A sequence such as $0.9, 0.99, 0.999, \ldots$ presses against $1$ without ever reaching it. If our only language for extremes were maximum and minimum, we would have to say that the set has no largest element and then stop, even though the number $1$ is controlling every estimate we want to make. The supremum is the device that lets analysis talk about the boundary value forced by a set, whether or not that boundary value is itself in the set.

[example: A Set With No Maximum] Let $A=\{x\in\mathbb{R}:0<x<1\}$. We first show that $A$ has no maximum. Choose any $x\in A$. Then $0<x<1$. From $x<1$, adding $x$ to both sides gives \begin{align*} x+x<x+1. \end{align*} Since $x+x=2x$, this is \begin{align*} 2x<x+1. \end{align*} Dividing by $2>0$ preserves the inequality, so \begin{align*} x<\frac{x+1}{2}. \end{align*} Now we check that $(x+1)/2$ is still in $A$. From $x<1$, adding $1$ to both sides gives \begin{align*} x+1<2. \end{align*} Dividing by $2>0$ gives \begin{align*} \frac{x+1}{2}<1. \end{align*} From $0<x$, adding $1$ to both sides gives \begin{align*} 1<x+1. \end{align*} Since $0<1$, transitivity gives \begin{align*} 0<x+1. \end{align*} Dividing by $2>0$ gives \begin{align*} 0<\frac{x+1}{2}. \end{align*} Thus $0<(x+1)/2<1$, so $(x+1)/2\in A$. Since $x<(x+1)/2$, the element $x$ is not a maximum of $A$. Because the choice of $x\in A$ was arbitrary, no element of $A$ is a maximum. We now show that $1$ is the least upper bound of $A$. If $a\in A$, then $0<a<1$, hence \begin{align*} a\le 1. \end{align*} Therefore $1$ is an upper bound for $A$. It remains to show that every real number below $1$ fails to be an upper bound. Let $b<1$. If $b<0$, then $1/2\in A$ because \begin{align*} 0<\frac{1}{2}. \end{align*} Also $1<2$, and dividing by $2>0$ gives \begin{align*} \frac{1}{2}<1. \end{align*} Since $b<0$ and $0<1/2$, transitivity gives \begin{align*} b<\frac{1}{2}. \end{align*} Thus $b$ is not an upper bound for $A$. Now suppose $0\le b<1$. Define \begin{align*} a=\frac{b+1}{2}. \end{align*} Since $0\le b$, adding $1$ gives \begin{align*} 1\le b+1. \end{align*} Because $0<1$, transitivity gives \begin{align*} 0<b+1. \end{align*} Dividing by $2>0$ gives \begin{align*} 0<\frac{b+1}{2}. \end{align*} Also, from $b<1$, adding $1$ to both sides gives \begin{align*} b+1<2. \end{align*} Dividing by $2>0$ gives \begin{align*} \frac{b+1}{2}<1. \end{align*} Therefore $0<a<1$, so $a\in A$. To compare $a$ with $b$, start from $b<1$ and add $b$ to both sides: \begin{align*} b+b<b+1. \end{align*} Since $b+b=2b$, this becomes \begin{align*} 2b<b+1. \end{align*} Dividing by $2>0$ gives \begin{align*} b<\frac{b+1}{2}. \end{align*} Because $a=(b+1)/2$, we have \begin{align*} b<a. \end{align*} Thus $b$ is not an upper bound for $A$. We have shown that $1$ is an upper bound for $A$, and that every real number $b<1$ fails to be an upper bound for $A$. Therefore $1$ is the least upper bound of $A$. Since $1\notin A$, this supremum is not attained as a maximum. [/example]

The example shows the central tension. Analysis studies limiting processes, but limiting processes often move toward values that are not attained. Supremum separates the question of being bounded above from the stronger question of attaining a largest element. This distinction is what makes the real line complete, what makes monotone convergence work, and what lets estimates pass to limits.

The word ``least'' in least upper bound is order-theoretic. Nothing in the first definition requires a metric, a topology, or a notion of distance. This is why the same idea appears in real analysis, order theory, measure theory, functional analysis, and probability. The real line then contributes one decisive extra feature: every nonempty subset bounded above has such a least upper bound.

## Definition

The boundary object we are trying to isolate must do more than dominate the set. The definition below says two things at once: $s$ lies above the whole set, and every other element lying above the whole set lies above $s$.

[definition: Supremum] Let $(P, \le)$ be a partially ordered set and let $A \subset P$. An element $s \in P$ is the supremum of $A$ if the following two conditions hold: \begin{align*} a \le s \quad \text{for every } a \in A, \end{align*} and whenever $u \in P$ satisfies $a \le u$ for every $a \in A$, we have $s \le u$. [/definition]

When such an element exists, it is denoted by $\sup A$. For a real-valued function $f: E \to \mathbb{R}$ on a set $E$, the notation $\sup_{x \in E} f(x)$ means $\sup f(E)$, where $f(E) = \{f(x) : x \in E\}$.

The definition has two jobs. The first condition says that $s$ is high enough to dominate every element of $A$. The second condition says that no smaller element of $P$ can do that job. The second condition is what distinguishes the supremum from a loose estimate.

## Basic Order Language

### Upper Bounds

Before we can use suprema in estimates, we need to separate two tasks that the definition of supremum combines. First we must know what it means for a candidate number or object to dominate every element of the set; only after that can we ask whether the candidate is the least such dominator. The phrase ``lies above the whole set'' is the reusable part of this test, so it deserves its own name.

[definition: Upper Bound] Let $(P, \le)$ be a partially ordered set and let $A \subset P$. An element $u \in P$ is an upper bound for $A$ if \begin{align*} a \le u \quad \text{for every } a \in A. \end{align*} [/definition]

Upper bounds form a region above the set. The supremum, when it exists, is the bottom edge of that region rather than an arbitrary point in it.

### Lower Bounds and Infima

Upper control is only half of the order picture. We need the downward version because many estimates are two-sided, and because reversing signs converts upper bounds into lower bounds.

[definition: Lower Bound] Let $(P, \le)$ be a partially ordered set and let $A \subset P$. An element $\ell \in P$ is a lower bound for $A$ if \begin{align*} \ell \le a \quad \text{for every } a \in A. \end{align*} [/definition]

Lower bounds give a region below the set. The infimum is the dual boundary value, and we need the definition below to name the top edge of that lower-bound region. Naming it separately keeps lower estimates from being treated as informal shadows of upper estimates.

[definition: Infimum] Let $(P, \le)$ be a partially ordered set and let $A \subset P$. An element $i \in P$ is the infimum of $A$ if the following two conditions hold: \begin{align*} i \le a \quad \text{for every } a \in A, \end{align*} and whenever $\ell \in P$ satisfies $\ell \le a$ for every $a \in A$, we have $\ell \le i$. [/definition]

The definition is phrased in a partially ordered set to expose the logical core. In the real line, the order is total and familiar, so the criterion can be written using strict inequalities. This form is often the version used in estimates.