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 $\forall n, m > N$:

\(\displaystyle \norm {\paren {x_n + y_n} - \paren {x_m + y_m} }\) \(=\) \(\displaystyle \norm {\paren {x_n - x_m} + \paren {y_n - y_m} }\)
\(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n - x_m} + \norm {y_n - y_m}\) Axiom (N3) of norm (Triangle Inequality)
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2 = \epsilon\)

Hence:

$\sequence {x_n + y_n}$ is a Cauchy sequences in $R$.

$\blacksquare$


Sources