Bounded Sets in a Metric Space

# Bounded Sets in a Metric Space

We will now extend the concept of boundedness to sets in a metric space.

 Definition: Let $(M, d)$ be a metric space. A subset $S \subseteq M$ is said to be Bounded if there exists a positive real number $r > 0$ such that $S \subseteq B(x, r)$ for some $x \in M$. The set $S$ is said to be Unbounded if it is not bounded.

By the definition above, we see that $S$ is bounded if there exists some open ball with a finite radius that contains $S$.

For example, consider the metric space $(M, d)$ where $d$ is the discrete metric defined for all $x, y \in M$ by:

(1)
\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \: x = y\\ 1 & \mathrm{if} \: x \neq y \end{matrix}\right. \end{align}

Let $S \subseteq M$. Then by the definition of the discrete metric, for all $x, y \in S$ we have that $d(x, y) \leq 1$. Therefore, if we consider any point $x \in S$ and take $r = 2$ then:

(2)
\begin{align} \quad S \subseteq B(x, 2) \end{align}

Therefore, $S$ is bounded. This shows that every subset of $M$ is bounded with respect to the discrete metric. In fact, the wholeset $M$ is also bounded and $M \subseteq B(x, 2)$ for any $x \in M$.

For another example, consider the metric space $(\mathbb{R}^3, d)$ where $d$ is the Euclidean metric. Consider the following set:

(3)
\begin{align} \quad S = \{ (x, y, z) \in \mathbb{R}^3 : x, y, z \geq 0 \} \subseteq \mathbb{R}^3 \end{align}

The set $S$ above is the first octant of $\mathbb{R}^3$, and is actually unbounded. To prove this, suppose that instead $S$ is bounded. Then there exists a maximal distance between some pair of points $\mathbf{x}, \mathbf{y} \in S$, say:

(4)
\begin{align} \quad d_{\mathrm{max}} = d(\mathbf{x}, \mathbf{y}) \end{align}

Then $S \subseteq B(x, d_{\mathrm{max}})$. For $k > 1$ consider the point $k\mathbf{y} = (kx_1, kx_2, kx_3) \in \mathbb{R}^3$. Then:

(5)
\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) = \sqrt{(x_1 - ky_1)^2 + (x_2 - ky_2)^2 + (x_3 - ky_3)^2} \end{align}

Since $(x_i - ky_1)^2 \geq (x_i - y_i)^2$ for each $i \in \{1, 2, 3 \}$ we see that:

(6)
\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) \geq d(\mathbf{x}, \mathbf{y}) = d_{\mathrm{max}} \end{align}

But then $\mathbf{y} \not \in B(\mathbf{x}, d_{\mathrm{max}})$ implies $\mathbf{y} \not \in S$ which is a contradiction. Therefore our assumption that $S$ was bounded is false.