Combination Theorem for Cauchy Sequences/Sum Rule
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Theorem
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Then:
- $\sequence {x_n + y_n}$ is a Cauchy sequence.
Proof
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:
- $\forall n, m > N_1: \norm{x_n - x_m} < \dfrac \epsilon 2$
Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N_2$ such that:
- $\forall n, m > N_2: \norm{y_n - y_m} < \dfrac \epsilon 2$
Now let $N = \max \set {N_1, N_2}$.
Then if $n, m > N$, both the above inequalities will be true.
Thus $\forall n, m > N$:
\(\ds \norm {\paren {x_n + y_n} - \paren {x_m + y_m} }\) | \(=\) | \(\ds \norm {\paren {x_n - x_m} + \paren {y_n - y_m} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - x_m} + \norm {y_n - y_m}\) | Axiom $\text N 3$ of norm: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2 = \epsilon\) |
Hence:
- $\sequence {x_n + y_n}$ is a Cauchy sequences in $R$.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.2$: Completions