Epsilon Definition of The Supremum and Infimum of a Bounded Set

# Epsilon Definition of The Supremum and Infimum of a Bounded Set

Recall from The Supremum and Infimum of a Bounded Set page the following definitions:

 Definition: Let $S$ be a set that is bounded above. We say that the supremum of $S$ denoted $\sup S = u$ is a number $u$ that satisfies the conditions that $u$ is an upper bound of $S$ and $u$ is the least upper bound of $S$, that is for any $v$ that is also an upper bound of $S$ then $u \leq v$.
 Definition: Let $S$ be a set that is bounded below. We say that the infimum of $S$ denoted $\inf S = w$ is a number $w$ that satisfies the conditions that $w$ is a lower bound of $S$ and $w$ is the greatest lower bound of $S$, that is for any $t$ that is also a lower bound of $S$ then $t \leq w$.

We will now reformulate these definitions with an equivalent statement that may be useful to apply in certain situations in showing that an upper bound $u$ is the supremum of a set, or showing that a lower bound $w$ is the infimum of a set.

 Theorem 1: Let $S$ be a nonempty subset of the real numbers that is bounded above. The upper bound $u$ is said to be the supremum of $S$ if and only if $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $u - \epsilon < x_{\epsilon}$.
• Proof: $\Rightarrow$ Let $S$ be a nonempty subset of the real numbers that is bounded above. We first want to show that if $u$ is an upper bound such that $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $u - \epsilon < x_{\epsilon}$, then $u = \sup S$. Let $u$ be an upper bound of $S$ that satisfies the condition stated above, and suppose that $v < u$. Then choose $\epsilon = u - v$, and so $u - v = \epsilon > 0$ and so there exists an element $x_{\epsilon} \in S$ such that $u - \epsilon < x_{\epsilon}$, and thus, $u = \sup S$.
• $\Leftarrow$ We now want to show that if $u = \sup S$ then $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $u - \epsilon < x_{\epsilon}$. Let $u = \sup S$ and let $\epsilon > 0$. We note that $u - \epsilon < u$ and so $u - \epsilon$ is not an upper bound of the set $S$. Therefore by the definition that $u = \sup S$, there exists some element $x_{\epsilon} \in S$ such that $u - \epsilon < x_{\epsilon}$. $\blacksquare$
 Theorem 2: Let $S$ be a nonempty subset of the real numbers that is bounded below. The lower bound $w$ is said to be the infimum of $S$ if and only if $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $x_{\epsilon} < w + \epsilon$.
• Proof: $\Rightarrow$ Let $S$ be a nonempty subset of the real numbers that is bounded below. We first want to show that if $w$ is a lower bound such that $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $x_{\epsilon} < w + \epsilon$, then $w = \inf S$. Let $w$ be a lower bound of $S$ that satisfies the condition stated above, and suppose that $w < t$. Then choose $\epsilon = t - w$, and so $t - w = \epsilon > 0$ and so there exists an element $x_{\epsilon} \in S$ such that $x_{\epsilon} < w + \epsilon$, and thus, $w = \inf S$.
• $\Leftarrow$ We now want to show that if $w = \inf S$ then $\forall \epsilon > 0$ there exists an element $x_{\epsilon} \in S$ such that $x_{\epsilon} < w + \epsilon$. Let $w = \inf S$ and let $\epsilon > 0$. We note that $w < w +\epsilon$ and so $w + \epsilon$ is not a lower bound of the set $S$. Therefore by the definition that $w = \inf S$, there exists some element $x_{\epsilon} \in S$ such that $x_{\epsilon} < w + \epsilon$. $\blacksquare$