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The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

Given a set $U$ and a family of its subsets, an exact cover selects members of that family so that each element of $U$ belongs to exactly one selected member. Thus the selected members are disjoint sets whose set union is $U$. Anchoring a recursive choice at the smallest uncovered element avoids counting a cover in different orders.

For fair division, let $J=\{1,\dots ,m\}$ and let the additive valuation of agent $i$ be $v_i(S)={\sum }_{j\in S}{v}_{ij}$. Use one variable $X_i$ per agent, with constraint domain $\mathcal{P}(J)$, the power set of the items. Empty bundles are permitted in this original formulation.

Require pairwise disjoint sets:Each such binary constraint involves exactly $X_i,X_{\ell }$. Require complete allocation by the single all-variable constraintFinally impose the envy-free allocation binary constraintseach on $X_i,X_{\ell }$. The two directions for a pair use different agents' additive valuations and must both hold. These conditions give precisely an ordered set partition of the items that is an envy-free allocation.

The events $B_1,B_2,\dots$ form a partition of a sample space $\Omega$ when they are pairwise disjoint sets and cover the whole sample space:Thus exactly one of these events occurs on every outcome. Under the usual set partition convention the parts are nonempty; empty parts can simply be discarded.

The set difference consists of elements belonging to the first set but not the second:The symmetric difference consists of elements belonging to exactly one of the two sets:In particular, the two sets in this union are disjoint sets.

Because $A\subseteq \mathcal{U}$, the set difference $A\setminus \mathcal{U}$ is empty and $\mathcal{U}\setminus A=\overline{A}$. Therefore the symmetric difference satisfiesThe disjoint sets $A$ and $\overline{A}$ have union $\mathcal{U}$, so

The fifty disjoint sets $\{1,100\},\{2,99\},\dots ,\{50,51\}$ form a partition of a set $\{1,\dots ,100\}$. If a chosen subset met each pair at most once, it would contain at most fifty elements. The pigeonhole principle therefore forces a $51$-element subset to contain both elements of at least one pair. Those two distinct elements sum to $101$.

The relative complement consists of the elements of the first set absent from the second, while the symmetric difference consists of elements in exactly one set:The two parts of the symmetric difference are disjoint sets.

A finite set $A$ of positive integers such that nonempty disjoint sets contained in $A$ have different sums, and the larger subset has the larger sum when their sizes differ. For $a_1<\cdots <a_n$, the size condition is equivalent to $a_1+\cdots +a_{k+1}>a_{n-k+1}+\cdots +a_n$ for $1\le k\le \lfloor (n-1)/2\rfloor$: the left side is the smallest possible sum of $k+1$ elements and the right side the largest sum of $k$ elements.