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. ### Real Suprema and Maxima For real sets, the abstract definition becomes useful only when we can test it numerically. We need an epsilon criterion because estimates in analysis are usually approximate rather than exact. [quotetheorem:8612] This theorem turns the order definition into the working language of analysis. To prove that a proposed upper bound is the least one, we do not inspect all upper bounds individually; we show that every smaller number misses some element of the set. A supremum may or may not be attained. This creates a practical distinction: an upper boundary can control the set without being a value of the set, as happens for an open interval at its missing endpoint. When the controlling value is actually present, arguments can use an element of the set rather than only an external bound. A maximum names precisely this attained case. It records both requirements at once: the candidate must belong to the set, and every element of the set must lie below it. [definition: Maximum] Let $(P, \le)$ be a partially ordered set and let $A \subset P$. An element $m \in P$ is a maximum of $A$ if $m \in A$ and \begin{align*} a \le m \quad \text{for every } a \in A. \end{align*} [/definition] A maximum is both an element of the set and an upper bound. The remaining question is whether this attained upper bound is automatically the least possible upper bound, or whether supremum could still be a different order-theoretic object. The point of the comparison is to justify replacing supremum calculations by largest-element arguments whenever an actual maximum is available. [quotetheorem:8613] The theorem explains why supremum is invisible in many finite computations: the largest element already does the work. Supremum becomes visible when the controlling value sits at a missing endpoint, a [limit point](/page/Limit%20Point), or an ideal boundary of the ordered set. ## Completeness of the Real Line The definition of supremum may be stated in any partially ordered set, but existence is a separate matter. The rational numbers have the same order symbols as the [real numbers](/page/Real%20Numbers), yet they do not contain all the least upper bounds suggested by their subsets. Completeness is the assertion that the real line has no such holes. A completeness axiom must rule out bounded sets whose best upper bound is missing from the ambient system. We need this theorem because it is the point where the real line becomes different from an arbitrary ordered field. [quotetheorem:3227] This property is not a decorative fact about $\mathbb{R}$; it is the engine behind many existence theorems. If a process produces better and better lower estimates or upper estimates, completeness provides a real number at which the estimates can settle. The failure over $\mathbb{Q}$ is the cleanest warning. We need a concrete comparison because the order on $\mathbb{Q}$ looks familiar, while its missing least upper bounds are hidden until a limiting boundary appears. [example: A Rational Set With an Irrational Boundary] Let \begin{align*} A = \{q \in \mathbb{Q} : q^2 < 2\}. \end{align*} Viewed as a subset of $\mathbb{R}$, we first show that $\sup_{\mathbb{R}} A=\sqrt{2}$. Let $q\in A$. Then $q\in\mathbb{Q}$ and $q^2<2$. If $q\le 0$, then $q<\sqrt{2}$ because $\sqrt{2}>0$. If $q>0$ and $q\ge \sqrt{2}$, multiplying $q\ge\sqrt{2}$ by the positive number $q$ gives \begin{align*} q^2\ge q\sqrt{2}. \end{align*} Multiplying $q\ge\sqrt{2}$ by the positive number $\sqrt{2}$ gives \begin{align*} q\sqrt{2}\ge(\sqrt{2})^2. \end{align*} Since $(\sqrt{2})^2=2$, transitivity gives \begin{align*} q^2\ge 2. \end{align*} This contradicts $q^2<2$. Hence $q<\sqrt{2}$ in all cases, so $\sqrt{2}$ is an upper bound for $A$ in $\mathbb{R}$. Now let $b<\sqrt{2}$. If $b<1$, then $1\in A$ because $1\in\mathbb{Q}$ and \begin{align*} 1^2=1<2. \end{align*} Since $b<1$, the element $1\in A$ satisfies $1>b$, so $b$ is not an upper bound for $A$. It remains to handle $1\le b<\sqrt{2}$. Set \begin{align*} \delta=\sqrt{2}-b. \end{align*} Since $b<\sqrt{2}$, subtracting $b$ gives $\delta>0$. By the *[Archimedean property](/theorems/737)*, choose $n\in\mathbb{N}$ such that \begin{align*} \frac{1}{n}<\delta. \end{align*} Let $m$ be the least integer greater than $nb$, and set \begin{align*} a=\frac{m}{n}. \end{align*} Since $m,n\in\mathbb{Z}$ and $n\ne 0$, we have $a\in\mathbb{Q}$. From $m>nb$ and $n>0$, division by $n$ gives \begin{align*} a=\frac{m}{n}>b. \end{align*} Because $m$ is the least integer greater than $nb$, the integer $m-1$ is not greater than $nb$, so \begin{align*} m-1\le nb. \end{align*} Adding $1$ to both sides gives \begin{align*} m\le nb+1. \end{align*} Dividing by $n>0$ gives \begin{align*} a=\frac{m}{n}\le b+\frac{1}{n}. \end{align*} Since $1/n<\delta=\sqrt{2}-b$, adding $b$ gives \begin{align*} b+\frac{1}{n}<\sqrt{2}. \end{align*} Thus \begin{align*} b<a<\sqrt{2}. \end{align*} Since $a>b\ge 1$, we have $a>0$. Multiplying $a<\sqrt{2}$ by the positive number $a$ gives \begin{align*} a^2<a\sqrt{2}. \end{align*} Multiplying $a<\sqrt{2}$ by the positive number $\sqrt{2}$ gives \begin{align*} a\sqrt{2}<(\sqrt{2})^2. \end{align*} Since $(\sqrt{2})^2=2$, transitivity gives \begin{align*} a^2<2. \end{align*} Thus $a\in A$, and $a>b$. Hence $b$ is not an upper bound for $A$. We have shown that $\sqrt{2}$ is an upper bound and that every real number below $\sqrt{2}$ fails to be an upper bound, so \begin{align*} \sup_{\mathbb{R}} A=\sqrt{2}. \end{align*} Inside $(\mathbb{Q},\le)$, the same set has no supremum. First, $\sqrt{2}$ is irrational. Suppose, for contradiction, that $\sqrt{2}=m/n$ with integers $m,n$ in lowest terms and $n\ne 0$. Squaring both sides gives \begin{align*} 2=\frac{m^2}{n^2}. \end{align*} Multiplying by $n^2$ gives \begin{align*} 2n^2=m^2. \end{align*} Thus $m^2$ is even. If $m$ were odd, then $m=2j+1$ for some integer $j$, and \begin{align*} m^2=(2j+1)^2. \end{align*} Expanding the square gives \begin{align*} m^2=4j^2+4j+1. \end{align*} Since $4j^2+4j=2(2j^2+2j)$ is even, $4j^2+4j+1$ is odd, contradicting that $m^2$ is even. Hence $m$ is even, so $m=2k$ for some integer $k$. Substituting into $2n^2=m^2$ gives \begin{align*} 2n^2=(2k)^2. \end{align*} Since $(2k)^2=4k^2$, this becomes \begin{align*} 2n^2=4k^2. \end{align*} Dividing by $2$ gives \begin{align*} n^2=2k^2. \end{align*} Thus $n^2$ is even. If $n$ were odd, then $n=2\ell+1$ for some integer $\ell$, and expanding gives \begin{align*} n^2=(2\ell+1)^2=4\ell^2+4\ell+1. \end{align*} Since $4\ell^2+4\ell=2(2\ell^2+2\ell)$ is even, $4\ell^2+4\ell+1$ is odd, contradicting that $n^2$ is even. Hence $n$ is even. Both $m$ and $n$ are divisible by $2$, contradicting that $m/n$ was in lowest terms. Therefore $\sqrt{2}\notin\mathbb{Q}$. Now let $r\in\mathbb{Q}$ be any rational upper bound for $A$. The real-valued computation above showed that every real number below $\sqrt{2}$ fails to be an upper