The Squeeze Theorem for Double Sequences of Real Numbers

The Squeeze Theorem for Double Sequences of Real Numbers

Recall that if $(a_n)_{n=1}^{\infty}$, $(b_n)_{n=1}^{\infty}$, and $(c_n)_{n=1}^{\infty}$ are sequences of real numbers that satisfy $a_n \leq b_n \leq c_n$ ultimately, i.e., there exists an $M \in \mathbb{N}$ such that this inequality chain holds if $n \geq N$, and if further, $(a_n)_{n=1}^{\infty}$ and $(c_n)_{n=1}^{\infty}$ converge to the same limit, $L$, then $(b_n)_{n=1}^{\infty}$ also converges to $L$.

This theorem was called the Squeeze theorem and there are many variants of it. Fortunately, such a theorem exists for double sequences of real number which we state and prove below.

 Theorem 1 (The Squeeze Theorem for Double Sequences of Real Numbers): Let $(a_{mn})_{m,n=1}^{\infty}$, $(b_{mn})_{m,n=1}^{\infty}$, and $(c_{mn})_{m,n=1}^{\infty}$ be double sequences of real numbers. Suppose that there exists an $M \in \mathbb{N}$ such that for all $m, n \geq M$ we have that $\displaystyle{a_{mn} \leq b_{mn} \leq c_{mn}}$, and suppose that $(a_{mn})_{m,n=1}^{\infty}$ and $(c_{mn})_{m,n=1}^{\infty}$ converge to $L \in \mathbb{R}$. Then $(b_{mn})_{m,n=1}^{\infty}$ converges to $L$.
• Proof: Suppose that there exists an $M \in \mathbb{N}$ such that for all $m, n \geq M$ we have that:
(1)
• Now, since $(a_{mn})_{m,n=1}^{\infty}$ converges to $L$ then for $\epsilon > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m, n \geq N_1$ then:
(2)
\begin{align} \quad \mid a_{mn} - L \mid < \epsilon \quad \Leftrightarrow \quad L - \epsilon \leq a_{mn} \leq L + \epsilon \quad (**) \end{align}
• Similarly, since $(c_{mn})_{m,n=1}^{\infty}$ converges to $L$ then for $\epsilon > 0$ there exists an $N_2 \in \mathbb{N}$ such that if $m, n \geq N_2$ then:
(3)
\begin{align} \quad \mid c_{mn} - L \mid < \epsilon \quad \Leftrightarrow \quad L - \epsilon \leq c_{mn} \leq L + \epsilon \quad (***) \end{align}
• Let $N = \max \{ M, N_1, N_2 \}$ so that for all $m, n \geq N$ we have that $(*)$, $(**)$, and $(***)$ hold. Then we have that:
(4)
\begin{align} \quad L - \epsilon < a_{mn} \leq b_{mn} \leq c_{mn} < L + \epsilon \end{align}
• But then this implies that for all $m, n \geq N$ that:
(5)
\begin{align} \quad L - \epsilon < b_{mn} < L + \epsilon \quad \Leftrightarrow \mid b_{mn} - L \mid < \epsilon \end{align}
• Therefore $(b_{mn})_{m,n=1}^{\infty}$ converges to $L$. $\blacksquare$