I was freshman when I was first introduced supremum and infimum. I could not understand a thing, yet everybody seemed to understand. I felt like stupid, inadequate and also insufficient person to learn mathematics. However, reality is not like that. In truth, I was the one to blame; I hadn’t dived deep into these topics deeply enough. I do not know if I am alone or not, but I think supremum and infimum are the topics which cannot by bypassed without learning and embracing. Today I want to show an interesting fact; if a set has least-upper-bound property and bounded below, then it has infimum. But first let us define supremum and infimum.

$\textbf{\small Definition 1.1:}$ Assume $S$ be an ordered set and $B$ is a nonempty, bounded above subset of $S$. Let $\alpha$ be a upper-bound of $B$. If there exists a $\beta$ that satisfies $\beta < \alpha$ then $\beta$ is not an upper-bound. Then $\alpha$ is called least-upper-bound of $B$.

or we can define supremum/ least-upper-bound as follows:

$\textbf{\small Definition 1.1.1:}$ Assume $S$ be an ordered set and $B$ is a nonempty, bounded above subset of $S$. If $\alpha \in S$ is an upper-bound of $B$ such that $\alpha \leq \alpha^{\prime}$ for every upper bound $\alpha^{\prime}$ of $B$, then $\alpha$ is called supremum of $B$, denoted $\alpha = \sup B$.

Those are exact same definitions of supremum. Both grasp the meaning firmly. It is important to see that we choose the least or smallest upper bound among every upper bounds from bounded (above) set. The second definition is more intuitive yet is a bit unconventional. First one is widely used it is from Baby Rudin and less intuitive for the first learners.

First definition states least property using a quite clever way. Namely, it says that if there would be an upper bound $\beta$ that is smaller than another upper bound $\alpha$ then $\beta$ should not be an upper bound. In other words, it guarantees that no one can find smaller upper bound than $\alpha$.

Similarly, we can state two different types of definition of infimum

$\textbf{\small Definition 1.2:}$ Assume $S$ be an ordered set and $B$ is a nonempty, bounded below subset of $S$. Let $\alpha$ be a lower-bound of $B$. If there exists a $\beta$ that satisfies $\beta > \alpha$ then $\beta$ is not an lower-bound. Then $\alpha$ is called greatest-upper-bound of $B$.

$\textbf{\small Definition 1.2.1:}$ Assume $S$ be an ordered set and $B$ is a nonempty, bounded below subset of $S$. If $\alpha \in S$ is a lower-bound of $B$ such that $\alpha \geq \alpha^{\prime}$ for every lower bound $\alpha^{\prime}$ of $B$, then $\alpha$ is called supremum of $B$, denoted $\alpha = \inf B$.

Now, let us state the theorem

$\textbf{\small Theorem 1.1:}$ Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is nonempty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then

$$ \begin{aligned} \alpha = \sup L \end{aligned} $$
$$ \begin{aligned} \alpha = \sup L \end{aligned} $$

exists in $S$, and $\alpha = \inf B$.

$\textbf{\small Proof:}$ First aim is to show $\alpha = \sup L$. Demonstration process is pretty straight-forward because it is enough for us to show $L$ is bounded. $L$ is nonempty since $B$ is bounded below and $L$ is the set of all lower bounds of $B$

$$ \begin{aligned} L = \{ y \mid y \leq x, x \in B \} \end{aligned} $$
$$ \begin{aligned} L = \{ y \mid y \leq x, x \in B \} \end{aligned} $$

Observe that $x$’s are upper bounds of $L$. We know ordered set $S$ has LUB property. Hence, $L$ has least-upper-bound, since $L$ is bounded above by $x$’s. Let this least-upper-bound be denoted as $\alpha$

$$ \begin{aligned} \alpha = \sup L \end{aligned} $$
$$ \begin{aligned} \alpha = \sup L \end{aligned} $$

The hard part is to show whether $\alpha = \inf B$. According to the definition: first we need to show $\alpha$ is a lower bound of $B$ and second we should show $\alpha$ is the greatest lower bound. If we rephrase “$\alpha$ is a lower bound of $B$” this sentence we end up with “$\alpha$ is in $L$” because $L$ is the set of all lower bounds of $B$.

Assume $\alpha \notin L$, then $\alpha > x$ where $x \in B$. Notice also that the elements of B are upper bounds for $L$, namely $x$’s are upper bounds of $L$. But we know that $\alpha$ is the least-upper-bound of $L$. Contradiction by definition of supremum: if $x < \alpha$, then $x$ is not an upper bound. Therefore, $\alpha \in L$.

Now let us show that this $\alpha$ is the greatest-lower-bound, in other words there is no strictly greater lower bound than $\alpha$ for $B$.

For the sake of reductio ad absurdum, let $\beta > \alpha$ and suppose $\beta$ is a lower bound. We know that $y \leq \alpha$ for all $y \in L$, since $\alpha$ is the supremum of $L$. Since $\beta \in L$ and $\alpha$ is supremum $\beta < \alpha$. Contradiction. Therefore $\beta$ is not a lower bound, namely $\beta \notin L$. Hence, $\alpha = \inf B$.

Magical is not it? I love this feeling.