bound for $A$, so $r\ge\sqrt{2}$. Since $r\in\mathbb{Q}$ and $\sqrt{2}\notin\mathbb{Q}$, we have $r\ne\sqrt{2}$. Hence \begin{align*} r>\sqrt{2}. \end{align*} Define \begin{align*} s=\frac{1}{2}\left(r+\frac{2}{r}\right). \end{align*} Since $r\in\mathbb{Q}$ and $r>\sqrt{2}>0$, we have $r\ne 0$. Therefore $2/r\in\mathbb{Q}$ and $s\in\mathbb{Q}$. Also $r>0$ and $2/r>0$, so \begin{align*} s>0. \end{align*} We compare $s$ with $r$. From the definition of $s$, \begin{align*} r-s=r-\frac{1}{2}\left(r+\frac{2}{r}\right). \end{align*} Distributing the factor $1/2$ gives \begin{align*} r-s=r-\frac{r}{2}-\frac{1}{r}. \end{align*} Since $r-r/2=r/2$, this is \begin{align*} r-s=\frac{r}{2}-\frac{1}{r}. \end{align*} Putting both terms over the common denominator $2r$ gives \begin{align*} r-s=\frac{r^2}{2r}-\frac{2}{2r}. \end{align*} Combining the numerators gives \begin{align*} r-s=\frac{r^2-2}{2r}. \end{align*} Because $r>\sqrt{2}>0$, multiplying $r>\sqrt{2}$ by the positive number $r$ gives \begin{align*} r^2>r\sqrt{2}. \end{align*} Multiplying $r>\sqrt{2}$ by the positive number $\sqrt{2}$ gives \begin{align*} r\sqrt{2}>(\sqrt{2})^2. \end{align*} Since $(\sqrt{2})^2=2$, we have \begin{align*} r\sqrt{2}>2. \end{align*} Thus $r^2>2$, so $r^2-2>0$. Also $2r>0$, and therefore \begin{align*} r-s>0. \end{align*} Hence $s<r$. It remains to show that $s$ is still an upper bound for $A$. Squaring the definition of $s$ gives \begin{align*} s^2=\left(\frac{1}{2}\left(r+\frac{2}{r}\right)\right)^2. \end{align*} Since $(1/2)^2=1/4$, this becomes \begin{align*} s^2=\frac{1}{4}\left(r+\frac{2}{r}\right)^2. \end{align*} Expanding the square gives \begin{align*} \left(r+\frac{2}{r}\right)^2=r^2+2r\frac{2}{r}+\left(\frac{2}{r}\right)^2. \end{align*} Since $r\ne 0$, \begin{align*} 2r\frac{2}{r}=4. \end{align*} Also, \begin{align*} \left(\frac{2}{r}\right)^2=\frac{4}{r^2}. \end{align*} Therefore \begin{align*} s^2=\frac{1}{4}\left(r^2+4+\frac{4}{r^2}\right). \end{align*} Subtracting $2=8/4$ gives \begin{align*} s^2-2=\frac{r^2+4+4/r^2}{4}-\frac{8}{4}. \end{align*} Combining the numerators gives \begin{align*} s^2-2=\frac{r^2+4+4/r^2-8}{4}. \end{align*} Since $4-8=-4$, this is \begin{align*} s^2-2=\frac{r^2-4+4/r^2}{4}. \end{align*} Because $r^2>0$, the sign of $r^2-4+4/r^2$ is the same as the sign of its product with $r^2$. Multiplying out gives \begin{align*} r^2\left(r^2-4+\frac{4}{r^2}\right)=r^4-4r^2+4. \end{align*} Factoring gives \begin{align*} r^4-4r^2+4=(r^2-2)^2. \end{align*} Since $r^2>2$, we have $r^2-2>0$, and therefore \begin{align*} (r^2-2)^2>0. \end{align*} It follows that \begin{align*} r^2\left(r^2-4+\frac{4}{r^2}\right)>0. \end{align*} Since $r^2>0$, division by $r^2$ gives \begin{align*} r^2-4+\frac{4}{r^2}>0. \end{align*} Dividing by $4>0$ gives \begin{align*} s^2-2>0. \end{align*} Thus \begin{align*} s^2>2. \end{align*} We already know $s>0$. If $s\le\sqrt{2}$, then multiplying by the nonnegative number $s$ gives \begin{align*} s^2\le s\sqrt{2}. \end{align*} Multiplying $s\le\sqrt{2}$ by the positive number $\sqrt{2}$ gives \begin{align*} s\sqrt{2}\le(\sqrt{2})^2. \end{align*} Since $(\sqrt{2})^2=2$, transitivity gives \begin{align*} s^2\le 2. \end{align*} This contradicts $s^2>2$. Therefore \begin{align*} s>\sqrt{2}. \end{align*} If $q\in A$, the first part of the example gives $q<\sqrt{2}$. Since $\sqrt{2}<s$, transitivity gives \begin{align*} q<s. \end{align*} Thus $s$ is a rational upper bound for $A$. Starting from an arbitrary rational upper bound $r$, we have constructed another rational upper bound $s$ with $s<r$. Therefore no rational upper bound can be least, so $A$ has no supremum in $(\mathbb{Q},\le)$. The same set has the real supremum $\sqrt{2}$, but that boundary point is missing from the rational ordered set. [/example] The example shows that ``bounded above'' and ``has a supremum'' are distinct unless the ambient ordered set has enough points. This is why completeness of $\mathbb{R}$ is a structural assumption, not a consequence of order axioms alone. A second way to see completeness is through monotone sequences. We need to isolate increasing sequences because each term improves the previous estimate, and a uniform upper bound prevents escape to infinity. [definition: Increasing Sequence] A sequence $a: \mathbb{N} \to \mathbb{R}$, written $(a_n)_{n=1}^{\infty}$, is increasing if \begin{align*} a_n \le a_{n+1} \quad \text{for every } n \in \mathbb{N}. \end{align*} [/definition] For an increasing sequence with an upper bound, the set of its values has a least upper bound. The possible obstruction is that this least upper bound might merely sit above all terms without being approached by them. The theorem rules out that obstruction for monotone sequences: once the sequence keeps moving upward and cannot pass its supremum, its terms must eventually lie arbitrarily close to it. [quotetheorem:743] This theorem is one of the first places where supremum becomes a construction tool. Instead of guessing the limit of an increasing sequence, we define it as the least upper bound of its range and then verify convergence by the epsilon characterisation. [example: Decimal Approximations from Below] Let $a_n=1-10^{-n}$ for $n\in\mathbb{N}$. We first verify that the sequence is increasing. For each $n$, \begin{align*} a_{n+1}-a_n=(1-10^{-(n+1)})-(1-10^{-n}). \end{align*} Distributing the minus sign gives \begin{align*} a_{n+1}-a_n=1-10^{-(n+1)}-1+10^{-n}. \end{align*} Since $1-1=0$, this becomes \begin{align*} a_{n+1}-a_n=10^{-n}-10^{-(n+1)}. \end{align*} Because $10^{-n}=10\cdot 10^{-(n+1)}$, \begin{align*} a_{n+1}-a_n=10\cdot 10^{-(n+1)}-10^{-(n+1)}. \end{align*} Factoring out $10^{-(n+1)}$ gives \begin{align*} a_{n+1}-a_n=(10-1)10^{-(n+1)}. \end{align*} Since $10-1=9$, \begin{align*} a_{n+1}-a_n=9\cdot 10^{-(n+1)}. \end{align*} Here $9>0$, and $10^{n+1}>0$, so its reciprocal $10^{-(n+1)}=1/10^{n+1}$ is positive. Therefore \begin{align*} a_{n+1}-a_n>0. \end{align*} Thus $a_n<a_{n+1}$ for every $n\in\mathbb{N}$, and $(a_n)$ is increasing. Also $10^n>0$, so $10^{-n}=1/10^n>0$. Subtracting $10^{-n}$ from $1$ gives \begin{align*} 1-10^{-n}<1. \end{align*} Since $a_n=1-10^{-n}$, we have $a_n<1$ for every $n\in\mathbb{N}$. Thus $1$ is an upper bound for $\{a_n:n\in\mathbb{N}\}$. Now let $b<1$. We show that $b$ is not an upper bound. Set \begin{align*} \varepsilon=1-b. \end{align*} Since $b<1$, subtracting $b$ from both sides gives $\varepsilon>0$. By the *Archimedean property*, choose $n\in\mathbb{N}$ such that \begin{align*} n>\frac{1/\varepsilon-1}{9}. \end{align*} Multiplying by $9>0$ gives \begin{align*} 9n>\frac{1}{\varepsilon}-1. \end{align*} Adding $1$ to both sides gives \begin{align*} 1+9n>\frac{1}{\varepsilon}. \end{align*} By *Bernoulli's inequality* applied to $10=1+9$, \begin{align*} 10^n=(1+9)^n\ge 1+9n. \end{align*} Combining $10^n\ge 1+9n$ with $1+9n>1/\varepsilon$ gives \begin{align*} 10^n>\frac{1}{\varepsilon}. \end{align*} Multiplying by $\varepsilon>0$ gives \begin{align*} \varepsilon 10^n>1. \end{align*} Since $10^n>0$, dividing by $10^n$ preserves the inequality: \begin{align*} \varepsilon>10^{-n}. \end{align*} Substituting $\varepsilon=1-b$ gives \begin{align*} 1-b>10^{-n}. \end{align*} Adding $b$ to both sides gives \begin{align*} 1>b+10^{-n}. \end{align*} Subtracting $10^{-n}$ from both sides gives \begin{align*} 1-10^{-n}>b. \end{align*} Since $a_n=1-10^{-n}$, this says $a_n>b$. Therefore every $b<1$ fails to be an upper bound for $\{a_n:n\in\mathbb{N}\}$. Since $1$ is an upper bound and no smaller real number is an upper bound, \begin{align*} \sup_{n\in\mathbb{N}}a_n=1. \end{align*} No term attains this supremum. If $a_n=1$, then \begin{align*} 1-10^{-n}=1. \end{align*} Subtracting $1$ from both sides gives \begin{align*} -10^{-n}=0. \end{align*} Multiplying by $-1$ gives \begin{align*} 10^{-n}=0. \end{align*} But $10^n>0$, so $10^{-n}=1/10^n>0$, contradicting $10^{-n}=0$. Thus the sequence climbs upward toward the boundary value $1$, but the supremum is never one of its terms. [/example] Suprema also describe unbounded behaviour by their absence. If an increasing sequence is not bounded above, there is no finite supremum of its range in $\mathbb{R}$, and the sequence diverges to infinity in the extended sense. ## Order Calculus for Suprema Once supremum is available, it behaves like an optimal version of an inequality. The basic rules say that order-preserving operations transport least upper bounds in predictable ways. These rules are used constantly when passing from pointwise estimates to estimates on suprema. ### Scaling and Inclusion Scaling and translation are the first operations to understand. We need these formulas because real analysis repeatedly changes variables and rescales inequalities. The point is not merely algebraic notation: translating a set should translate every upper bound by the same amount, and multiplying by a positive scalar should preserve the order relation that defines upper bounds. The least upper bound should therefore move with the set under these order-preserving transformations. [quotetheorem:8614] The positivity of $\lambda$ matters because multiplying by a negative number reverses the order. We need a separate rule for negative scaling because reversing order changes an upper problem into a lower problem. [quotetheorem:8615] The next rule explains how supremum responds to inclusion. We need it because many exact suprema are found by trapping a difficult set between simpler sets. [quotetheorem:8616] Monotonicity is often the quickest way to trap an unknown supremum between two known quantities. In applications, the set $B$ is usually easier to estimate, while $A$ is the set that actually matters. [example: Bounding an Oscillatory Set] Let \begin{align*} A=\left\{\frac{\sin n}{n}:n\in\mathbb{N}\right\}. \end{align*} We bound $\sup A$ from below and above without computing its exact value. Fix $n\in\mathbb{N}$. Since $n\ge 1$, we have $n>0$. By the *standard bound for sine*, \begin{align*} -1\le \sin n\le 1. \end{align*} Dividing each part of this inequality by the positive number $n$ preserves the order, so \begin{align*} -\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}. \end{align*} Also $1\le n$. Dividing by $n>0$ gives \begin{align*} \frac{1}{n}\le \frac{n}{n}. \end{align*} Since $n>0$, we have $n/n=1$, hence \begin{align*} \frac{1}{n}\le 1. \end{align*} Combining $\frac{\sin n}{n}\le \frac{1}{n}$ with $\frac{1}{n}\le 1$ gives \begin{align*} \frac{\sin n}{n}\le 1. \end{align*} Since this holds for every $n\in\mathbb{N}$, every element of $A$ is at most $1$. Thus $1$ is an upper bound for $A$. The set $A$ is nonempty because $1\in\mathbb{N}$. Substituting $n=1$ into the defining formula gives \begin{align*} \frac{\sin 1}{1}=\sin 1. \end{align*} Therefore $\sin 1\in A$. Since $A$ is nonempty and bounded above, the *Least Upper Bound Property of $\mathbb{R}$* gives the existence of $\sup A$. By the definition of supremum, $\sup A$ is an upper bound for $A$. Applying this upper-bound property to the element $\sin 1\in A$ gives \begin{align*} \sin 1\le \sup A. \end{align*} Since $1$ is an upper bound for $A$ and $\sup A$ is the least upper bound, the defining minimality property of the supremum gives \begin{align*} \sup A\le 1. \end{align*} Therefore \begin{align*} \sin 1\le \sup A\le 1. \end{align*} The exact supremum is not needed here: pointwise estimates already force it to lie between the concrete value $\sin 1$ and the uniform upper bound $1$. [/example] ### Suprema of Sums Sums of sets introduce another pattern. We need a name for the set of all pairwise sums because it converts a two-variable upper-bound problem into a single supremum problem. [definition: Minkowski Sum] Let $A, B \subset \mathbb{R}$. The Minkowski sum of $A$ and $B$ is \begin{align*} A + B = \{a + b : a \in A,\ b \in B\}. \end{align*} [/definition] The definition packages a two-variable optimisation problem as the supremum of a single set. We need the corresponding theorem to say that independent choices can be optimised independently. [quotetheorem:8617] The theorem is the set version of the inequality $a+b \le \sup A + \sup B$, with the added assertion that the bound is optimal. The optimality uses the fact that each supremum can be approached from below by elements of its set. ## Supremum of Functions Many suprema in analysis are not taken over visible lists of numbers but over the values of a function. This shifts attention from a set $A \subset \mathbb{R}$ to the image of a map $f: E \to \mathbb{R}$. ### Function Values and Attainment When a function need not attain a largest value, its supremum records the best uniform upper estimate. We need this definition because function notation keeps track of which domain is being tested. [definition: Supremum of a Function] Let $E$ be a nonempty set and let $f: E \to \mathbb{R}$ be a function. The supremum of $f$ over $E$ is \begin{align*} \sup_{x \in E} f(x) = \sup f(E), \end{align*} provided the supremum of the image set $f(E) \subset \mathbb{R}$ exists. [/definition] This is not a new order concept; it is the same supremum applied to a set of function values. The notation keeps the domain variable visible, which matters when the same function is restricted to different domains. [example: Supremum Without Attainment for a Function] Let $f: \mathbb{R} \to \mathbb{R}$ be given by \begin{align*} f(x)=\frac{x^2}{1+x^2}. \end{align*} We show that $\sup f(\mathbb{R})=1$, but that no real input attains the value $1$. Fix $x\in\mathbb{R}$. Since $x^2\ge 0$, adding $1$ gives \begin{align*} 1+x^2\ge 1. \end{align*} Because $1>0$, we have $1+x^2>0$. Also $x^2\ge 0$, so division by the positive number $1+x^2$ gives \begin{align*} \frac{x^2}{1+x^2}\ge 0. \end{align*} Thus $f(x)\ge 0$. To compare $f(x)$ with $1$, write $1$ with denominator $1+x^2$: \begin{align*} 1-f(x)=1-\frac{x^2}{1+x^2}. \end{align*} Since $1+x^2>0$, we have \begin{align*} 1=\frac{1+x^2}{1+x^2}. \end{align*} Substituting