Cofinite topology (${\tau }_{\mathrm{cof}}=\{\varnothing \}\cup \{X\setminus F:F\subseteq X\text{ finite}\}$)

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]

Let $Z$ be the underlying set of real numbers with the cofinite topology. The diagonal $W=\{(t,t):t\in \mathbb{R}\}$ in $Z\times \mathbb{R}$ with its subspace topology is homeomorphic to the usual real line. Indeed, $t\mapsto (t,t)$ is a continuous bijection: its first coordinate is continuous because usual finite sets are closed, and its second coordinate is the identity. The inverse is the second coordinate projection restricted to $W$. Consequently $W$ is not compact, whereas $Z$ is compact by the fact that every subset of a cofinite space is compact. Thus $W$ is not homeomorphic to $Z$.

For a set $X$ with its cofinite topology, a subset $S$ is a connected subset exactly when $S$ is empty, a singleton, or infinite. Indeed, its subspace topology is the cofinite topology on $S$: the traces of open sets are precisely $S\setminus F$ for finite $F\subseteq S$, together with the empty set. For infinite $S$, two nonempty open sets intersect, because their complements have finite union. Hence $S$ is a hyperconnected space and therefore a connected space. A finite $S$ has the discrete topology, which admits a separation of a topological space as soon as $S$ has at least two points.

For an infinite set $X$ with its cofinite topology, every infinite subset is connected, but $X\times X$ has an infinite disconnected topological subspace. Choose $a\ne b$ and use $D=\{a,b\}\times X$. The open sets $(X\setminus \{b\})\times X$ and $(X\setminus \{a\})\times X$ have disjoint nonempty traces $\{a\}\times X$ and $\{b\}\times X$ on $D$, covering it. Thus $D$ is a product of a discrete two-point subspace with an infinite cofinite factor. The ambient open rectangles may intersect; only their traces must be disjoint for a separation of $D$.

By convergence of a sequence in a topological space, $x_n\to a$ means that every open neighbourhood of $a$ contains all sufficiently late terms. In the cofinite topology, this is equivalent to the condition that, for every $b\ne a$, the value $b$ occurs only finitely often. Necessity uses the open neighbourhood $X\setminus \{b\}$; sufficiency chooses an index beyond every last occurrence of the finitely many points excluded by a given open neighbourhood.

Consequently a sequence with distinct terms converges to every point of $X$. For the original choice $x_n=n$ on $X={\mathbb{N}}_{>0}$, take the two open neighbourhoodsAll terms with $n\ge 9$ lie in $U_1$ and all terms with $n\ge 8$ lie in $U_2$. More generally, each finite excluded set has a last visit, so $x_n\to 1$ and $x_n\to 2$, and in fact $x_n\to a$ for every $a\in X$. No numerical distance is tending to zero here; the open sets specify the meaning of convergence.

Distinct terms are sufficient, not necessary. The sequence $1,1,2,2,3,3,\dots$ also converges to every point, since each value occurs only twice. By contrast, the constant sequence $x_n=1$ converges only to $1$: an open neighbourhood of any $a\ne 1$ can exclude $1$. The alternating sequence $1,2,1,2,\dots$ converges to no point, since some value different from any proposed limit occurs infinitely often.

Every subset $A$ of a cofinite topological space is a compact subset. To see the finite reduction directly, take an open cover of a nonempty $A$ by open sets of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite complement, the set $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a finite subcover. The empty set is already compact.

Equivalently, the subspace topology on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus (F\cap A)$. The argument applies again inside $A$. A concrete open cover $\{X\setminus \{k\}:k\in X\}$ of $X={\mathbb{N}}_{>0}$ needs only the two members $X\setminus \{1\}$ and $X\setminus \{2\}$ when these labels are distinct: each fills the other's single gap.

On $X={\mathbb{N}}_{>0}$, the subset of even integers is compact but not closed, because it is infinite and proper. A compact subset of a Hausdorff space is guaranteed to be closed, whereas compactness without the Hausdorff space assumption does not suffice. Changing the topology changes compactness: the same infinite set with its discrete topology has the singleton open cover, which has no finite subcover. The finite-exception argument is special to finite complements; in the cocountable topology, compact subsets of a cocountable space must instead be finite.

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

Let $Z$ be a nonempty infinite set with the cofinite topology, let $T$ be any topological space, and let $Y$ be Hausdorff. For a continuous map $F:Z\times T\to Y$, every slice $z\mapsto F(z,t)$ is constant. Thus $F(z,t)=g(t)$, where $g(t)=F(z_0,t)$ for any fixed $z_0$. The map $g$ is continuous, since $t\mapsto (z_0,t)$ is continuous into the product topology. It is unique because the coordinate projection ${\pi }_T$ is a surjection. If $F$ is a surjection, so is $g$.

For $T=[0,1]$ and $Y$ homeomorphic to the circle, a surjective $g$ cannot be injective: otherwise the compact-to-Hausdorff continuous bijection theorem would contradict the fact that the circle and interval are not homeomorphic. Hence distinct $a,b\in [0,1]$ satisfy $F(z,a)=F(z,b)$ simultaneously for all $z\in Z$.

A continuous map from a nonempty finite product space of infinite sets with their cofinite topologies to a Hausdorff space is a constant function. Fixing all but one coordinate gives a continuous map from an infinite cofinite topology, so changing that coordinate does not change the value. Any two points in a finite Cartesian product can be joined by finitely many one-coordinate changes. This proves the assertion. In particular every real-valued continuous function on $Z\times Z$ is constant when $Z$ is the real line equipped with the cofinite topology.