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Glue two copies of the affine line over a base along the principal open subset $D(t)$ by the identity. This scheme has two copies of the zero section and a canonical morphism of schemes to the original affine line. Over $\mathbb{Z}$, its functor of points assigns to a commutative ring $R$ the pairs $(r,e)$ with $e$ an idempotent in $R/(r)$. Indeed, $r$ gives the map to the affine line, while $e$ partitions $V(r)$ into the two clopen sets assigned to the zero sections. Complementing the opposite part gives two open subsets covering $\mathrm{Spec}R$, whose intersection is $D(r)$; the coordinate $r$ defines maps on them which glue. This also proves that no lifting of $e$ to an idempotent of $R$ is necessary.

Allow terminal positions of an infinite game of perfect information to carry a losing-player label. A clopen set of infinite winning plays still gives a determined infinite game. At a position where all infinite continuations have one winner, the other player can win only by forcing a terminal loss for that winner. The usual winning-position attractor in an infinite game decides this reachability game. The tree of remaining undecided positions is well-founded: every infinite branch would eventually have its clopen payoff decided by a prefix. Assign the already decided winners upwards by well-founded recursion. Terminal labels are essential; a terminal node created by a covering need not be lost by the player whose turn would have come next.

A clopen up-set is both clopen in the topology and upward closed in the order. The clopen up-sets of a Priestley space form a distributive lattice under intersection and union.

For $A\subseteq \mathbb{N}$, the basic set $\hat{A}=\{U:A\in U\}$ in the Stone-Čech compactification of the natural numbers is a clopen set, since its complement is $\widehat{\mathbb{N}\setminus A}$. Distinct ultrafilters are separated by complementary basic sets. A basic cover with no finite subcover would give complements with the finite intersection property, whose generated filter on a set extends to an ultrafilter missing the entire cover. Thus the space is a compact Hausdorff space.

For an infinite coinfinite set $A\subseteq \mathbb{N}$, let $E_A=[A{]}^{\omega }$ in the space of infinite subsets of the natural numbers. This cone is a nonempty clopen set in the Ellentuck topology, because it is the basic neighbourhood $[\varnothing ,A]$, and any $X\notin E_A$ has a finite initial segment containing a point outside $A$ and hence a basic neighbourhood disjoint from $E_A$.

The cone is a closed set with empty interior in the ordinary topology on infinite subsets: membership of a point outside $A$ witnesses its complement as an open set, while every ordinary finite-stem neighbourhood can be extended by an arbitrarily large point outside $A$. Thus $E_A$ is an ordinary nowhere dense set, but it is neither ordinary open nor Ellentuck nowhere dense. It provides both distinctions between these two topologies.

In the space of infinite subsets of the natural numbers, the families $C_n=\{X:n\notin X\}$ are clopen sets in the ordinary topology on infinite subsets and thus in the finer Ellentuck topology. Their union is $[\mathbb{N}{]}^{\omega }\setminus \{\mathbb{N}\}$. Every basic Ellentuck topology neighborhood of $\mathbb{N}$ contains $\mathbb{N}\setminus \{n\}$ for some sufficiently large $n$, so this union is not a closed set.

A normalized positive linear functional on $C(\Omega )$ for Cantor space gives a finitely additive set function on the algebra of clopen sets. If a countable disjoint union of such sets is itself clopen, compactness gives a finite subcover; all other terms are empty. Finite additivity therefore gives the required countable additivity, making $\ell$ a premeasure. The Caratheodory extension theorem produces a unique Borel probability measure.

A cylinder set in a product of spaces restricts a finite collection of coordinates $F$ and leaves all other coordinates free. It has the form ${\pi }_F^{-1}(A)$, where ${\pi }_F$ is the coordinate projection onto the finite subproduct. Taking $A$ to be a product of open sets gives the basic open cylinders for the product topology. In a finite-alphabet full shift, specifying exact letters at finitely many coordinates gives a clopen set.

For a finite set $s$ and an infinite set $A$ lying above $\max s$, putThese sets form the basic neighborhoods of the Ellentuck topology, also called the star topology. It is finer than the ordinary topology on infinite subsets. Every basic neighborhood is a clopen set in the Ellentuck topology, and a closed set in the ordinary topology.

Every convergent sequence in the Stone-Čech compactification of the natural numbers is eventually constant. If distinct ultrafilters $U_n$ converged, isolate each from the other sequence terms using convergence and finitely many separating clopen sets. Finite disjointification gives disjoint $B_n\in U_n$. The union of the even-indexed $B_n$ then belongs exactly to the even-indexed ultrafilters, contradicting convergence of membership in a clopen set.

