# Bounded Sets

Consider the set $\mathbb{N} = \{ 1, 2, 3, ... \}$. As you might imagine, there is no "largest" element in this set. If we were to claim that some element $u \in \mathbb{N}$ was the largest element in $\mathbb{N}$, then we note that $u + 1 \in \mathbb{N}$ and $u < u + 1$. There is a least element in this set though, namely $1$. We can say that every element $n \in \mathbb{N}$ is such that $1 \leq n$. We could alternatively say that every element $n \in \mathbb{N}$ is such that $0 \leq n$ or $-3234 \leq n$. We will now formally define this sort of idea.

Definition: Let $S \subseteq \mathbb{R}$ where $S \neq \emptyset$. We say that $S$ is bounded above by $M$ if for all $x \in \mathbb{S}$, $x \leq M$. We say that $S$ is bounded below by $m$ if for all $x \in \mathbb{S}$, $m \leq x$. We say that $S$ is bounded if it is both bounded above and bounded below, and we say that $S$ is unbounded if it is not both bounded above and below. |

From this definition, we can intuitively say that the set of natural numbers $\mathbb{N}$ is bounded below by any number $m \leq 1$. However, we note that $\mathbb{N}$ is not bounded above since such as $M$ does not exist. Therefore in general we say that the set of natural numbers is not bounded.

Another example is the set $S := \{ x \in \mathbb{R} : -2 \leq x \leq 2 \}$. As we can guess, this set $S$ is bounded below by $m = -2$ and bounded above by $M = 2$. Therefore we say that $S$ is bounded.

Sometimes a set might not be bounded above and might also not be bounded below. For example consider the set of integers $\mathbb{Z}$ which is clearly a subset of $\mathbb{R}$. This set is not bounded below and not bounded above.

We will now look at some theorems regarding bounded sets.

Theorem 1: If $A \subseteq B$ and $B$ is a bounded set then $A$ is a bounded set. Further, if $A \subseteq B$ and $A$ is an unbounded set then $B$ is an unbounded set. |

**Proof:**Let $A \subseteq B$.

- For the first part of the proof, let $B$ be a bounded set and let $m$ be any lower bound to $B$ and let $M$ be any upper bound to $B$. Then if $x \in A$, then $x \in B$, but $\forall x \in B$ we have that $m ≤ x ≤ M$, and so $\forall x \in A$, $m ≤ x ≤ M$ so $A$ is a bounded set.

- For the second part of the proof, let $A$ be an unbounded set and suppose instead that $B$ is a bounded set. Let $m$ be any lower bound to $B$ and let $M$ be any upper bound to $B$. Then if $x \in A$, then $x \in B$, but $\forall x \in B$ we have that $m ≤ x ≤ M$, and so $\forall x \in A$, $m ≤ x ≤ M$, so $A$ is bounded, but that contradicts the fact that $A$ is an unbounded set, so our assumption that $B$ was bounded is false. Therefore $B$ is an unbounded set. $\blacksquare$

Note that in the proof of Theorem 1, we could have omitted the second part of the proof since it is the contrapositive of the first part.