Cauchy Convergence Criterion for Double Sequences

# Cauchy Convergence Criterion for Double Sequences

Recall that a normal sequence of real numbers $(a_n)_{n=1}^{\infty}$ 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 \mid a_m - a_n \mid < \epsilon \end{align}

We noted that since $\mathbb{R}$ is a complete metric space that every Cauchy sequence converges in $\mathbb{R}$.

We will now introduce a similar concept to double sequences of real numbers.

 Definition: A double sequence $(a_{mn})_{m,n=1}^{\infty}$ is said to be a Cauchy Double Sequence if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m_1, n_1, m_2, n_2 \geq N$ then $\mid a_{m_1n_1} - a_{m_2n_2} \mid < \epsilon$.

We now show that a double sequence converges in $\mathbb{R}$ if and only if it is a Cauchy double sequence.

 Theorem 1 (Cauchy's Convergence Criterion for Double Sequnces): Let $(a_{mn})_{m, n=1}^{\infty}$ be a double sequence of real numbers. Then $(a_{mn})_{m,n=1}^{\infty}$ converges if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m_1, n_1, m_2, n_2 \geq N$ then $\mid a_{m_1n_1} - a_{m_2n_2} \mid < \epsilon$.
• Proof: $\Rightarrow$ Let $\epsilon > 0$ and suppose that the double sequence $(a_{mn})_{m,n=1}^{\infty}$ converges, say to $A \in \mathbb{R}$. Then for $\frac{\epsilon_1} = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m_1, n_1, m_2, n_2 \geq N^*$ then both:
(2)
\begin{align} \quad \mid a_{m_1n_1} - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
(3)
\begin{align} \quad \mid a_{m_2n_2} - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (**) \end{align}
• So let $N = N^*$. Then for all $m_1, n_1, m_2, n_2 \geq N$ we have that $(*)$ and $(**)$ both hold and so:
(4)
\begin{align} \quad \mid a_{m_1n_1} - a_{m_2n_2} \mid =\mid a_{m_1n_1} - A + A - a_{m_2n_2} \mid \leq \mid a_{m_1n_1} - A \mid + \mid A - a_{m_2n_2} \mid < \epsilon_1 + \epsilon_1 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• So Cauchy's condition is satisfied.
• $\Rightarrow$ Now suppose that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m_1, n_1, m_2, n_2 \geq N$ then:
(5)
\begin{align} \quad \mid a_{m_1n_1} - a_{m_2n_2} \mid < \epsilon \end{align}
• Then for $\epsilon_1 = \frac{\epsilon}{2}$ there exists an $N_1 \in \mathbb{N}$ such that if $m_1, n_1, m_2, n_2 \geq N_1$ then:
(6)
\begin{align} \quad \mid a_{m_1n_1} - a_{m_2n_2} \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (***) \end{align}
• Now note that if we let $m_1 = n_1 = m$ and $m_2 = n_2 = n$ then for all $m, n \geq N$ we have that that the single sequence $a_{nn} = b_n$ is such that $\mid b_m - b_n \mid < \epsilon$ for all $m, n \geq N$. So $(b_n)_{n=1}^{\infty}$ is a Cauchy sequence and since $\mathbb{R}$ is a complete metric space, $(b_n)_{n=1}^{\infty}$ converges to some $A \in \mathbb{R}$.
• So, for $\epsilon_2 = \frac{\epsilon}{2}$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:
(7)
\begin{align} \quad \mid b_n - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (****) \end{align}
• Let $N = \max \{ N_1, N_2 \}$. Then if $m, n \geq N$ we have that both $(***)$ and $(****)$ hold, and so:
(8)
\begin{align} \quad \quad \mid a_{mn} - A \mid = \mid a_{mn} - a_{nn} + a_{nn} - A \mid \leq \mid a_{mn} - a_{nn} \mid + \mid a_{nn} - A \mid = \mid a_{mn} - b_n \mid + \mid b_n - A \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore the double sequence $(a_{mn})_{m,n=1}^{\infty}$ converges. $\blacksquare$