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Since $X_0=0$, its supremum $X^{\star }$ is nonnegative. For $b>0$, continuity givesThe events increase with $n$. Part (iii) therefore yields $\mathbb{P}(X^{\star }>b)={\lim }_n{e}^{-(b+1/n)}=e^{-b}$. Letting $b\downarrow 0$ gives $\mathbb{P}(X^{\star }>0)=1$; letting $b\to \infty$ gives finiteness almost surely. ConsequentlyIts probability density function is $e^{-x}{1}_{\{x>0\}}$, with no atom at zero. Using strict exceedances above also avoids presupposing that the infinite-horizon supremum is attained.

An element $m$ is a maximum of $S$ if $m\in S$ and $s\le m$ for every $s\in S$. Thus it is an upper bound that is itself a member. Every maximum is the least upper bound: any upper bound must lie above $m$ because $m\in S$. A supremum outside $S$ need not be a maximum.

An element $u\in A$ is a least upper bound of $S$ if it is an upper bound, meaning $s\le u$ for every $s\in S$, and $u\le v$ for every upper bound $v\in A$ of $S$. The candidate need not belong to $S$. Because a partial order is an antisymmetric relation, two least upper bounds must be equal. A least upper bound is also called the supremum.

Under the product order, every point of the open unit disc has both coordinates below one, so $(1,1)$ is an upper bound. Any upper bound $(u,v)$ must have $u\ge 1$ and $v\ge 1$, since the two coordinate projections of the open unit disc are $(-1,1)$. Therefore $(1,1){\le }_P(u,v)$, proving ${\sup }_{{\le }_P}S=(1,1)$.

Under the lexicographic order, every point $(1,v)$, with arbitrary real $v$, is an upper bound, because every point of $S$ has first coordinate strictly below one. Conversely, an upper bound cannot have first coordinate $u<1$: a point $(x,0)\in S$ with $x>u$ would exceed it. If an upper bound has $u>1$, the upper bound $(1,0)$ is strictly smaller. If it is $(1,v)$, the upper bound $(1,v-1)$ is strictly smaller. Hence there is no least upper bound under the lexicographic order. This illustrates the supremum of an open disc under product and lexicographic orders.

For $n$ equal strips and a uniform bound $|f^{\prime \prime }(x)|\le M$ on $[a,b]$, the composite trapezoidal error bound isOn $[1,25]$, part (b) gives $M=1/2$: $f^{\prime \prime }$ is positive and decreases from $1/2$ after $x=1$. HenceTo certify error at most $10^{-4}$, choose $n\ge 2400$; equality gives exactly $10^{-4}$ in this bound. This is a sufficient certificate, not a claim that every smaller $n$ fails in actual numerical error.

Strictly, the maximum specified over the open interval $(1,25)$ is not attained; its supremum is $1/2$. Using the maximum on the closed interval, or simply the uniform bound $M=1/2$, gives the valid error estimate above.

A risk function can decay rapidly for each fixed regression function while its supremum over a function class decays more slowly or fails to decay. Bounds depending on each function's local differentiability remainder need not be uniform. In fixed-design nonparametric regression, a class with no common smoothness bound permits smooth bump functions whose supports shrink between design points. Such sequences obstruct uniform estimation without contradicting fixed-function mean squared error bounds.

Suppose $|Z(u)-Z(v)|\le L\Vert u-v\Vert$ with a random finite $L$. On $L\le \ell$, a finite net with radius $\eta$ bounds the full supremum by the grid maximum plus $\ell \eta$. The Markov inequality controls the exceptional event using $\mathbb{E}L$, and a union bound controls the grid maximum. Choosing the net spacing balances these two costs.

If every member of a set $a$ belongs to some cumulative-hierarchy stage, replacement collects the least such stages. Their ordinal supremum gives a common stage $V_{\beta }$ containing every element of $a$, so $a\in V_{\beta +1}$. The argument also works without foundation and bounds a set of well-founded elements inside the hierarchy.

For a BV space function with $Du=\nabla udx+D^{s}u$, relaxed graph area is $A(u)=\int \sqrt{1+|\nabla u{|}^2}dx+|D^{s}u|(\Omega )$. Its dual supremum uses the coupled constraint ${\phi }_0^2+|\phi {|}^2\le 1$ in the pairing $\int ({\phi }_0+u\mathrm{div}\phi )$. It is convex and lower semicontinuous in $L^1$. Independent bounds on the two test fields define a different functional.

For centered multivariate normal distributions represented by vectors $U,V$ with equal corresponding variances and $\mathbb{E}[U_i{U}_j]\ge \mathbb{E}[V_i{V}_j]$ for all distinct indices, $\mathbb{P}(U_i\le u_i\text{ for all }i)\ge \mathbb{P}(V_i\le u_i\text{ for all }i)$ for every threshold vector. In particular the supremum of $U$ is stochastically smaller than the supremum of $V$. Equal corresponding variances are essential to this formulation. The related Sudakov-Fernique inequality compares expected values of maxima using variances of increments and does not require equal pointwise variances.

A boundary of phase space approached as a massless outgoing particle's energy tends to zero. A zero four-vector is not a physical positive-energy one-particle state. An energy bound requiring such a particle is a supremum or infimum, even when the closed kinematic interval is conventionally described using minimum and maximum. This distinction can matter when applying three-body formulas at mixed massive-massless recoil thresholds.

