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A tail of a sequence consists of its terms from some index $N$ onward, retaining their order and repetitions. Discarding finitely many initial terms does not change its limit of a sequence or whether it is a Cauchy sequence. A tail can also be considered as the set of values $\{x_n:n\ge N\}$ when forming an infimum, supremum, or tail diameter of a sequence; that set alone forgets the original order and repetitions.

Total variation distance asks how much two probability distributions can disagree about the same event. For probability measures $P,Q$ on one measurable space $(S,\mathcal{F})$, its definition isThe supremum ranges over all measurable sets, so $d_{\mathrm{TV}}(P,Q)\le \epsilon$ guarantees that every yes-or-no prediction has probability error at most $\epsilon$. This metric takes values between zero and one; it is zero exactly when the two probability measures agree. This makes it useful for comparing a sampling algorithm's output with its target probability distribution, and for defining the mixing time of a Markov chain.

On a finite set, add the excess probability mass wherever $P$ exceeds $Q$. That excess is also half the sum of all absolute differences:The last expression reveals the common probability mass that the two probability distributions can share.^[ref] The coupling characterization of total variation distance turns this overlap into an actual joint construction; minimum sum of errors in a simple hypothesis test gives it a statistical interpretation. One quantity connects probability bookkeeping, comparison of sampling methods, and the limits of a statistical test.

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

For a path of finite variation, its total-variation process iswhere the supremum is over every partition of an interval of $[0,t]$. It is an increasing process, and the signed measure $dA$ satisfies $|dA|=dV(A)$.

A sequence of functions $f_n:X\to Y$ converges uniformly to $f:X\to Y$, where $Y$ is a metric space with metric $d$, precisely whenFor real numbers, use the absolute value $d(u,v)=|u-v|$. Compare the universal quantification and existential quantification:The decisive change is whether $N$ can depend on $x$. Uniform convergence implies pointwise convergence.^[ref]

For functions taking values in real numbers on a nonempty domain of a function, the supremum error is $E_n={\sup }_{x\in X}|f_n(x)-f(x)|$, allowing $+\infty$. Then uniform convergence is equivalent to $E_n\to 0$. The supremum need not be a maximum: pointwise strict bounds below $\epsilon$ only give $E_n\le \epsilon$, so use tolerance $\epsilon /2$ when proving this equivalence. This single error bound controls every possible input, including one chosen after seeing the index.

For $f_n(x)=x^n$ and its pointwise limit on $[0,1]$, the error is zero at one but approaches one as $x\to 1$ from below. Consequentlyfor every positive integer $n$. This supremum is never attained: below one the error is $x^n<1$, and at one it is zero. Deleting the endpoint leaves the same supremum error on $[0,1)$. To disprove uniform convergence directly, choose $\epsilon =1/2$: for any proposed $N$, take $n=N$ and $x=2^{-1/N}<1$. The error equals $1/2$, violating the strict tolerance. Thus a moving input defeats every proposed common threshold.

On the smaller interval $[0,a]$, for a fixed $0\le a<1$, the pointwise limit is zero andFor $0<a<1$ and $0<\epsilon <1$, the exact first successful common threshold isThe logarithm and floor function turn an infinite approximation statement into a finite stopping rule. At $a=0.8$, $0.8^{20}=0.01152921504606846976$ fails tolerance $0.01$, while $0.8^{21}=0.009223372036854775808$ succeeds. At $a=1/2$ and $\epsilon =1/8$, the ratio is exactly three: $N=4$ is necessary because equality is insufficient. For $a=0$, every error is zero. The domain of a function determines whether one threshold can certify every input.

For another explicit stopping rule, let $h_n(x)=x/(n+1)$ on $[0,1]$. Its uniform limit is zero and its supremum error is $1/(n+1)$. For every $\epsilon >0$, $N=\max \{1,\lceil 1/\epsilon \rceil \}$ works, using the ceiling function. Unlike $x^n$ on $[0,1]$, it controls even the endpoint.

The practical beauty of uniform convergence is that one quantitative bound can survive further operations: the uniform limit theorem preserves continuity, and uniform convergence preserves integrals on a compact interval controls accumulated error. Differentiation needs additional hypotheses, such as uniform derivative convergence with an anchored value. These results distinguish an approximation that looks accurate at selected inputs from one that justifies replacing a whole function in a calculation.

For a bounded function $f$ and a partition of an interval $P$, the upper Darboux sum multiplies the supremum of $f$ on each subinterval by its length and adds the results.

A value function assigns each dynamic programming state and time the best objective function value attainable over the remaining decisions. It is a score, whereas a control policy specifies the actions producing that score. In the 0-1 knapsack problem above, $V_3(5)=7$ is the score, and choosing the items of weights $2$ and $3$ attains it. In sequential mathematical optimization, the Bellman equation computes a value function by comparing immediate reward plus continuation value. Use a supremum or infimum when the best score need not be attained by any admissible control policy.

For a nonempty bounded array of real numbers, each column infimum is at most every row supremum, because ${\inf }_k{x}_{k,n}\le x_{m,n}\le {\sup }_j{x}_{m,j}$ for all $m,n$. Taking the supremum over columns and then the infimum over rows proves the displayed inequality. Equality need not hold: for $x_{m,n}=1_{\{m=n\}}$ on positive integer indices, every column infimum is zero and every row supremum is one. This elementary bound is the order-theoretic starting point for stronger minimax theorems with additional hypotheses.