Proofs Regarding The Supremum or Infimum of a Bounded Set

# Proofs Regarding The Supremum or Infimum of a Bounded Set

We will now look at some proofs regarding the supremum/infimum of a bounded set.

Before we do though, let's first recall that for a bounded set $A$, to prove that $\sup A = u$ for some $u \in \mathbb{R}$ we must show that:

• 1) $u$ is an upper bound to the set $A$. That is $\forall a \in A$, $a ≤ u$.
• 2) $u$ is the least upper bound to the set $A$. That is if $b < u$ then $\exists a \in A$ such that $b < a$ Similarly, to show that $\inf A = w$ for some $w \in \mathbb{R}$ we must show that:

• 1) $w$ is a lower bound to the set $A$. That is $\forall a \in A$, $w ≤ a$.
• 2) $w$ is the greatest lower bound to the set $A$. That is if $w < b$ then $\exists a \in A$ such that $a < b$. We will now look at some proofs regarding the supremum and infimum of a bounded set.

## Example 1

Prove that for the set $A := [m, n] = \{ x \in \mathbb{R} : m ≤ x ≤ n \}$, that $\sup(A) = n$.

We first show that $n$ is an upper bound to the set $A$. Since $A$ is defined such that $m ≤ x ≤ n$, then clearly $x ≤ n$ for all $x \in A$.

We must now proceed to show that $n$ is the least upper bound to the set $A$, that is if $b < n$, then there exists an $a \in A$ such that $b < a$. There are two cases to consider.

First, consider the case where $m ≤ b ≤ n$. Then choose $a = \frac{b + n}{2}$ and so $b < a$ and $a \in A$. The diagram below illustrates this argument geometrically. Now consider the case where $b ≤ m$. Then choose $a = \frac{m + n}{2}$ and so once again $b < a$ and $a \in A$. The diagram below illustrates this. Therefore $\sup A = n$.

## Example 2

Prove that for the set $A := (m, n) = \{ x \in \mathbb{R} : m < x < n \}$ that $\inf (A) = m$.

This example is very similar to example 1. We first show that $m$ is a lower bound to the set $A$. Since $A$ is defined such that $m < x < n$, then clearly $m ≤ x$ for all $x \in A$.

We must now proceed to show that $m$ is the greatest lower bound to the set $A$, that is if $m < b$ then there exists an $a \in A$ such that $a < b$. Once again there are two cases to consider.

First consider the case where $m < b < n$. Then choose $a = \frac{m + b}{2}$ and so $a < b$ and $a \in A$.

Now consider the case where $n ≤ b$. Then choose $a = \frac{m + n}{2}$ and once again $a < b$ and $a \in A$.

Therefore $\inf A = m$.

## Example 3

Prove that every finite nonempty has a supremum.

We will do this proof by the principle of mathematical induction. Let $A$ be a finite set such that $A \neq \emptyset$. Since $A$ is finite, there exists a natural number $n$ such that $\mid A \mid = n$.

Let $P(n)$ be the statement that $\mid A_n \mid = n$ and $A_n$ has a supremum. Our base step is to check $P(1)$. Clearly, any set $A_1$ containing only one element has a supremum, namely that single element and so $P(1)$ is true.

Now suppose for some $k \in \mathbb{N}$ that $P(k)$ is true, that is $\mid A_k \mid = k$ and that $A_k$ has a supremum. We want to show the statement $P(k+1)$ to be true, that is $\mid A_{k+1} \mid = k + 1$ and $A_{k+1}$ has a supremum. We know that $\sup A_k = a_j$ for some $j \in \mathbb{N}$ such that $1 ≤ j ≤ k$. Consider the set $A_{k+1}$. Removing the element in $A_{k+1}$ that is not in $A_k$, let's call it $b$, produces some set $A_k$ which has a supremum $a_j$, and so $\sup A_{k+1} = \max \{ a_j, b \}$. Therefore $P(k+1)$ is true.

By the principle of mathematical induction $P(n)$ is true for all $n \in \mathbb{N}$, that is if $A$ is a finite nonempty set containing $n$ elements then $A$ has a supremum.

## Example 4

Let $S = \{ x \in \mathbb{R} : 2 < x < 3 \}$. Prove that $\sup S = 3$ and that $\inf S = 2$.

We will first prove that $\sup S = 3$. To do so, we need to show that $3$ is an upper bound of the set $S$ and that $3$ is the least upper bound for $S$. From the definition of $S$ we have that $\forall x \in S$, $x < 3$, and so $3$ is an upper bound of $S$, that is $\sup S ≤ 3$.

Suppose that $u < 3$. If $3 = \sup S$ then there exists an element $s \in S$ such that $u < s$. We need to consider two cases. The first case is when $u ≤ 2$. Then choose $s = 2.5$ and so $u < 2.5 \in S$. The second case is when $2 < u < 3$. In such a case choose $s = \frac{u + 3}{2}$ and so $u < s \in S$. Therefore $\sup S = 3$.

We now want to prove that $\inf S = 2$. To do so, we need to show that $2$ is a lower bound of the set $S$ and that $2$ is the greatest lower bound for $S$. From the definition of $S$ we have that $\forall x \in S$, $2 < x$ and so $2$ is a lower bound of $S$, that is $2 ≤ \inf S$.

Now suppose that $2 < w$. If $2 = \inf S$ then there exists an element $s \in S$ such that $s < w$. We need to consider two cases. The first case is when $3 < w$. Then choose $s = 2.5$ and so $S \ni 2.5 < w$. The second case is when $2 < w < 3$. In such a case choose $s = \frac{2 + w}{2}$ and so $S \ni s < w$. Therefore $\inf S = 2$.