Additional Cauchy Sequence Proofs

# Additional Cauchy Sequence Proofs

Recall the following definition of a Cauchy sequence from the Cauchy Sequences page:

Definition: A sequence $(a_n)$ is called a Cauchy Sequence if $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\forall m, n ≥ N$, then $\mid x_n - x_m \mid < \epsilon$. A sequence is not Cauchy if $\exists \epsilon_0 > 0$ such that $\forall N \in \mathbb{N}$ there exists at least one $m$ and one $n$ where $m, n > N$ such that $\mid x_n - x_m \mid ≥ \epsilon_0$. |

We will now look at some rather trivial proofs regarding Cauchy sequences.

Theorem 1: If $(a_n)$ and $(b_n)$ are Cauchy sequences then $(a_n + b_n)$ is a Cauchy sequence. |

**Proof:**Let $(a_n)$ and $(b_n)$ be Cauchy sequences. We want to show that $(a_n + b_n)$ is then a Cauchy sequence, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid (a_n + b_n) - (a_m + b_m) \mid < \epsilon$.

- Now since $(a_n)$ is a Cauchy sequence, we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N _1\in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \epsilon_1 = \frac{\epsilon}{2}$.

- Similarly since $(b_n)$ is a Cauchy sequence, we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $m, n ≥ N_2$ then $\mid b_n - b_m \mid < \epsilon_2 = \frac{\epsilon}{2}$.

- Now let $N = \mathrm{max} \{ N_1, N_2 \}$, and so for $m, n ≥ N$ we have that:

\begin{align} \quad \quad \mid (a_n + b_n) - (a_m + b_m) \mid = \mid a_n - a_m + b_n - b_m \mid ≤ \mid a_n - a_m \mid + \mid b_n - b_m \mid < \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: If $(a_n)$ and $(b_n)$ are Cauchy sequences then $(a_n - b_n)$ is a Cauchy sequence. |

**Proof:**Let $(a_n)$ and $(b_n)$ be Cauchy sequences. We want to show that $(a_n - b_n)$ is then a Cauchy sequence, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid (a_n - b_n) - (a_m - b_m) \mid < \epsilon$.

- Now since $(a_n)$ is a Cauchy sequence, we have that for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N _1\in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \epsilon_1 = \frac{\epsilon}{2}$.

- Similarly since $(b_n)$ is a Cauchy sequence, we have that for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $m, n ≥ N_2$ then $\mid b_n - b_m \mid < \epsilon_2 = \frac{\epsilon}{2}$.

- Now let $N = \mathrm{max} \{ N_1, N_2 \}$, and so for $m, n ≥ N$ we have that:

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

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

Theorem 3: If $(a_n)$ is a Cauchy sequence that for any $k \in \mathbb{R}$, $(ka_n)$ is also a Cauchy sequence. |

**Proof:**Let $(a_n)$ be a Cauchy sequence and let $k = 0$. Then the sequence $(ka_n) = (0)$ which is clearly Cauchy. Suppose that $k \in \mathbb{R}$ but $k \neq 0$. We want to show that the sequence $(ka_n)$ is also a Cauchy sequence, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid ka_n - ka_m \mid < \epsilon$.

- Since $(a_n )$ is a Cauchy sequence, then for $\epsilon_1 = \frac{\epsilon}{\mid k \mid} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \epsilon_1 = \frac{\epsilon}{\mid k \mid}$.

- Now choose $N = N_1$, and so for $m, n ≥ N$ we have that:

\begin{align} \quad \mid ka_n - ka_m \mid = \mid k \mid \mid a_n - a_m \mid < \mid k \mid \epsilon_1 = \mid k \mid \frac{\epsilon}{\mid k \mid} = \epsilon \end{align}

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

Theorem 4: If $(a_n)$ and $(b_n)$ are Cauchy sequences then the sequence $(a_nb_n)$ is also Cauchy. |

**Proof:**Let $(a_n)$ and $(b_n)$ be Cauchy sequences. We want to show that $(a_nb_n)$ is also a Cauchy sequence, that is $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid a_nb_n - a_mb_m \mid < \epsilon$. Doing some algebraic manipulation we get that:

\begin{align} \quad \quad \mid a_nb_n - a_mb_m \mid = \mid a_nb_n - a_mb_n + a_mb_n - a_mb_m \mid = \mid b_n(a_n - a_m) + a_m(b_n - b_m) \mid < \mid b_n \mid \mid a_n - a_m \mid + \mid a_m \mid \mid b_n - b_m \mid \end{align}

- Now recall that every Cauchy sequence is bounded. Since $(a_n)$ and $(b_n)$ are Cauchy sequences it follows that for $0 < M_1, M_2 \in \mathbb{R}$ that $\mid a_n \mid < M_1$ and $\mid b_n \mid < M_2$ for all $n \in \mathbb{N}$ and so:

\begin{align} \quad \quad \quad \mid a_nb_n - a_mb_m \mid = ... < \mid b_n \mid \mid a_n - a_m \mid + \mid a_m \mid \mid b_n - b_m \mid < M_2 \mid a_n - a_m \mid + M_1 \mid b_n - b_m \mid \end{align}

- Also, since $(a_n)$ is Cauchy there exists an $N_1 \in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \frac{\epsilon}{2M_2}$.

- Similarly since $(b_n)$ is Cauchy there exists an $N_2 \in \mathbb{N}$ such that if $m, n ≥ N_2$ then $\mid b_n - b_m \mid < \frac{\epsilon}{2M_1}$.

- Choose $N = \mathrm{max} \{ N_1, N_2 \}$ and so for $m, n ≥ N$:

\begin{align} \quad \quad \mid a_nb_n - a_mb_m \mid = ... < M_2 \mid a_n - a_m \mid + M_1 \mid b_n - b_m \mid < M_2 \frac{\epsilon}{2M_2} + M_1 \frac{\epsilon}{2M_1} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

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