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A point of a topological space is closed when its singleton is a closed set. In a T1 space every point is closed. The closed points of an affine scheme $\mathrm{Spec}A$ are its maximal ideals; the closure of the point corresponding to $\mathfrak{p}$ is $V(\mathfrak{p})$.

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 $X$ be a Hausdorff space, let $K_n$ be a nested sequence of nonempty compact subsets of $X$, and let $f:X\to Y$ be a continuous map into a T1 space. Then the displayed equality holds. The forward inclusion is true for every map. For the reverse, fix $y\in {\bigcap }_{n}f(K_n)$. Each $F_n=K_n\cap f^{-1}(\{y\})$ is nonempty and closed in the compact set $K_1$: the $K_n$ are closed by the Hausdorff assumption, while the fiber is closed because $Y$ is T1 and $f$ is continuous. The nested sets $F_n$ have the finite intersection property, so compactness gives a point in their intersection. Its image is $y$.

The proof only needs the relevant fibers to be closed in $K_1$, rather than all of $Y$ to be T1. A Hausdorff space as codomain is a familiar sufficient special case. Mere continuity into an arbitrary space does not suffice.

The extension. Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal topological space can be separated by a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$.

For the bounded Tietze extension theorem, first suppose $g:C\to [-M,M]$ is continuous. The subsets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are closed in $X$, because $C$ is closed. Urysohn's lemma supplies a continuous function $h:X\to [-M/3,M/3]$ with $h=-M/3$ on $A$ and $h=M/3$ on $B$. Directly checking the two outer ranges and the middle range of $g$ givesEmpty level sets cause no difficulty: take the appropriate constant, or zero if both are empty. This is the Urysohn two-thirds approximation.

Apply this successively to the residual on $C$, beginning with $f$ and $M_0=1$. We get $h_n$ with $\Vert h_n{\Vert }_{\infty }\le \frac{1}{3}(\frac{2}{3}{)}^n$ and residual after the first $n$ terms at most $(2/3{)}^n$. The Weierstrass M-test gives uniform convergence of $F={\sum }_{n\ge 0}{h}_n$. By the uniform limit theorem, $F$ is continuous; the residual bound proves $F{|}_C=f$, and the sum of the norm bounds is one. Thus

Closed sets as exact zero sets. If $X$ is a perfectly normal space, write a closed $C$ as ${\bigcap }_{n\ge 0}{U}_n$ with each $U_n$ open. Apply Urysohn's lemma to $C$ and $X\setminus U_n$ to obtain $f_n:X\to [0,1]$ vanishing on $C$ and equal to one on $X\setminus U_n$. The uniformly convergent seriesIndeed all summands vanish on $C$; outside $C$, at least one summand is strictly positive. The uniform limit theorem proves continuity. This realizes the closed G-delta set as a zero set construction.

Conversely, suppose each closed set is the zero set of a continuous function. For disjoint closed $C,D$, select $f,g:X\to [0,1]$ with zero sets $C,D$. Their sum never vanishes, andThe quotient is continuous. This also proves normality by inverse images of disjoint endpoint neighborhoods. Each zero set is a G-delta set, since $f^{-1}(0)={\bigcap }_{n\ge 1}{f}^{-1}([0,1/n))$; thus this property characterizes perfectly normal spaces among T1 spaces.

Subspaces. Let $Y\subseteq X$ and let $A,B$ be disjoint closed subsets of $Y$. Put $C={\overline{A}}^X$ and $D={\overline{B}}^X$. Since $C\cap Y=A$ and $D\cap Y=B$, the intersection $C\cap D$ misses $Y$. Choose continuous functions $f,g$ on $X$ whose zero sets are $C,D$. On the open neighborhood $W=X\setminus (C\cap D)$ of $Y$, the same quotient $f/(f+g)$ is continuous and is zero on $A$, one on $B$. Its inverse images of $[0,1/3)$ and $(2/3,1]$, restricted to $Y$, are disjoint open neighborhoods. Therefore every subspace is a normal topological space: perfect normality implies hereditary normality.

A strict counterexample to the converse. Let $D$ be an uncountable discrete space, and let $X=D\cup \{\infty \}$ be its one-point compactification. Its neighborhoods of $\infty$ have finite complement in $D$, because compact subsets of a discrete space are finite. The space is compact Hausdorff, hence a T1 space and normal. A subspace missing $\infty$ is discrete. A subspace containing $\infty$ is closed in $X$, since its complement is a subset of the open discrete part; it is therefore compact Hausdorff and normal. Thus all subspaces are normal.

But $\{\infty \}$ is not a G-delta set. A countable intersection of its open neighborhoods omits at most countably many elements of $D$, and cannot omit all of uncountable $D$. Hence $X$ is not a perfectly normal space. This is the uncountable discrete one-point compactification counterexample.

A zero-dimensional space has a base of clopen sets. A totally disconnected space has no connected subset with more than one point, equivalently all its connected components are singletons. We use these definitions in the stated Hausdorff setting.

If a Hausdorff space has a clopen base, any two different points can be separated by a clopen set. A connected subset containing both would be split into two nonempty disjoint relatively open sets. Thus zero-dimensional implies totally disconnected here.

The compact converse, including the component lemma. For a point $x$ of a compact Hausdorff $K$, putThis is its quasicomponent, a closed compact set. We prove the Šura-Bura component lemma that $Q_x$ equals the connected component of $x$.

