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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]

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 a map $f:\mathbb{N}\to \mathbb{N}$ from the cofinite topology to the tail topology on the natural numbers, continuity is equivalent to every set $\{n:f(n)\ge k\}$ being empty or cofinite. If the image is unbounded, every such set is nonempty and cofinite, so $f(n)\to \infty$. If it is bounded, its maximum $M$ exists, and the inverse image of the tail at $M$ forces $f(n)=M$ eventually, with $f(n)\le M$ always. Conversely either condition makes every tail preimage empty or cofinite.

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.

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.

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 $Z$.

On an infinite set, the family consisting of all finite subsets and the whole set need not be closed under arbitrary unions. On $\mathbb{Z}$, the union of the singleton subsets of $2\mathbb{Z}$ is infinite and proper, so this family is not a topology. It must not be confused with the cofinite topology, whose proper open sets have finite complements.

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 topological space is hyperconnected if any two nonempty open sets intersect. Every continuous map from a nonempty hyperconnected space to a Hausdorff space is constant: two distinct image points would have disjoint open neighbourhoods whose nonempty inverse images contradict hyperconnectedness. Every infinite set with the cofinite topology is hyperconnected.

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 $D$.

For the cofinite topology ${\tau }_1$, the empty set and $\mathbb{N}$ are included. A nonempty union of cofinite sets is cofinite because its complement is contained in the finite complement of any one member. A finite intersection is cofinite because its complement is a finite union of finite sets; intersections involving the empty set are empty. This proves all topology axioms.

For the tail topology on the natural numbers ${\tau }_2$, $\mathbb{N}=I_1$. Any nonempty collection of tails has a smallest starting index, so its union is that tail; a finite intersection of tails is the tail with largest starting index. Empty sets and empty-family conventions supply the remaining cases. Thus ${\tau }_2$ is also a topology.

We classify continuous maps between cofinite and tail topologies. Continuity is equivalent toIf $f$ has unbounded image, each $E_k$ is nonempty, hence cofinite. For every threshold $k$, eventually $f(n)\ge k$, which is exactly $f(n)\to \infty$. If the image is bounded, it is a nonempty finite subset of $\mathbb{N}$ and has a maximum $N$. Then $E_N$ is nonempty and cofinite, but $f(n)\le N$ always, so $f(n)=N$ except at finitely many indices. These exhaust the two possibilities and prove necessity. Their sufficiency is checked separately below.

There is a small defect in the printed definition: the listed family omits $\mathbb{R}$ and is not a topology, since ${\bigcup }_{n\ge 1}(-n,\infty )=\mathbb{R}$. We use the intended upper-ray topology on the real line, obtained by adjoining $\mathbb{R}$ to that family.

Take $f(n)=n$. For every real $a$,has finite complement and is therefore open in the cofinite topology. The inverse images of the empty set and the whole line are also open. Hence this nonconstant function is a continuous map for the corrected upper-ray topology on the real line. The same inverse-image calculation verifies the condition for every ray literally listed in the question.

Any two nonempty open sets in the cofinite topology on $\mathbb{N}$ intersect: their complements have finite union, which cannot exhaust $\mathbb{N}$. Thus this is a hyperconnected space. Suppose a continuous map $f$ took two distinct values $a$ and $b$. The usual real line is a Hausdorff space, so there are disjoint open intervals $U$ and $V$ containing $a$ and $b$. Their inverse images under the continuous map are disjoint, nonempty open sets, contradicting hyperconnectedness. Every such $f$ is constant.

Include the empty set, and take all unions of intervals $[a,b)$ with $a<b$. These intervals cover the real line, and the intersection of two is either empty or $[\max (a,c),\min (b,d))$. Distributing intersections over unions therefore makes finite intersections of open sets open; arbitrary unions are open by construction. The whole line is open, so the axioms of a topology hold. This is the lower limit topology, whose space is the Sorgenfrey line.

If $x<y$, the neighborhoods $[x,y)$ and $[y,y+1)$ are disjoint. Hence $\boxed{(\mathbb{R},{\tau }_1)\text{ is Hausdorff}}$.

For the identity into the cofinite topology, an inverse image is the same subset of $\mathbb{R}$. Let $F$ be finite and $x\notin F$. If $F$ has a point greater than $x$, choose $b$ to be its least such point; otherwise take $b=x+1$. Then $[x,b)$ avoids every point of $F$, including $b$ itself. Taking these intervals over all $x\notin F$ writes $\mathbb{R}\setminus F$ as a ${\tau }_1$-open set. The inverse image of every cofinite open set, and of the empty set, is open. The identity map is continuous, because ${\tau }_2\subseteq {\tau }_1$.

This does not define a topology on $\mathbb{Z}$. Every singleton $\{2k\}$ is declared open, but their union over $k\in \mathbb{Z}$ is $2\mathbb{Z}$, an infinite proper subset. It is therefore not declared open, violating closure under arbitrary unions. This is the example that finite subsets do not define a topology on an infinite set. The empty set, whole space and finite-intersection requirements do hold, but they do not compensate for the failed union axiom. This is different from the cofinite topology, which uses finite complements rather than finite sets.

