Combination Theorem for Sequences/Normed Division Ring/Sum Rule
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Theorem
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
- $\ds \lim_{n \mathop \to \infty} y_n = m$
Then:
- $\sequence {x_n + y_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$
Proof
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
Since $\sequence {x_n}$ is convergent to $l$, we can find $N_1$ such that:
- $\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon 2$
Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:
- $\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon 2$
Now let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true.
Thus $\forall n > N$:
\(\ds \norm {\paren {x_n + y_n} - \paren {l + m} }\) | \(=\) | \(\ds \norm {\paren {x_n - l} + \paren {y_n - m} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - l} + \norm {y_n - m}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2 = \epsilon\) |
Hence:
- $\sequence {x_n + y_n}$ is convergent to $l + m$.
$\blacksquare$