Properties of The Supremum and Infimum of a Bounded Set

We will now begin to look at some very important properties of the supremum and infimum of a nonempty bounded set.

Theorem 1: If $S$ is a bounded set then any subset $T \subseteq S$ is also a bounded set.

• Proof: Suppose that $S$ is a nonempty bounded set. Then there exists real numbers $m, M \in \mathbb{R}$ where $m ≤ M$ such that $m ≤ x ≤ M$ for all $x \in S$.

• Let $T \subseteq S$ be a nonempty set. Then by transitivity $m ≤ x ≤ M$ for all $x \in T$, and so $T$ is bounded as well.

Theorem 2: Let $S$ be a nonempty bounded subset of the real numbers. If $T \subseteq S$, then $\sup T ≤ \sup S$.

• Proof: Let $S$ be a nonempty subset of the real numbers and let $u_1 = \sup S$. Then for all $x \in S$ we have that $x ≤ u_1$. Now let $T \subseteq S$ be nonempty. Since $S$ is bounded, any subset $T$ of $S$ must also be bounded by theorem 1, and so $u_2 = \sup T$ exists.

• Now suppose that $\sup T > \sup S$, that is $u_2 > u_1$. By the definition that $u_2$ is the least upper bound of the set $T$, we have that since $u_1 < u_2$, then there exists an element $a \in T$ such that $u_1 < a$. But then if $a \in T$, then since $T \subseteq S$ we have that $a \in S$, and thus since $u_1 < a$ we have a contradiction since then $u_1$ isn't the supremum of $S$. Therefore $\sup T ≤ \sup S$. $\blacksquare$

Theorem 3: Let $S$ be a nonempty bounded subset of the real numbers. If $T \subseteq S$ then $\inf S ≤ \inf T$.

• Proof: Let $S$ be a nonempty subset of the real numbers and let $w_1 = \inf S$. Then for all $x \in S$ we have that $w_1 ≤ x$. Now let $T \subseteq S$ be nonempty. Since $S$ is bounded, any subset $T$ of $S$ must also be bounded by theorem 1, and so $w_2 = \inf T$ exists.

• Now suppose that $\inf S > \inf T$, that is $w_2 < w_1$. By the definition that $w_2$ is the greatest lower bound of the set $T$, we have that since $w_2 < w_1$ then there exists an element $a \in T$ such that $a < w_1$. But then if $a \in T$, then since $T \subseteq S$ we have that $a \in S$, and thus since $a < w_1$ we have a contradiction since then $w_1$ isn't the infimum of $S$. Therefore $\inf S ≤ \inf T$. $\blacksquare$

We can merge theorems 2 and 3 to get that if $S$ is a nonempty bounded subset of the real numbers, then if $T \subseteq S$ then $\inf S ≤ \inf T ≤ sup T ≤ \sup S$.

Theorem 4: Let $S$ and $T$ be nonempty bounded subsets of the real numbers. If $x ≤ y$ for all $x \in S$ and for all $y \in T$ then $\sup S ≤ \inf T$.

• Proof: Since $x ≤ y$ for all $x \in S$ and for all $y \in T$ then we have that any $y$ is an upper bound for the set $S$, that is $\sup S ≤ y$. Since $\sup S ≤ y$, then $\sup S$ is a lower bound for the set $T$ and so $\sup S ≤ \inf T$. $\blacksquare$