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:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {y_n}$ be a subsequence of $\sequence {x_n}$.
Suppose $l \ne 0$, and for all $n: y_n \ne 0$.
Then:
- $\ds \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$
Proof
By Limit of Subsequence equals Limit of Sequence then $\sequence {y_n}$ is convergent with:
- $\ds \lim_{n \mathop \to \infty} y_n = l$
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:
| \(\ds \norm { {y_n }^{-1} - l^{-1} }\) | \(<\) | \(\ds \dfrac {2 \norm {y_n} } {\norm l} \norm { {y_n}^{-1} - l^{-1} }\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\norm l^2} \paren {\norm {y_n} \norm { {y_n}^{-1} - l^{-1} } \norm l}\) | multiplying and dividing by $\norm l$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\norm l^2} \norm {y_n \paren { {y_n}^{-1} - l^{-1} } l}\) | Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\norm l^2} \norm {y_n y_n^{-1} l - y_n l^{-1} l}\) | Ring Axiom $\text D$: Distributivity of Product over Addition | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\norm l^2} \norm {l - y_n}\) | Inverse Property of Division Ring | |||||||||||
| \(\ds \) | \(<\) | \(\ds \dfrac 2 {\norm l^2} \epsilon'\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\norm l^2} \paren {\dfrac {\epsilon \norm l^2 } 2}\) | Definition of $\epsilon'$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) | cancelling terms |
Hence:
- $\sequence { {y_n}^{-1} }$ is convergent with $\ds \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$.
$\blacksquare$