Boundedness of Double Sequences of Real Numbers

# Boundedness of Double Sequences of Real Numbers

Recall that a sequence of real numbers $(a_n)_{n=1}^{\infty}$ is said to be bounded above if there exists an $M \in \mathbb{R}$ such that for all $n \in \mathbb{N}$ we have that:

(1)
\begin{align} \quad a_n \leq M \end{align}

Similarly, a sequence of real numbers $(a_n)_{n=1}^{\infty}$ is said to be bounded below if there exists an $m \in \mathbb{R}$ such that for all $n \in \mathbb{N}$ we have that:

(2)
\begin{align} \quad m \leq a_n \end{align}

More generally, $(a_n)_{n=1}^{\infty}$ is said to be bounded if there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ we have that:

(3)
\begin{align} \quad \mid a_n \mid \leq M \end{align}

We will now give analogous definitions in defining a bounded double sequence of real numbers.

 Definition: A double sequence of real numbers, $(a_{mn})_{m,n=1}^{\infty}$ is said to be Bounded Above if there exists an $M' \in \mathbb{R}$ (called an Upper Bound) such that $a_{mn} \leq M'$ for all $m, n \in \mathbb{N}$. Similarly, this double sequence is said to be Bounded Below if there exists an $m' \in \mathbb{R}$ (called a Lower Bound) such that $m' \leq a_{mn}$ for all $m, n \in \mathbb{N}$. We said that this double sequence is Bounded if there exists an $M^* \in \mathbb{R}$, $M^* > 0$ such that $\mid a_{mn} \mid \leq M'$ for all $m, n \in \mathbb{N}$. If a double sequence is not bounded we say that it is **Unbounded.

For example, consider the following double sequence of real numbers:

(4)
\begin{align} \quad \left ( a_{mn} \right )_{m,n=1}^{\infty} = \left ( \frac{1}{m + n} \right )_{m, n= 1}^{\infty} \end{align}

We claim that this double sequence is bounded above by $\frac{1}{2}$ and below by $0$. To show this, we first note that for all $m, n \in \mathbb{N}$ that since $m, n \geq 1$ that $m + n \geq 2$. So for all $m, n \in \mathbb{N}$:

(5)
\begin{align} \quad \frac{1}{m + n} \leq \frac{1}{2} \end{align}

So $\frac{1}{2}$ is an upper bound for this double sequence. Moreover, since $m, n \geq 1 > 0$ we see that $m + n > 0$ and so for all $m, n \in \mathbb{N}$:

(6)
\begin{align} \quad 0 < \frac{1}{m + n} \end{align}

Therefore $0$ is a lower bound for this double sequence.

Since this double sequence has both an upper and a lower bound we conclude that $(a_{mn})_{m,n=1}^{\infty}$ is bounded.

For another example, consider the following double sequence:

(7)
\begin{align} \quad (a_{mn})_{m,n=1}^{\infty} = m + n \end{align}

Then it is not hard to see that this sequence is not bounded above (but it is bounded below by $2$)!