Cofinite topology Created 2026-09-29 Updated 2026-10-14
The cofinite topology on a set 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 and chooseBoth are open sets, and their intersection is . Thus 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.
Two open sets with finite exclusions still overlap in an infinite cofinite topology
. Original sets on . Only the first sixteen labels are drawn; the dots continue indefinitely and are all included after the last excluded label. Spacing is for readability and defines no metric. Filled circles belong to the displayed set; crosses are excluded. Details.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]
Chapter 1: Topological spaces. Example 1.5.10 and Proposition 1.5.11: closed points, overlapping neighbourhoods, non-repeating sequences and uniqueness in Hausdorff spaces
Finite union Created 2026-10-14 Updated 2026-10-25
A finite union is a union of finitely many sets. The indexing set is finite; its member sets need not be finite sets. The union of an empty indexed family is the empty set.
A metric space is a set with a function satisfying , symmetry and the triangle inequality. A subset is open if every has some open open ball contained in .
For finitely many open sets and in their intersection, choose radii with . The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is , also open.
For distinct points put . The balls and are disjoint: a common point would make by the triangle inequality. This proves the Hausdorff condition.
A singleton is closed because each has a ball of radius avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.
Any two nonempty open sets of are cofinite by part (iii). Their complements have finite union, so the two sets must intersect because is infinite. Thus is a hyperconnected space.
A Hausdorff space with at least two points has disjoint nonempty open neighbourhoods of two distinct points. This is impossible here. Therefore is not Hausdorff.
Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is a finite union of finite sets and cannot be all of . Thus is a hyperconnected space.
If the continuous map took distinct values at , choose disjoint open neighbourhoods of in the Hausdorff space . By continuity, and would be nonempty disjoint open sets of , a contradiction. ThereforeThe proof establishes the general fact that continuous maps from an infinite cofinite space to a Hausdorff space are constant.
Pasting lemma for closed subspaces Created 2026-10-03 Updated 2026-10-31
If a topological space is the union of finitely many closed subspaces and continuous maps on those subspaces agree on every overlap, they combine into a continuous map on the whole space. For a closed set in the target, its preimage is a finite union of closed sets, proving continuity.
