Combination Theorem for Sequences/Normed Division Ring/Sum Rule

From ProofWiki
Jump to: navigation, search


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:

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


$\sequence {x_n + y_n}$ is convergent and $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$


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 $\forall n > N$:

\(\displaystyle \norm {\paren {x_n + y_n} - \paren {l + m} }\) \(=\) \(\displaystyle \norm {\paren {x_n - l} + \paren {y_n - m} }\) $\quad$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - l} + \norm {y_n - m}\) $\quad$ Axiom (N3) of norm (Triangle Inequality) $\quad$
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2 = \epsilon\) $\quad$ $\quad$


$\sequence {x_n + y_n}$ is convergent to $l + m$.