this expression for $1$ gives \begin{align*} 1-f(x)=\frac{1+x^2}{1+x^2}-\frac{x^2}{1+x^2}. \end{align*} Combining fractions with the same denominator gives \begin{align*} 1-f(x)=\frac{1+x^2-x^2}{1+x^2}. \end{align*} Since $x^2-x^2=0$, this becomes \begin{align*} 1-f(x)=\frac{1}{1+x^2}. \end{align*} The numerator $1$ is positive and the denominator $1+x^2$ is positive, so \begin{align*} 1-f(x)>0. \end{align*} Therefore $f(x)<1$ for every $x\in\mathbb{R}$, so $1$ is an upper bound for $f(\mathbb{R})$. Now let $b<1$. Set \begin{align*} \varepsilon=1-b. \end{align*} Subtracting $b$ from $b<1$ gives $\varepsilon>0$, so $\sqrt{\varepsilon}>0$. Choose \begin{align*} x=\frac{1}{\sqrt{\varepsilon}}. \end{align*} Then $x\in\mathbb{R}$ and \begin{align*} x^2=\left(\frac{1}{\sqrt{\varepsilon}}\right)^2. \end{align*} Squaring numerator and denominator gives \begin{align*} x^2=\frac{1^2}{(\sqrt{\varepsilon})^2}. \end{align*} Since $1^2=1$ and $(\sqrt{\varepsilon})^2=\varepsilon$, we obtain \begin{align*} x^2=\frac{1}{\varepsilon}. \end{align*} Substituting this into $f$ gives \begin{align*} f(x)=\frac{1/\varepsilon}{1+1/\varepsilon}. \end{align*} Since $\varepsilon>0$, we have $\varepsilon\ne 0$, and \begin{align*} 1=\frac{\varepsilon}{\varepsilon}. \end{align*} Hence \begin{align*} 1+\frac{1}{\varepsilon}=\frac{\varepsilon}{\varepsilon}+\frac{1}{\varepsilon}. \end{align*} Combining fractions with denominator $\varepsilon$ gives \begin{align*} 1+\frac{1}{\varepsilon}=\frac{\varepsilon+1}{\varepsilon}. \end{align*} Therefore \begin{align*} f(x)=\frac{1/\varepsilon}{(\varepsilon+1)/\varepsilon}. \end{align*} Dividing by $(\varepsilon+1)/\varepsilon$ means multiplying by its reciprocal, so \begin{align*} f(x)=\frac{1}{\varepsilon}\cdot\frac{\varepsilon}{\varepsilon+1}. \end{align*} Since $\varepsilon>0$, cancellation gives \begin{align*} f(x)=\frac{1}{1+\varepsilon}. \end{align*} Because $\varepsilon=1-b$, subtracting $\varepsilon$ from $1$ gives \begin{align*} b=1-\varepsilon. \end{align*} Now subtract $b$ from $f(x)$: \begin{align*} f(x)-b=\frac{1}{1+\varepsilon}-(1-\varepsilon). \end{align*} Put the second term over the positive denominator $1+\varepsilon$: \begin{align*} f(x)-b=\frac{1}{1+\varepsilon}-\frac{(1-\varepsilon)(1+\varepsilon)}{1+\varepsilon}. \end{align*} Combining fractions with the same denominator gives \begin{align*} f(x)-b=\frac{1-(1-\varepsilon)(1+\varepsilon)}{1+\varepsilon}. \end{align*} Expanding the product gives \begin{align*} (1-\varepsilon)(1+\varepsilon)=1+\varepsilon-\varepsilon-\varepsilon^2. \end{align*} Since $\varepsilon-\varepsilon=0$, this is \begin{align*} (1-\varepsilon)(1+\varepsilon)=1-\varepsilon^2. \end{align*} Substituting into the expression for $f(x)-b$ gives \begin{align*} f(x)-b=\frac{1-(1-\varepsilon^2)}{1+\varepsilon}. \end{align*} Since $1-(1-\varepsilon^2)=\varepsilon^2$, we obtain \begin{align*} f(x)-b=\frac{\varepsilon^2}{1+\varepsilon}. \end{align*} Here $\varepsilon>0$, so $\varepsilon^2>0$. Also $1+\varepsilon>0$. Therefore \begin{align*} f(x)-b>0. \end{align*} Thus $f(x)>b$. Since every real number $b<1$ is exceeded by some value of $f$, no number below $1$ is an upper bound for $f(\mathbb{R})$. Together with the fact that $1$ is an upper bound, this proves \begin{align*} \sup_{x\in\mathbb{R}} f(x)=1. \end{align*} Finally, suppose that some $x\in\mathbb{R}$ satisfied $f(x)=1$. Then \begin{align*} \frac{x^2}{1+x^2}=1. \end{align*} As shown above, $1+x^2>0$, so multiplying both sides by $1+x^2$ gives \begin{align*} x^2=1+x^2. \end{align*} Subtracting $x^2$ from both sides gives \begin{align*} 0=1, \end{align*} which is impossible. Hence the supremum is approached by taking $|x|$ large, but it is not attained at any real input. [/example] For continuous functions on compact sets, this failure disappears. We need the extreme value theorem because it gives a structural condition under which supremum becomes maximum again. [quotetheorem:182] The theorem explains why closed and bounded intervals feel easier than open intervals. On compact domains, supremum and maximum coincide for continuous real-valued functions. On noncompact or nonclosed domains, the supremum may still exist but remain unattained. ### Uniform Size and Uniform Convergence To measure the size of a bounded function, we need to control positive and negative values at the same time. The supremum norm does this by taking the supremum of the absolute value. [definition: Supremum Norm] Let $E$ be a nonempty set and let $f: E \to \mathbb{R}$ be bounded. The supremum norm of $f$ on $E$ is \begin{align*} \|f\|_\infty = \sup_{x \in E} |f(x)|. \end{align*} [/definition] The absolute value in this definition makes the norm insensitive to sign. It asks how large the function can be in magnitude, not how high it rises above the horizontal axis. The point of a norm is that it should behave well under addition. We need the triangle inequality to justify treating $\|\cdot\|_\infty$ as a genuine measure of function size. [quotetheorem:8257] This theorem is a prototype for many arguments in function spaces: prove a pointwise inequality first, then take a supremum over the domain. The supremum is the passage from local control at each point to uniform control on the whole set. [Uniform convergence](/page/Uniform%20Convergence) can also be expressed by suprema. We need a definition based on worst-case error because pointwise convergence lets the bad point depend too freely on the stage $n$. [definition: Uniform Convergence] Let $E$ be a nonempty set, and let $f_n: E \to \mathbb{R}$ and $f: E \to \mathbb{R}$ be functions for every $n \in \mathbb{N}$. The sequence $(f_n)_{n=1}^{\infty}$ converges uniformly to $f$ on $E$ if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for every $n \ge N$ and every $x \in E$, \begin{align*} |f_n(x)-f(x)| < \varepsilon. \end{align*} [/definition] Uniform convergence is built from a worst-case bound because it is a global statement. Equivalently, when the suprema are finite for all sufficiently large $n$, uniform convergence is the condition \begin{align*} \sup_{x \in E} |f_n(x)-f(x)| \to 0. \end{align*} Every point of $E$ must be controlled by the same error bound at the same stage of the sequence. [example: Pointwise Convergence That Is Not Uniform] Let $f_n:(0,1)\to\mathbb{R}$ be given by \begin{align*} f_n(x)=x^n. \end{align*} We first show that $f_n(x)\to 0$ for each fixed $x\in(0,1)$. Fix $x\in(0,1)$. Since $0<x<1$, dividing $1>x$ by the positive number $x$ gives \begin{align*} \frac{1}{x}>1. \end{align*} Set $c=1/x$ and $h=c-1$. Then $c>1$, so $h>0$, and $c=1+h$. By *Bernoulli's inequality*, for every $n\in\mathbb{N}$, \begin{align*} c^n=(1+h)^n\ge 1+nh. \end{align*} Let $\varepsilon>0$. Since $h>0$, the *Archimedean property* gives $N\in\mathbb{N}$ such that \begin{align*} Nh>\frac{1}{\varepsilon}-1. \end{align*} Adding $1$ gives \begin{align*} 1+Nh>\frac{1}{\varepsilon}. \end{align*} If $n\ge N$, then multiplying by $h>0$ gives \begin{align*} nh\ge Nh. \end{align*} Adding $1$ gives \begin{align*} 1+nh\ge 1+Nh. \end{align*} Since $c^n\ge 1+nh$, transitivity gives \begin{align*} c^n\ge 1+Nh. \end{align*} Combining this with $1+Nh>1/\varepsilon$ yields \begin{align*} c^n>\frac{1}{\varepsilon}. \end{align*} Both sides are positive, so taking reciprocals reverses the inequality: \begin{align*} \frac{1}{c^n}<\varepsilon. \end{align*} Also $c^n>0$, hence $1/c^n>0$, so \begin{align*} 0<\frac{1}{c^n}<\varepsilon. \end{align*} Since $c=1/x$, we have $1/c=x$. Therefore \begin{align*} \frac{1}{c^n}=\left(\frac{1}{c}\right)^n=x^n. \end{align*} Thus $0<x^n<\varepsilon$ whenever $n\ge N$, so $f_n(x)\to 0$ for each fixed $x\in(0,1)$. Now fix $n\in\mathbb{N}$ and compute the supremum of $\{x^n:x\in(0,1)\}$. If $0<x<1$, then $x^n>0$ because it is a product of $n$ positive factors. Also, for each $k=1,\ldots,n$, the number $x^{k-1}$ is positive, so multiplying $x<1$ by $x^{k-1}$ gives \begin{align*} x^k<x^{k-1}. \end{align*} Applying this inequality for $k=n,n-1,\ldots,1$ gives $x^n<1$. Hence \begin{align*} 0<x^n<1. \end{align*} So $1$ is an upper bound for $\{x^n:x\in(0,1)\}$. To show that no smaller number is an upper bound, let $b<1$ and set $\varepsilon=1-b$. Then $\varepsilon>0$. First suppose $0<\varepsilon<1$. Choose \begin{align*} x=\left(1-\frac{\varepsilon}{2}\right)^{1/n}, \end{align*} using the positive $n$th root. Since $0<\varepsilon<1$, division by $2>0$ gives \begin{align*} 0<\frac{\varepsilon}{2}<\frac{1}{2}. \end{align*} Subtracting from $1$ gives \begin{align*} 0<1-\frac{\varepsilon}{2}<1. \end{align*} Taking positive $n$th roots preserves order on positive numbers, so \begin{align*} 0<x<1. \end{align*} By the definition of $x$, \begin{align*} x^n=1-\frac{\varepsilon}{2}. \end{align*} Since $\varepsilon/2<\varepsilon$, multiplying by $-1$ reverses the inequality: \begin{align*} -\frac{\varepsilon}{2}>-\varepsilon. \end{align*} Adding $1$ gives \begin{align*} 1-\frac{\varepsilon}{2}>1-\varepsilon. \end{align*} Because $\varepsilon=1-b$, we have $1-\varepsilon=b$, and therefore \begin{align*} x^n>b. \end{align*} Now suppose $\varepsilon\ge 1$. Choose $x=1/2$. Then $x\in(0,1)$. Since $1/2>0$, the product $(1/2)^n$ is positive, so \begin{align*} x^n>0. \end{align*} From $\varepsilon\ge 1$, multiplying by $-1$ gives \begin{align*} -\varepsilon\le -1. \end{align*} Adding $1$ gives \begin{align*} 1-\varepsilon\le 0. \end{align*} Because $b=1-\varepsilon$, we obtain \begin{align*} b\le 0. \end{align*} Thus \begin{align*} x^n>0\ge b. \end{align*} In every case, for each $b<1$ there is some $x\in(0,1)$ with $x^n>b$. Thus no number below $1$ is an upper bound. Since $1$ is an upper bound and every smaller number fails to be an upper bound, \begin{align*} \sup_{x\in(0,1)}x^n=1. \end{align*} For every $x\in(0,1)$, we have $x^n>0$, so the absolute value satisfies \begin{align*} |f_n(x)-0|=|x^n|=x^n. \end{align*} Hence, for every $n\in\mathbb{N}$, \begin{align*} \sup_{x\in(0,1)}|f_n(x)-0|=\sup_{x\in(0,1)}x^n=1. \end{align*} This prevents uniform convergence to $0$. Indeed, taking $\varepsilon=1/2$, uniform convergence to $0$ would require some $N\in\mathbb{N}$ such that for every $n\ge N$ and every $x\in(0,1)$, \begin{align*} |f_n(x)-0|<\frac{1}{2}. \end{align*} Fix any $n\ge N$. Since $\sup_{x\in(0,1)}x^n=1$, the number $1/2$ is not an upper bound for $\{x^n:x\in(0,1)\}$. Therefore there exists $x\in(0,1)$ such that \begin{align*} x^n>\frac{1}{2}. \end{align*} For this $x$, \begin{align*} |f_n(x)-0|=x^n>\frac{1}{2}, \end{align*} which contradicts the required uniform bound. Thus $f_n\to 0$ pointwise on $(0,1)$, but not uniformly on $(0,1)$. [/example] The example shows why supremum is stronger than checking values point by point. The location where the function is large moves toward the endpoint as $n$ grows, but the supremum still detects the persistent worst-case error. ## Suprema, Limits, and Limsup Supremum is also the raw material for describing eventual behaviour. A sequence may oscillate and fail to converge, so a single ordinary limit may not exist. The useful replacement is to ignore each finite initial segment and measure the largest values still possible in the remaining tail. Taking these tail suprema separates temporary early spikes from upper values that persist arbitrarily far out. The definition packages this tail-by-tail measurement into one number. Its hypotheses ensure that the suprema of the tails exist and that their eventual lower boundary is meaningful. [definition: Limit Superior] Let $(a_n)_{n=1}^{\infty}$ be a sequence in $\mathbb{R}$. Suppose each tail set is bounded above and the set of tail suprema is bounded below: \begin{align*} \{a_m : m \ge n\}, \qquad \{\sup_{m \ge n} a_m : n \in \mathbb{N}\}. \end{align*} The limit superior of $(a_n)$ is \begin{align*} \limsup_{n \to \infty} a_n = \inf_{n \in \mathbb{N}} \sup_{m \ge n} a_m. \end{align*} [/definition] The definition uses a supremum inside an infimum. The inner supremum asks how high the sequence can still get after time $n$; the lower-boundedness assumption ensures that the outer infimum is a real number rather than an escape to $-\infty$. [example: Oscillation Detected by Limsup] Let $a_n=(-1)^n+1/n$. We compute the supremum of each tail $\{a_m:m\ge n\}$ and then take the infimum of those tail suprema. If $m=2k$ is even, then $(-1)^m=(-1)^{2k}=1$, so \begin{align*} a_{2k}=1+\frac{1}{2k}. \end{align*} If $m=2k-1$ is odd, then $(-1)^m=(-1)^{2k-1}=-1$, so \begin{align*} a_{2k-1}=-1+\frac{1}{2k-1}. \end{align*} Since $k\in\mathbb{N}$, we have $2k-1\ge 1$. Dividing $1\le 2k-1$ by the positive number $2k-1$ gives \begin{align*} \frac{1}{2k-1}\le 1. \end{align*} Adding $-1$ to both sides gives \begin{align*} -1+\frac{1}{2k-1}\le 0. \end{align*} Thus every odd-indexed term satisfies $a_{2k-1}\le 0$. Fix $n\in\mathbb{N}$, and let $e_n$ be the least even integer with $e_n\ge n$. Since $e_n$ is even, $e_n=2j$ for some $j\in\mathbb{N}$, and therefore \begin{align*} a_{e_n}=(-1)^{2j}+\frac{1}{e_n}. \end{align*} Because $(-1)^{2j}=1$, this becomes \begin{align*} a_{e_n}=1+\frac{1}{e_n}. \end{align*} Now let $m\ge n$. If $m$ is even, then $e_n\le m$ because $e_n$ is the least even integer at least $n$. Since $e_n>0$ and $m>0$, taking reciprocals reverses the inequality: \begin{align*} \frac{1}{m}\le \frac{1}{e_n}. \end{align*} Adding $1$ gives \begin{align*} 1+\frac{1}{m}\le 1+\frac{1}{e_n}. \end{align*} For even $m$, we have $a_m=1+1/m$, while $a_{e_n}=1+1/e_n$. Hence \begin{align*} a_m\le a_{e_n}. \end{align*} If $m$ is odd, then $a_m\le 0$. Since $e_n>0$, we have $1/e_n>0$, so \begin{align*} 0<1+\frac{1}{e_n}. \end{align*} Using $a_{e_n}=1+1/e_n$, this gives \begin{align*} 0<a_{e_n}. \end{align*} Combining $a_m\le 0$ with $0<a_{e_n}$ gives \begin{align*} a_m\le a_{e_n}. \end{align*} Thus every element of the tail $\{a_m:m\ge n\}$ is at most $a_{e_n}$. Since $e_n\ge n$, the term $a_{e_n}$ itself belongs to the tail, so it is a maximum of the