_n} } {\norm l }$.

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

Then $\forall n > N$:

1. $\quad \norm {y_n - l} < \epsilon'$.
2. $\quad 1 \lt \dfrac {2 \norm {y_n} } {\norm l }$.

Hence:

 $\displaystyle \norm { {y_n }^{-1 } - l^{-1 } }$ $\lt$ $\displaystyle \dfrac {2 \norm {y_n} } {\norm l } \norm { {y_n}^{-1} - l^{-1} }$ Point 1. above. $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 { \norm {l }^2 } \paren { \norm { {y_n } } \norm { {y_n}^{-1} - l^{-1} } \norm { l } }$ Multiply and divide 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). $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 { \norm {l }^2 } \norm { y_n y_n^{-1} l - y_n l^{-1} l }$ Axion (D) of a ring (Distributive). $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 { \norm {l }^2 } \norm { l - y_n }$ Inverse property of a division ring $\displaystyle$  $\displaystyle$ $\displaystyle$ $\lt$ $\displaystyle \dfrac { 2 } { \norm { l }^2 } \epsilon'$ Point 2. above $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \dfrac {2} { {\norm l}^2 } \paren { \dfrac { \epsilon {\norm l}^2 } {2} }$ Definition of $\epsilon'$ $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \epsilon$ Cancelling terms.

Hence:

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

$\blacksquare$