If an open set meets $\overline{A}$, it meets $A$. Thus if two open sets meet the closure of an irreducible set, they both meet $A$, their intersection meets $A$ by irreducibility, and therefore it also meets $\overline{A}$. This proves irreducibility of the closure. A singleton is irreducible, so its closure is an irreducible closed subset.

In a Hausdorff space, two distinct points have disjoint open neighborhoods. An irreducible set cannot contain both, so every nonempty irreducible closed subset is a singleton, which is closed and has a unique generating point. Hence every Hausdorff space is a sober topological space. An infinite set with the cofinite topology is itself irreducible, because any two nonempty open sets meet, but every singleton is closed. Its whole space is therefore not the closure of a singleton, proving that it is not sober.

For the proposed sobrification, let $\hat{X}$ consist of nonempty irreducible closed subsets. The basic operations areThe last identity uses irreducibility of each member of $\hat{X}$. Hence these sets already form a topology. The map $U\mapsto \hat{U}$ is injective: $\overline{\{x\}}$ meets an open $U$ exactly when $x\in U$, so equality of the hats forces equality of the original opens. It is surjective onto the topology just constructed by definition.

Now let $C$ be a nonempty irreducible closed subset of $\hat{X}$. Its complement is $\hat{W}$ for a unique open $W\subseteq X$. Put $F=X\setminus W$. ThenIf opens $U,V$ meet $F$, choose points in those intersections. Their singleton closures lie in $C$, so $\hat{U},\hat{V}$ meet $C$. Irreducibility of $C$ forces $\widehat{U\cap V}$ to meet $C$, and hence $U\cap V$ meets $F$. Thus $F$ is nonempty, closed and irreducible, so $F$ is itself a point of $\hat{X}$.

The closure of this point in $\hat{X}$ is exactly $\{A:A\subseteq F\}$: inclusion makes every open meeting $A$ meet $F$, whereas a point of $A\setminus F$ is separated by the open $X\setminus F$. Hence $C=\overline{\{F\}}$. Any other point $G$ with the same closure satisfies $G\subseteq F$ and $F\subseteq G$, forcing $F=G$. This proves the constructed space is sober.

Write $D(f)=\mathbb{C}\setminus f^{-1}(0)$. The zero and constant-one polynomials give $D(0)=\varnothing$ and $D(1)=\mathbb{C}$. Also,so $\tau$ is closed under finite intersections. Every nonzero one-variable complex polynomial has only finitely many roots, and every finite subset $\{a_1,\dots ,a_m\}$ is the zero set of ${\prod }_j(z-a_j)$. Consequently $\tau$ is exactly the cofinite topology: its open sets are the empty set and the complements of finite sets. An arbitrary union of such sets is again empty or has finite complement, so $\tau$ is a topology.

The product topology on $X\times Y$ is the topology with basisThe proposed complement need not be open. TakeThen $g^{-1}(0)$ is the diagonal. If its complement were open, a point such as $(0,1)$ would have a basic neighbourhood $U\times V$ contained in that complement. But $U$ and $V$ are nonempty cofinite subsets of $\mathbb{C}$, so $U\cap V\ne \varnothing$. For $t\in U\cap V$, the point $(t,t)$ belongs both to $U\times V$ and to the diagonal, a contradiction. Hence

A proper closed subset in the cofinite topology is finite, so cannot contain the infinite subset $A$. Its only closed superset is therefore the whole space:In particular, every infinite subset is dense in this topology.

For a nonempty open subset $U$ of the quotient, $p^{-1}(U)$ is nonempty and open in the cofinite space. Thereforeis finite. Surjectivity implies that $(X/\mathcal{R})\setminus U$ is finite as well: it is the image of that finite preimage. Hence every quotient-open set is cofinite or empty, proving the desired inclusion.

Equality is not necessary. Give $X=\mathbb{Z}$ its cofinite topology and use two equivalence classes, the even and odd integers. Neither singleton of the quotient is open, since its preimage and its complement are both infinite. The quotient is the two-point indiscrete topology, whereas the cofinite topology on a finite two-point set is discrete.This is the quotient topology of a cofinite space phenomenon, not a failure of continuity of the projection.

Every topological subspace $S\subseteq X$ again has the cofinite topology: its nonempty open subsets have finite complements in $S$, and every finite exclusion is obtained by intersecting with a cofinite open subset of $X$. As usual, the empty set is also open; this exception is implicit in the paper's shorthand definition.

If $S$ is infinite, any two nonempty relatively open sets intersect, since their complements are finite. Hence $S$ is hyperconnected space and therefore connected. If $S$ is finite, its cofinite topology is discrete topology. It is then connected exactly when it has at most one point. ThusThere are also infinite disconnected subsets of the product. Choose distinct $a,b\in X$ and put $D=\{a,b\}\times X$. This is infinite. The traces on $D$ of the ambient open setsare respectively $\{a\}\times X$ and $\{b\}\times X$. They are disjoint, nonempty and cover $D$, giving a separation. This answers the infinite-subset question, rather than only producing two isolated product points.