Suppose $Q_x=A\cup B$ were a separation into nonempty disjoint relatively closed sets, with $x\in A$. Both are compact, hence closed in $K$. Normality of compact Hausdorff spaces gives disjoint open neighborhoods $U\supseteq A$, $V\supseteq B$. The compact set $H=K\setminus (U\cup V)$ misses $Q_x$. For each $y\in H$, choose a clopen neighborhood $N_y$ of $x$ missing $y$; these exist by the definition of $Q_x$. The open complements cover $H$, so finitely many suffice. Their intersection $N$ is clopen, contains $Q_x$, and lies in $U\cup V$. If $H$ is empty take $N=K$.

Now $N\cap U$ is clopen in $K$: it is open, and its complement is $(K\setminus N)\cup (N\cap V)$, also open. It contains $x$ and misses $B$, contradicting that $B\subseteq Q_x$ lies in every clopen neighborhood of $x$. Hence $Q_x$ is connected. Conversely every connected subset containing $x$ lies in every clopen neighborhood of $x$, so the connected component is contained in $Q_x$. Equality follows.

If $K$ is totally disconnected, then $Q_x=\{x\}$. Given an open neighborhood $O$ of $x$, for each $y\in K\setminus O$ choose a clopen neighborhood of $x$ missing $y$. Compactness again gives a finite intersection $N$ with $x\in N\subseteq O$. These $N$ form a clopen base. Therefore

The Stone-Čech characterization. The stated clopen separation property is strong zero-dimensionality in this normal Hausdorff setting. A normal T1 space is a Tychonoff space, so its Stone-Čech compactification exists. We use its extension property for maps into compact Hausdorff spaces and the density of $X$.

First assume $\beta X$ is a zero-dimensional space, and let $C,D$ be disjoint closed subsets of $X$. Urysohn's lemma gives $f:X\to [0,1]$ equal to zero on $C$ and one on $D$. Its continuous extension to $\beta X$ puts ${\overline{C}}^{\beta X}$ and ${\overline{D}}^{\beta X}$ in disjoint zero and one level sets. Using a clopen base, cover the first compact set by finitely many clopen sets missing the second; their union $H$ is clopen and separates them. Thus $U=H\cap X$ and $V=X\setminus U$ are disjoint clopen sets containing $C,D$.

Conversely, assume the stated clopen separation property on $X$. Let $A,B$ be disjoint closed subsets of $\beta X$. Normality supplies a continuous function $f:\beta X\to [0,1]$ with $f{|}_A=0$, $f{|}_B=1$. The setsare disjoint and closed in $X$. Select clopen $U\subseteq X$ containing $C$ and missing $D$; the given disjoint pair of clopen neighborhoods supplies such a $U$. Its characteristic function, one on $U$ and zero outside, extends from $X$ to a continuous map $\beta X\to \{0,1\}$. Let $H$ be its one level set, a clopen set.

Every $a\in A$ is in ${\overline{C}}^{\beta X}$: any neighborhood of $a$, intersected with $f^{-1}([0,1/3))$, is an open neighborhood meeting the dense set $X$. Likewise $B\subseteq {\overline{D}}^{\beta X}$. Since $C\subseteq H$ and $D\subseteq \beta X\setminus H$, it follows that $A\subseteq H$ and $B\cap H=\varnothing$.

Apply this to $A=\{z\}$ and $B=\beta X\setminus O$, for any point $z$ and open neighborhood $O$. It yields a clopen $H$ with $z\in H\subseteq O$, proving that $\beta X$ has a clopen base. Hence the strong zero-dimensionality and Stone-Čech compactification criterion is

The claim is false for an arbitrary codomain $Y$. Let $X=[0,1]$ with its usual topology, let $Y$ be the Sierpiński space with open sets $\varnothing ,\{1\},Y$, and define $f(0)=0$, $f(x)=1$ for $x>0$. This is a continuous map: the only nontrivial open-set preimage is $f^{-1}(\{1\})=(0,1]$, which is relatively open in $X$.

The sets $K_n=[0,1/n]$ are nonempty, nested and compact. ButThus the extra generality in the printed codomain changes the assertion materially. This is the failure of nested compact image intersection without closed fibers.

If the codomain is a T1 space, the continuous images of nested compact intersections theorem does give the desired equality. The forward inclusion holds for every map. For the reverse, take $y\in {\bigcap }_{n}f(K_n)$ and set $F_n=K_n\cap f^{-1}(\{y\})$. The fibers are closed by continuity and the T1 property, so the $F_n$ are nested nonempty closed subsets of the compact set $K_1$. The finite intersection property gives $x\in {\bigcap }_n{F}_n$, with $f(x)=y$. It is enough that these particular fibers be closed in $K_1$; a Hausdorff codomain, as in the native self-map formulation, is a familiar sufficient assumption.

A T1 space is perfectly normal when it is normal and every closed subset is a G-delta set. Equivalently every closed subset is the zero set of a continuous function into $[0,1]$, using the closed G-delta set as a zero set theorem. If disjoint closed sets are the exact zero sets of $f,g$, then $f/(f+g)$ has those sets as its exact zero and one levels. Every subspace is normal, although the converse fails.

A separation axiom is a condition on a topological space expressing how points or closed subsets can be distinguished by open sets. The Kolmogorov space condition distinguishes distinct points by some open set; the T1 space condition makes every point closed; the stronger Hausdorff space condition gives disjoint neighbourhoods to distinct points.

A T3 space is a regular topological space which is also a T1 space. Every metric space has this property: singletons are closed, and a sufficiently small ball about a point outside a closed set has closure avoiding that set. A second-countable T3 space is a metrizable space by Urysohn's metrization theorem.