A nonrelativistic particle accepted only at a fixed flight time through an ideal uniform-field tube has fixed parallel speed $v_{\parallel }=l/t_0$. On-axis entry and wall clearance bound the independent perpendicular speed by $|q|Ba/(2m)$, provided a full gyration occurs before exit. Adding parallel and perpendicular kinetic energies gives the displayed ideal energy window, where $K_0=m{v}_{\parallel }^2/2$. The upper endpoint is a grazing limit; strict wall exclusion makes it a supremum.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

Let $X$ be a complete metric space and let $F_n$ be nonempty closed bounded subsets with $F_{n+1}\subseteq F_n$ and Hausdorff distance $d_H(F_n,F_{n+1})\le a_n$, where $a_n\ge 0$ and ${\sum }_n{a}_n<\infty$. Then $G={\bigcap }_n{F}_n$ is nonempty and closed, and $d_H(F_n,G)\le {\sum }_{k=n}^{\infty }{a}_k\to 0$.

To prove this, start with any $x_n\in F_n$ and choose $x_{k+1}\in F_{k+1}$ with $d(x_k,x_{k+1})<a_k+\eta 2^{n-k-1}$. The triangle inequality makes the resulting sequence a Cauchy sequence. Its limit exists by completeness and belongs to every $F_j$ by nesting and closedness. Its distance from the starting point is at most ${\sum }_{k=n}^{\infty }{a}_k+\eta$. Let $\eta \downarrow 0$ and take the supremum over starting points. The reverse directed distance is zero since $G\subseteq F_n$. Unlike the usual shrinking-diameter form of Cantor's intersection theorem, the sets need not have diameters tending to zero.

For a nonempty set $A\subseteq \mathbb{R}$, a real number $s$ is its supremum precisely when it is an upper bound and no smaller real number is an upper bound. It is unique: two least upper bounds must each be no greater than the other.

For $A=(0,1)$, every number at least one is an upper bound, and none smaller than one bounds the whole set. Thus $\sup A=1$, although $1\notin A$. For $A=[0,1]$, that same supremum is also the maximum. The approximation property of a supremum supplies witnesses arbitrarily close to an unattained upper edge.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.

For $D=\{(x,y):x^2+y^2<1\}$, the product order gives least upper bound $(1,1)$, because each coordinate has supremum one. The lexicographic order instead has no least upper bound. Every point $(1,v)$ is an upper bound, but $(1,v-1)$ is a strictly smaller upper bound; a point with first coordinate below one cannot bound the disc. The same geometric set therefore has different order-completeness behavior under the two partial orders. Neither conclusion is determined by the usual drawing alone.

Let $A,B$ be nonempty sets of real numbers bounded above. Their Minkowski sum is $A+B=\{a+b:a\in A,b\in B\}$, allowing every pair from the Cartesian product $A\times B$. ThenThe sum of the two suprema is an upper bound for every permitted sum. For the reverse bound, use the approximation property of a supremum to choose $a>\sup A-\epsilon /2$ and $b>\sup B-\epsilon /2$. Their sum is within $\epsilon$ of the claimed value. This reasoning does not require either supremum to be a maximum.

Separate choices matter. On $X=[0,1]$, choose the functions $f(x)=x$ and $g(x)=1-x$. Each has supremum one, but their sum at a shared input is always one:The maximum of $f$ occurs at $x=1$, while that of $g$ occurs at $x=0$. A common input cannot realize both choices. With separate inputs, choosing $x=1$ and $y=0$ gives $f(x)+g(y)=2$. The allowed pairs of output values are the entire unit square for separate inputs, but only the line segment $a+b=1$ for the shared input.

For any two bounded functions on the same nonempty domain of a function, the useful general estimate isThe inequality can be strict, as the example shows. Equality does not require attained maxima, but it does require that the two large values can be approached at compatible inputs. For example, $f(x)=g(x)=x$ on $(0,1)$ gives equality with value two, although neither function reaches its supremum. This distinction is useful when an objective function combines terms sharing a decision variable: adding the best possible termwise scores gives an upper bound, not necessarily a feasible combined score.

For $x(t)=A\cos {\omega }_{1}t+B\cos {\omega }_{2}t$ with irrational frequency ratio and nonzero amplitudes, $|x|\le |A|+|B|$. The phase trajectory is dense on the two-torus, so this bound is a supremum. The required exact phase alignment may never occur at finite time. Thus an exam's maximum-amplitude estimate can be a limiting bound rather than an attained displacement; finite observation intervals have smaller maxima.

The tail diameter of a sequence in a metric space is $D_N=\sup \{d(x_m,x_n):m,n\ge N\}$, allowing $+\infty$. The bounds are nonincreasing as $N$ grows because fewer pairs remain. A sequence is a Cauchy sequence exactly when $D_N\to 0$. The supremum need not be attained: for $q_n=1-2^{-n}$, $D_N=2^{-N}$ even though no two finite terms achieve that difference. For $(-1{)}^n$, $D_N=2$ for every $N$. For the harmonic numbers, $D_N=+\infty$ for every $N$, because each remaining tail is unbounded.