# Bounded Subsets in Euclidean Space

Definition: Let $S \subseteq \mathbb{R}^n$. The set $S$ is said to be Bounded if there exists a $\mathbf{x} \in \mathbb{R}^n$ and a positive real number $r > 0$ such that $S \subseteq B(\mathbf{x}, r)$ and $S$ is said to be Unbounded otherwise. |

*In other words, a subset $S$ of $\mathbb{R}^n$ is bounded if $S$ is the subset of some ball in $\mathbb{R}^n$.*

In $\mathbb{R}^2$, a set $S \subseteq \mathbb{R}^2$ that is bounded might look something like:

Meanwhile, a set $S$ that is unbounded might look something like:

For a more concrete example, consider the subset $S = [0, 1) \times [0, 1) \subseteq \mathbb{R}^2$. Then the ball $B(\mathbf{0}, 2)$ is such that $S \subseteq B(\mathbf{0}, 2)$, so $S$ is bounded. However, the subset $S = \{ (x, y) \in \mathbb{R}^2 : x \geq 0, y \geq 0 \} \subseteq \mathbb{R}^2$ is unbounded. To prove this, set $\mathbf{x} = \mathbf{0}$. Suppose that there exists an $r > 0$ such that $S \subseteq B(\mathbf{0}, r)$. now consider the point $\mathbf{y} = (r+1, r+1) \in S$. Then:

(1)Therefore $(r + 1, r + 1) \not \in B(\mathbf{0}, r)$ for all $r > 0$, but $(r+1, r+1) \in S$ for all $r > 0$, so $S \not \subseteq B(\mathbf{0}, r)$ for all $r > 0$, so $S$ is unbounded.