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If distinct points $x,y$ of a dynamical system on a compact metric space satisfy $d(T^{n}x,T^{n}y)\to 0$ in one compatible metric, no compatible metric can make $T$ an isometry. On a compact space, all compatible metrics give the same asymptotic-pair property by uniform continuity, whereas an isometry preserves the strictly positive distance between distinct points. In a full shift, a constant sequence and a sequence differing at just one coordinate converge to each other under forward shifts, proving this obstruction directly in the product topology.

A bad sequence in a preorder is an infinite sequence with no indices $i<j$ satisfying $x_i\le x_j$. A preorder is a well-quasi-ordering precisely when no such sequence exists.

The set-theoretic Baire space is ${\omega }^{\omega }$, the set of all infinite sequences of natural numbers, with the product topology obtained from discrete $\omega$.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

If $(a_n)$ is a nonnegative decreasing sequence, then ${\sum }_{n\ge 1}{a}_n$ converges if and only if the condensed series ${\sum }_{k\ge 0}{2}^k{a}_{2^k}$ converges. Comparing each dyadic block with its first and last terms proves both directions.

If a sequence converges, its Cesaro means have the same limit. Given a small tolerance, split the average error into a fixed finite initial segment and a tail whose individual errors are below that tolerance. The first contribution is a fixed constant divided by $n$, while the second is bounded by the tolerance. The converse fails: the alternating sequence $(-1{)}^n$ has means tending to zero while its terms do not converge.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

A real sequence converges if and only if the displayed condition holds. A Cauchy sequence makes all future differences uniformly small, so the forward implication is immediate. Conversely, failure of the Cauchy sequence criterion produces a fixed tolerance and, for each $n$, a future term differing from $a_n$ by at least half that tolerance. Choosing the least such positive lag defines one function that contradicts the displayed condition. The completeness of the real numbers then gives convergence. Testing only every fixed integer lag is weaker: $a_n=\log n$ passes those tests but diverges.

If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.

An exponential generating function encodes a sequence using the displayed factorial denominators. It is especially useful for counting labelled combinatorial structures: taking an unordered set of structures with positive size corresponds to exponentiating their exponential generating function. The rooted-tree generating function illustrates this through $T=z{e}^T$.

Given a size bound with finitely many possible objects at each position, put finite bad sequences in a tree ordered by extension. The tree is finitely branching. Arbitrarily long bad sequences would give an infinite bad sequence by König infinity lemma. This compactness argument converts an infinite well-quasi-ordering theorem into a uniform finite length bound.

A finite repetition-free sequence in a set $X$ is an injective function from an initial segment $\{0,\dots ,k-1\}$ of the natural numbers into $X$, including the empty sequence for $k=0$. For a finite set of size $m$, the number of these sequences isThe order of each list is part of the data, so it can be enumerated without choosing an order on its underlying support.

If $X$ is a Dedekind-finite set, its set of finite repetition-free sequences is also Dedekind-finite. An injective sequence of distinct finite lists would, by countable union of explicitly ordered finite lists without choice, either produce an injection $\mathbb{N}\to X$ or use only finitely many entries. The latter possibility is impossible because a fixed finite pool supports only finitely many repetition-free lists. If $X$ is infinite, the one-entry lists also show that the resulting set is infinite.

For every natural parameter $k$, there is a finite length $N$ such that every sequence of $N$ finite rooted trees satisfying $|T_i|\le k+i$ has an earlier tree homeomorphically embedding into a later tree. It is a true finite-combinatorial principle unprovable in Peano arithmetic and in arithmetical transfinite recursion theory. Each fixed-parameter instance can still be verified by a sufficiently large finite search.