Properties Of Cauchy Sequences Sum And Multiple Laws

# Properties of Cauchy Sequences - Sum and Multiple Laws

Recall from the Cauchy Sequences of Real Numbers page that a sequence $(a_n)$ of real numbers is said to be Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(1)\begin{align} \quad |a_m - a_n| < \epsilon \end{align}

We will now prove some basic laws regarding the sum of Cauchy sequences and multiples of Cauchy sequences.

Theorem 1: Let $(a_n)$ and $(b_n)$ be Cauchy sequences. Then $(a_n + b_n)$ is a Cauchy sequence. |

**Proof:**Let $\epsilon > 0$ be given.

- Since $(a_n)$ is a Cauchy sequence, for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:

\begin{align} \quad |a_m - a_n| < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}

- Since $(b_n)$ is a Cauchy sequence, for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:

\begin{align} \quad |b_m - b_n| < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}

- Let $N = \max \{ N_1, N_2 \}$. Then if $n \geq N$ we have that both $(*)$ and $(**)$ hold, so:

\begin{align} \quad |(a_m + b_m) - (a_n + b_n)| = |(a_m - a_n) + (b_m - b_n)| \leq |a_m - a_n| + |b_m - b_n| < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $(a_n + b_n)$ is a Cauchy sequence. $\blacksquare$

Theorem 2: Let $(a_n)$ be a Cauchy sequence and let $c \in \mathbb{R}$. Then $(ca_n)$ is a Cauchy sequence. |

**Proof:**Let $\epsilon > 0$ be given.

- If $c = 0$ then $(ca_n) = (0)$ is clearly a Cauchy sequence. So suppose that $c \neq 0$.

- Since $(a_n)$ is a Cauchy sequence we have that for $\epsilon_1 = \frac{\epsilon}{|c|} > 0$ that there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:

\begin{align} \quad |a_m - a_n| < \epsilon_1 = \frac{\epsilon}{|c|} \quad (*) \end{align}

- So let $N = N_1$. Then if $n \geq N$ we have that $(*)$ holds and so:

\begin{align} \quad |ca_m - ca_n| = |c||a_m - a_n| < |c| \cdot \epsilon_1 = |c| \cdot \frac{\epsilon}{|c|} = \epsilon \end{align}

- Therefore $(ca_n)$ is a Cauchy sequence. $\blacksquare$