Uniqueness of Double Limits of Double Sequences of Real Numbers

# Uniqueness of Double Limits of Double Sequences of Real Numbers

Recall from the Double Limits and Iterated Limits of Double Sequences of Real Numbers page that a double sequence $(a_{mn})_{m,n=1}^{\infty}$ is said to converge to a limit $A \in \mathbb{R}$ 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_{mn} - A \mid < \epsilon \end{align}

If such a number $A \in \mathbb{R}$ exists we say that $A$ is the limit of the double sequence $(a_{mn})_{m,n=1}^{\infty}$.

Like with limits of regular sequences of real numbers, if a double sequences of real numbers converges, then the limit is unique as we prove below.

 Theorem 1: If $(a_{mn})_{m,n=1}^{\infty}$ is a double sequence that converges to both $A$ and $B$, $A, B \in \mathbb{R}$, then $A = B$.
• Proof: Suppose that $(a_{mn})_{m,n=1}^{\infty}$ converges to both $A$ and $B$. Let $\epsilon > 0$ be given. Then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m, n \geq N_1$ then:
(2)
\begin{align} \quad \mid a_{mn} - A \mid < \epsilon_1 = \frac{\epsilon}{2} \quad (*) \end{align}
• Similarly, for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $m, n \geq N_2$ then:
(3)
\begin{align} \quad \mid a_{mn} - B \mid < \epsilon_2 = \frac{\epsilon}{2} \quad (**) \end{align}
• Let $N = \max \{N_1, N_2 \}$. Then for all $m, n \geq N$ we have that both $(*)$ and $(**)$ hold so:
(4)
\begin{align} \quad \: \mid A - B \mid = \mid A - a_{mn} + a_{mn} - B \mid \leq \mid A - a_{mn} \mid + \mid a_{mn} - B \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• So $\mid A - B \mid < \epsilon$ for all $\epsilon > 0$ which implies that $\mid A - B \mid = 0$, i.e., $A - B = 0$. Thus $A = B$ and the limit of a convergent double sequence of real numbers is unique. $\blacksquare$