The Boundedness of Convergent Double Sequences of Real Numbers

Recall from the Boundedness of Double Sequences of Real Numbers page that a double sequence $(a_{mn})_{m,n=1}^{\infty}$ is said to be bounded above if there exists an $M' \in \mathbb{R}$ such that for all $m, n \geq N$ we have that:

(1)

\begin{align} \quad a_{mn} \leq M' \end{align}

We said that $(a_{mn})_{m,n=1}^{\infty}$ is bounded below if there exists an $m' \in \mathbb{R}$ such that for all $m, n \geq N$ we have that:

(2)

\begin{align} \quad m' \leq a_{mn} \end{align}

We said that $(a_{mn})_{m,n=1}^{\infty}$ is bounded if it is both bounded above and bounded below, i.e., there exists an $M^* \in \mathbb{R}$, $M^* > 0$ such that for all $m, n \geq N$ we have that:

(3)

\begin{align} \quad \mid a_{mn} \mid \leq M^* \end{align}

Now recall that if a regular sequence $(a_n)_{n=1}^{\infty}$ is convergent then it is also bounded. We prove an analogous theorem for double sequences below.

Theorem 1: If $(a_{mn})_{m,n=1}^{\infty}$ is a convergent double sequence of real numbers then $(a_{mn})_{m,n=1}^{\infty}$ is bounded.

• Proof: Let $a_{mn})_{m,n=1}^{\infty}$ be a convergent double sequence of real numbers that converges to some $A \in \mathbb{R}$. Then for $\epsilon_0 = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(4)

\begin{align} \quad \mid a_{mn} - A \mid < \epsilon_0 = 1 \end{align}

• So for all $m, n \geq N$ we equivalently have that:

(5)

\begin{align} \quad A - 1 < a_{mn} < A + 1 \end{align}

• Define $M$ as follows:

(6)

\begin{align} \quad M = \max_{1 \leq j \leq N}_{1 \leq k \leq N} \{ a_{jk} \} \end{align}

• So if $M^* = \max \{ \mid M \mid, \mid A \mid + 1 \}$ then $M^* > 0$ and is such that $\mid a_{mn} \mid \leq M^*$ for all $m, n \in \mathbb{N}$. So $(a_{mn})_{m,n=1}^{\infty}$ is bounded. $\blacksquare$