tail. Therefore \begin{align*} \sup_{m\ge n}a_m=a_{e_n}=1+\frac{1}{e_n}. \end{align*} By the definition of limit superior, \begin{align*} \limsup_{n\to\infty}a_n=\inf_{n\in\mathbb{N}}\left(1+\frac{1}{e_n}\right). \end{align*} We now compute this infimum. Since $e_n>0$, we have $1/e_n>0$. Adding $1$ gives \begin{align*} 1<1+\frac{1}{e_n}. \end{align*} So $1$ is a lower bound for the set $\{1+1/e_n:n\in\mathbb{N}\}$. Let $c>1$ and set $\varepsilon=c-1$. Then $\varepsilon>0$. By the *Archimedean property*, choose $N\in\mathbb{N}$ such that \begin{align*} N>\frac{1}{\varepsilon}. \end{align*} If $N$ is even, set $e=N$; if $N$ is odd, set $e=N+1$. Then $e$ is even and $e\ge N$. Combining $e\ge N$ with $N>1/\varepsilon$ gives \begin{align*} e>\frac{1}{\varepsilon}. \end{align*} Multiplying by $\varepsilon>0$ gives \begin{align*} \varepsilon e>1. \end{align*} Dividing by $e>0$ gives \begin{align*} \varepsilon>\frac{1}{e}. \end{align*} Now take $n=e$. Since $e$ is even, the least even integer at least $n=e$ is $e$ itself, so $e_n=e$. Therefore \begin{align*} 1+\frac{1}{e_n}=1+\frac{1}{e}. \end{align*} From $1/e<\varepsilon$, adding $1$ gives \begin{align*} 1+\frac{1}{e}<1+\varepsilon. \end{align*} Since $\varepsilon=c-1$, adding $1$ gives \begin{align*} 1+\varepsilon=c. \end{align*} Hence \begin{align*} 1+\frac{1}{e_n}<c. \end{align*} Thus every $c>1$ fails to be a lower bound. Since $1$ is a lower bound and no number greater than $1$ is a lower bound, \begin{align*} \inf_{n\in\mathbb{N}}\left(1+\frac{1}{e_n}\right)=1. \end{align*} It follows that \begin{align*} \limsup_{n\to\infty}a_n=1. \end{align*} The ordinary limit does not exist. Along even indices, \begin{align*} a_{2k}=1+\frac{1}{2k}. \end{align*} Subtracting $1$ gives \begin{align*} a_{2k}-1=\frac{1}{2k}. \end{align*} Given $\varepsilon>0$, the *Archimedean property* gives $K\in\mathbb{N}$ such that \begin{align*} K>\frac{1}{2\varepsilon}. \end{align*} If $k\ge K$, then transitivity gives \begin{align*} k>\frac{1}{2\varepsilon}. \end{align*} Multiplying by $2\varepsilon>0$ gives \begin{align*} 2\varepsilon k>1. \end{align*} Dividing by $2k>0$ gives \begin{align*} \varepsilon>\frac{1}{2k}. \end{align*} Since $1/(2k)>0$, we have \begin{align*} \left|a_{2k}-1\right|=\left|\frac{1}{2k}\right|=\frac{1}{2k}<\varepsilon. \end{align*} Thus $a_{2k}\to 1$. Along odd indices, \begin{align*} a_{2k-1}=-1+\frac{1}{2k-1}. \end{align*} Subtracting $-1$ gives \begin{align*} a_{2k-1}-(-1)=\frac{1}{2k-1}. \end{align*} Given $\varepsilon>0$, the *Archimedean property* gives $K\in\mathbb{N}$ such that \begin{align*} K>\frac{1+1/\varepsilon}{2}. \end{align*} If $k\ge K$, then transitivity gives \begin{align*} k>\frac{1+1/\varepsilon}{2}. \end{align*} Multiplying by $2>0$ gives \begin{align*} 2k>1+\frac{1}{\varepsilon}. \end{align*} Subtracting $1$ gives \begin{align*} 2k-1>\frac{1}{\varepsilon}. \end{align*} Multiplying by $\varepsilon>0$ gives \begin{align*} \varepsilon(2k-1)>1. \end{align*} Dividing by $2k-1>0$ gives \begin{align*} \varepsilon>\frac{1}{2k-1}. \end{align*} Since $1/(2k-1)>0$, we have \begin{align*} \left|a_{2k-1}-(-1)\right|=\left|\frac{1}{2k-1}\right|=\frac{1}{2k-1}<\varepsilon. \end{align*} Thus $a_{2k-1}\to -1$. If $(a_n)$ converged to some real number $L$, then every subsequence would converge to $L$. The even subsequence converges to $1$, while the odd subsequence converges to $-1$, and $1\ne -1$. Therefore $(a_n)$ does not converge. The limit superior records the upper limiting height that remains visible in every tail, even though the sequence itself oscillates. [/example] The limsup is not merely notation. We need to compare it with ordinary limits because it should extend the familiar notion of limit when convergence actually occurs. [quotetheorem:8618] This result says that limsup extends the ordinary limit rather than replacing it with an unrelated construction. When convergence holds, the eventual upper ceiling collapses to the same value as the eventual lower floor. The same idea governs radii of convergence for [power series](/page/Power%20Series), growth rates of sequences, and many compactness arguments. Supremum appears there because eventual upper bounds are often easier to establish than exact limits. ## Suprema in Sets, Measures, and Essential Bounds In measure theory and analysis of functions, the ordinary supremum can be too sensitive to negligible exceptional sets. A function that is unbounded at a single point has infinite ordinary supremum, even if that point has measure zero and does not affect any integral. The measure-theoretic replacement ignores behaviour on null sets. We need the essential supremum because $L^\infty$ spaces identify functions that agree almost everywhere. [definition: Essential Supremum] Let $(E, \mathcal{E}, \mu)$ be a [measure space](/page/Measure%20Space) and let $f: E \to \mathbb{R}$ be measurable. Suppose the set \begin{align*} \{M \in \mathbb{R} : f \le M \text{ } \mu\text{-a.e.}\} \end{align*} is nonempty and bounded below. The essential supremum of $f$ is \begin{align*} \operatorname{ess\,sup} f = \inf\{M \in \mathbb{R} : f \le M \text{ } \mu\text{-a.e.}\}. \end{align*} [/definition] This definition is still built from the order idea of least upper bound, but the word ``upper'' has been weakened to hold almost everywhere. The nonemptiness and lower-boundedness assumptions are the finite real-valued case; without them, measure theory usually records the essential supremum as an extended-real value. The resulting quantity is stable under [changing a function on a null set](/theorems/4915). [example: Ordinary Supremum Versus Essential Supremum] Let $f: [0,1] \to \mathbb{R}$ be defined by $f(0)=100$ and $f(x)=1$ for $x\in(0,1]$. We first compute the ordinary supremum. If $x\in[0,1]$, then either $x=0$ or $x\ne 0$. If $x=0$, then \begin{align*} f(x)=f(0)=100. \end{align*} If $x\ne 0$, then $x\in[0,1]$ gives $0\le x\le 1$. Combining $0\le x$ with $x\ne 0$ gives $0<x$, so $x\in(0,1]$. Hence \begin{align*} f(x)=1. \end{align*} Thus every value of $f$ is either $100$ or $1$. Since $1\le 100$ and $100\le 100$, we have \begin{align*} f(x)\le 100 \quad \text{for every } x\in[0,1]. \end{align*} Therefore $100$ is an upper bound for $f([0,1])$. Also $0\in[0,1]$ and $f(0)=100$, so $100\in f([0,1])$. If $u$ is any upper bound for $f([0,1])$, then every element of $f([0,1])$ is at most $u$. Applying this to the element $100\in f([0,1])$ gives \begin{align*} 100\le u. \end{align*} Thus $100$ is itself an upper bound and lies below every upper bound. Therefore \begin{align*} \sup_{x\in[0,1]} f(x)=100. \end{align*} Now compute the essential supremum with respect to [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^1$. The inequality $f(x)\le 1$ holds for every $x\in(0,1]$ because $f(x)=1$ there, and it fails at $x=0$ because \begin{align*} f(0)=100>1. \end{align*} Hence \begin{align*} \{x\in[0,1]:f(x)>1\}=\{0\}. \end{align*} For every $\eta>0$, the interval $(-\eta/2,\eta/2)$ contains $0$, and its length is \begin{align*} \frac{\eta}{2}-\left(-\frac{\eta}{2}\right)=\frac{\eta}{2}+\frac{\eta}{2}=\eta. \end{align*} Thus the singleton $\{0\}$ has Lebesgue outer measure at most $\eta$ for every $\eta>0$. Since Lebesgue measure is nonnegative, \begin{align*} 0\le \mathcal{L}^1(\{0\})\le \eta \quad \text{for every } \eta>0. \end{align*} If $\mathcal{L}^1(\{0\})>0$, then choosing $\eta=\mathcal{L}^1(\{0\})/2$ gives \begin{align*} \mathcal{L}^1(\{0\})\le \frac{\mathcal{L}^1(\{0\})}{2}. \end{align*} Subtracting $\mathcal{L}^1(\{0\})/2$ from both sides gives \begin{align*} \frac{\mathcal{L}^1(\{0\})}{2}\le 0, \end{align*} contradicting $\mathcal{L}^1(\{0\})>0$. Therefore \begin{align*} \mathcal{L}^1(\{0\})=0. \end{align*} Since the set where $f>1$ is null, $f\le 1$ $\mathcal{L}^1$-almost everywhere. Thus $1$ is an admissible almost-everywhere upper bound. If $M\ge 1$, then for every $x\in(0,1]$, \begin{align*} f(x)=1\le M. \end{align*} Therefore any point where $f(x)>M$ must be $x=0$, so \begin{align*} \{x\in[0,1]:f(x)>M\}\subset \{0\}. \end{align*} By monotonicity of measure and $\mathcal{L}^1(\{0\})=0$, \begin{align*} \mathcal{L}^1(\{x\in[0,1]:f(x)>M\})\le \mathcal{L}^1(\{0\})=0. \end{align*} Measures are nonnegative, so this measure is also at least $0$. Hence \begin{align*} \mathcal{L}^1(\{x\in[0,1]:f(x)>M\})=0. \end{align*} Thus $f\le M$ holds $\mathcal{L}^1$-almost everywhere, and every $M\ge 1$ is an admissible almost-everywhere upper bound. If $M<1$, then for every $x\in(0,1]$, \begin{align*} f(x)=1>M. \end{align*} Therefore \begin{align*} (0,1]\subset \{x\in[0,1]:f(x)>M\}. \end{align*} The interval $[0,1]$ has Lebesgue measure $1$. Since $[0,1]$ is the disjoint union of $\{0\}$ and $(0,1]$, finite additivity gives \begin{align*} \mathcal{L}^1([0,1])=\mathcal{L}^1(\{0\})+\mathcal{L}^1((0,1]). \end{align*} Substituting $\mathcal{L}^1([0,1])=1$ and $\mathcal{L}^1(\{0\})=0$ gives \begin{align*} 1=0+\mathcal{L}^1((0,1]). \end{align*} Hence \begin{align*} \mathcal{L}^1((0,1])=1. \end{align*} By monotonicity of measure, \begin{align*} \mathcal{L}^1(\{x\in[0,1]:f(x)>M\})\ge \mathcal{L}^1((0,1])=1. \end{align*} Thus the set where $f>M$ is not null, so $f\le M$ does not hold almost everywhere for any $M<1$. The admissible almost-everywhere upper bounds are exactly the real numbers $M$ with $M\ge 1$. The number $1$ is a lower bound for this set because $1\le M$ for every $M\ge 1$, and $1$ itself belongs to the set. Therefore the infimum of the admissible almost-everywhere upper bounds is $1$, so \begin{align*} \operatorname{ess\,sup} f=1. \end{align*} The ordinary supremum sees the value at the single point $0$, while the essential supremum ignores it because that point has Lebesgue measure zero. [/example] The comparison shows that essential supremum is not a weaker estimate for pointwise questions. It is a different estimate, designed for spaces where functions agreeing almost everywhere are regarded as the same object. A related order-theoretic idea is the supremum of a family of sets under inclusion. We need this example because it shows that supremum means least object above a family in the relevant order, not necessarily largest number below a ceiling. [quotetheorem:8619] This theorem reminds us that supremum is not tied to numerical size. It is the least object above a family in the relevant order. For sets ordered by inclusion, being above means containing every set in the family. The same language is often expressed by saying that a supremum is a join. A partially ordered set in which every pair of elements has both a join and a meet is a lattice, and one in which every subset has both is a complete lattice. The power set $\mathcal{P}(X)$ ordered by inclusion is the guiding example: joins are unions and meets are intersections. ## Failure Modes and Common Pitfalls Most mistakes with suprema come from silently replacing them with maxima. This replacement is valid only when the least upper bound is attained. Open domains, noncompact domains, and discontinuities are frequent sources of failure. The first pitfall is to infer existence of a point attaining the supremum from existence of the supremum itself. We need examples where the value escapes in different ways so the distinction becomes stable. [example: A Supremum Escaping to Infinity in the Domain] Let $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=\arctan x$. By definition, $\arctan x$ is the unique number $y\in(-\pi/2,\pi/2)$ such that $\tan y=x$. Therefore, for every $x\in\mathbb{R}$, \begin{align*} \arctan x<\frac{\pi}{2}. \end{align*} Thus $\pi/2$ is an upper bound for $f(\mathbb{R})$. Now let $b<\pi/2$. We show that $b$ is not an upper bound for $f(\mathbb{R})$. Set \begin{align*} \varepsilon=\frac{\pi}{2}-b. \end{align*} Since $b<\pi/2$, subtracting $b$ from both sides gives \begin{align*} 0<\frac{\pi}{2}-b. \end{align*} Hence $\varepsilon>0$. First suppose $0<\varepsilon<\pi$. Define \begin{align*} y=\frac{\pi}{2}-\frac{\varepsilon}{2}. \end{align*} Since $\varepsilon>0$, division by $2>0$ gives \begin{align*} 0<\frac{\varepsilon}{2}. \end{align*} Subtracting $\varepsilon/2$ from $\pi/2$ gives \begin{align*} \frac{\pi}{2}-\frac{\varepsilon}{2}<\frac{\pi}{2}. \end{align*} Thus $y<\pi/2$. Since $\varepsilon<\pi$, division by $2>0$ gives \begin{align*} \frac{\varepsilon}{2}<\frac{\pi}{2}. \end{align*} Subtracting $\varepsilon/2$ from both sides gives \begin{align*} 0<\frac{\pi}{2}-\frac{\varepsilon}{2}. \end{align*} Thus $0<y$. Combining $-\pi/2<0$ with $0<y<\pi/2$ gives \begin{align*} -\frac{\pi}{2}<y<\frac{\pi}{2}. \end{align*} So $y\in(-\pi/2,\pi/2)$. Take \begin{align*} x=\tan y. \end{align*} Then $x\in\mathbb{R}$, and the defining inverse relation for $\arctan$ gives \begin{align*} \arctan x=y. \end{align*} Substituting the value of $y$ gives \begin{align*} \arctan x=\frac{\pi}{2}-\frac{\varepsilon}{2}. \end{align*} Because $\varepsilon>0$, we have $\varepsilon/2<\varepsilon$. Multiplying by $-1$ reverses the inequality, so \begin{align*} -\frac{\varepsilon}{2}>-\varepsilon. \end{align*} Adding $\pi/2$ gives \begin{align*} \frac{\pi}{2}-\frac{\varepsilon}{2}>\frac{\pi}{2}-\varepsilon. \end{align*} Since $\varepsilon=\pi/2-b$, subtracting $\varepsilon$ from $\pi/2$ gives \begin{align*} \frac{\pi}{2}-\varepsilon=b. \end{align*} Therefore \begin{align*} \arctan x>b. \end{align*} So if $0<\varepsilon<\pi$, the value $f(x)=\arctan x$ exceeds $b$. Now suppose $\varepsilon\ge \pi$. Take $x=0$. Since $0\in(-\pi/2,\pi/2)$ and $\tan 0=0$, the defining inverse relation gives \begin{align*} \arctan 0=0. \end{align*} From $\varepsilon\ge \pi$, multiplying by $-1$ reverses the inequality: \begin{align*} -\varepsilon\le -\pi. \end{align*} Adding $\pi/2$ gives \begin{align*} \frac{\pi}{2}-\varepsilon\le \frac{\pi}{2}-\pi. \end{align*} Since $\pi/2-\pi=-\pi/2$, this becomes \begin{align*} \frac{\pi}{2}-\varepsilon\le -\frac{\pi}{2}. \end{align*} Because $b=\pi/2-\varepsilon$, we obtain \begin{align*} b\le -\frac{\pi}{2}. \end{align*} Since $-\pi/2<0$, transitivity gives \begin{align*} b<0. \end{align*} Using $\arctan 0=0$, this says \begin{align*} b<\arctan 0. \end{align*} So if $\varepsilon\ge\pi$, the value $f(0)=\arctan 0$ exceeds $b$. In both cases, every real number $b<\pi/2$ fails to be an upper bound for $f(\mathbb{R})$. Since $\pi/2$ is an upper bound and every smaller real number is not an upper bound, \begin{align*} \sup_{x\in\mathbb{R}}\arctan x=\frac{\pi}{2}. \end{align*} No real input attains this supremum. Indeed, for every $x\in\mathbb{R}$, the defining range condition gives \begin{align*} \arctan x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \end{align*} Hence \begin{align*} \arctan x<\frac{\pi}{2}. \end{align*} Therefore \begin{align*} \arctan x\ne \frac{\pi}{2}. \end{align*} The supremum is approached by taking inputs whose arctangent lies closer and closer to $\pi/2$, so the missing value comes from escape to infinity in the domain rather than from a missing endpoint of $\mathbb{R}$. [/example] This example explains why compactness appears in the extreme value theorem. Compactness prevents both missing boundary points and escape to infinity. Another pitfall is to assume that a supremum exists in every ordered set. We need a finite partial-order example because it separates nonexistence of a supremum from analytic complications such as irrational numbers or infinite sets. [example: Upper Bounds Without a Least Upper Bound] Let $P=\{a,b,u,v\}$, ordered by equality together with the four additional relations $a\le u$, $a\le v$, $b\le u$, and $b\le v$. Thus, for $x,y\in P$, the relation $x\le y$ holds exactly when $x=y$ or when $(x,y)$ is one of $(a,u)$, $(a,v)$, $(b,u)$, and $(b,v)$. Take $A=\{a,b\}$. The element $u$ is an upper bound for $A$. Indeed, the only elements of $A$ are $a$ and $b$, and the defining relations include \begin{align*} a\le u. \end{align*} They also include \begin{align*} b\le u. \end{align*} Therefore every element of $A$ is less than or equal to $u$. Similarly, $v$ is an upper bound for $A$ because the defining relations include \begin{align*} a\le v. \end{align*} and \begin{align*} b\le v. \end{align*} We now determine all upper bounds of $A$. Let $s\in P$. For $s$ to be an upper bound for $A$, it must satisfy both \begin{align*} a\le s \end{align*} and \begin{align*} b\le s. \end{align*} There are four possible values of $s$. If $s=a$, then the required condition $b\le s$ becomes \begin{align*} b\le a. \end{align*} But $b\ne a$, and the ordered pair $(b,a)$ is not one of the four additional defining relations. Hence $b\le a$ is false, so $a$ is not an upper bound for $A$. If $s=b$, then the required condition $a\le s$ becomes \begin{align*} a\le b. \end{align*} But $a\ne b$, and the ordered pair $(a,b)$ is not one of the four additional defining relations. Hence $a\le b$ is false, so $b$ is not an upper bound for $A$. If $s=u$, then the two required inequalities are \begin{align*} a\le u \end{align*} and \begin{align*} b\le u, \end{align*} both of which are defining relations. Hence $u$ is an upper bound for $A$. If $s=v$, then the two required inequalities are \begin{align*} a\le v \end{align*} and \begin{align*} b\le v, \end{align*} both of which are defining relations. Hence $v$ is an upper bound for $A$. Therefore the upper bounds of $A$ are exactly $u$ and $v$. Now test whether either upper bound is least. If $u$ were the supremum of $A$, then $u$ would have to be less than or equal to every upper bound of $A$. Since $v$ is an upper bound, this would require \begin{align*} u\le v. \end{align*} But $u\ne v$, and the ordered pair $(u,v)$ is not one of the four additional defining relations. Hence $u\le v$ is false, so $u$ is not the supremum of $A$. If $v$ were the supremum of $A$, then $v$ would have to be less than or equal to every upper bound of $A$. Since $u$ is an upper bound, this would require \begin{align*} v\le u. \end{align*} But $v\ne u$, and the ordered pair $(v,u)$ is not one of the four additional defining relations. Hence $v\le u$ is false, so $v$ is not the supremum of $A$. The set $A$ has upper bounds, namely $u$ and $v$, but neither upper bound lies below the other. Thus $A$ has no least upper bound in $P$. [/example] This finite example isolates the role of the ambient order. Total orders avoid incomparable upper bounds, but partial orders need lattice properties before arbitrary suprema can be expected. The final common mistake concerns estimates. Knowing that every element of $A$ is at most $M$ proves only $\sup A \le M$. To prove equality, one must also show that values of $A$ come arbitrarily close to $M$ from below. [remark: Proving an Exact Supremum] For a nonempty set $A \subset \mathbb{R}$ and a proposed value $s \in \mathbb{R}$, the standard two-step method is: first prove $a \le s$ for every $a \in A$; then prove that for every $\varepsilon > 0$ there exists $a \in A$ with $s-\varepsilon < a$. [/remark] This method is more reliable than trying to find a maximum. It works whether or not the supremum is attained, and it is the template behind many approximation arguments. ## Beyond and Connected Topics Supremum is the entry point to the completeness theory of the real numbers. The least upper bound property is equivalent, in standard developments, to several other completeness principles: Cauchy completeness, nested interval principles, and monotone convergence. The [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) are the natural place to continue with these foundational equivalences. In topology and metric spaces, supremum appears through boundedness, compactness, and continuous real-valued functions. The extreme value theorem connects least upper bounds to compact sets, and the [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) develop the surrounding compactness language. In analysis of functions, the supremum norm leads to uniform convergence, Banach spaces of bounded functions, and approximation theorems. This direction continues naturally into the [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions), where pointwise control is often upgraded to norm control. In complex analysis, suprema enter maximum modulus principles and estimates on holomorphic functions. Although the complex numbers are not ordered, the real-valued modulus $|f(z)|$ is, so suprema of moduli become a central tool. The [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis) provides the standard continuation of that theme. In measure theory, ordinary supremum gives way to essential supremum, producing the norm on $L^\infty$. This shift is indispensable whenever functions are identified almost everywhere, since pointwise exceptional values should not change the size of an $L^\infty$ class. ## References Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Walter Rudin, *Principles of Mathematical Analysis* (1976). Tom M. Apostol, *Mathematical Analysis* (1974). Terence Tao, *Analysis I* (2016).