Combination Theorem for Cauchy Sequences/Sum Rule

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 $\forall n, m > N$:

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

Hence:

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

$\blacksquare$