Every identity neighbourhood of a compact Hausdorff totally disconnected topological group contains an open normal subgroup. Choose a clopen set $U$ containing the identity inside that neighbourhood. Compactness and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$, hence $VU=U$. The setwise stabilizer $H=\{g:gU=U\}$ is an open subgroup contained in $U$. Its index is finite, so its normal core is an open normal subgroup contained in $U$. These subgroups recover the profinite group as the inverse limit of its finite quotients.

For a profinite group $G$ and a complete discrete valuation ring $I$ with a continuous ${\mathbb{Z}}_p$-algebra structure, an $I$-valued measure is a bounded $I$-linear functional on continuous $I$-valued functions, with the uniform norm. Equivalently it assigns finitely additive values in $I$ to clopen sets. Locally constant functions are uniformly dense, so integration extends uniquely from finite sums. Countable additivity in the usual real-measure sense is not imposed. The coefficient ring may have infinite residue field and need not be compact.

The rational cut is both open and closed in $\mathbb{Q}$. It is relatively open as $\mathbb{Q}\cap (-\infty ,\sqrt{2})$. Since $\sqrt{2}$ is irrational, its relative complement is $\mathbb{Q}\cap (\sqrt{2},\infty )$, also relatively open. Thus this cut is a clopen set in the subspace topology of the rationals.

For $x$ in a compact Hausdorff space $K$, its quasi-component $Q$ is the intersection of all clopen sets containing $x$. It is closed and therefore compact. Suppose $Q=A\bigsqcup B$ is a separation of a topological space into two nonempty relatively closed sets, with $x\in A$. The sets $A,B$ are compact and hence closed in $K$. Normality of a compact Hausdorff space gives disjoint open neighborhoods $U,V$ of $A,B$.

The compact set $K\setminus (U\cup V)$ is disjoint from $Q$. For each point $y$ of that set choose a clopen $C_y$ containing $x$ but not $y$. Finitely many complements cover $K\setminus (U\cup V)$, so their corresponding intersection $C$ is clopen, contains $Q$, and is contained in $U\cup V$. Now $C\cap U$ is clopen in $C$, hence clopen in $K$, and contains $x$ while missing $B$. This contradicts the definition of $Q$. Thus $Q$ is connected. Any connected set containing $x$ lies in every clopen neighborhood of $x$, so the connected component of $x$ equals $Q$.

For the second assertion, $D$ is closed in the ambient Hausdorff space $X$, and is itself compact Hausdorff. Let $C$ be a connected component of $D$ disjoint from ${\partial }_{X}D$. Choose $p\in C$. By the first assertion $C$ is the quasi-component of $p$ in $D$. For each $y\in {\partial }_{X}D$ choose a clopen subset of $D$ containing $p$ but not $y$. Compactness of ${\partial }_{X}D\subseteq D$ gives a finite intersection $H$ of these sets which contains $C$ and avoids the whole boundary. Thus $H\subseteq {\mathrm{int}}_{X}D$. It is closed in $X$, because it is closed in the closed set $D$, and is open in $X$, because it is relatively open in $D$ and contained in its interior. It is nonempty and proper, contradicting connectedness of $X$. Hence every connected component of $D$ meets its boundary.

Finally suppose a closed subset $A$ of the square has the point-separating clopen property specified in the paper. Every nonempty relatively open subset of the square contains a nontrivial line segment: it contains an interior point of the square and a small ball about that point. If such a segment lay in $A$, a clopen subset of $A$ separating its endpoints would induce a separation of a topological space of the connected segment. This is impossible. Therefore $A$ contains no nonempty relatively open subset of the square. Being closed, it is nowhere dense in the square, and also nowhere dense in ${\mathbb{R}}^2$.

Let $\mathcal{B}$ be a topological basis consisting of clopen sets. For $U\in \mathcal{B}$ its characteristic function ${\chi }_U:X\to \{0,1\}$ is continuous, since the preimages of both singleton values are open. DefineThe coordinate criterion for product topology makes this map continuous. If $x\ne y$, the $T_0$ property gives an open set containing one and not the other. A topological basis element contained in that open set still distinguishes them, so their corresponding characteristic coordinates differ. Therefore $e$ is injective.

Finally, for each topological basis element $U$,The right-hand side is the preimage of an open cylinder in the product of discrete two-point topological spaces. These cylinders recover a topological basis for the topology of $X$, proving continuity of the inverse onto the image. Hence

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

Let $T=\mathrm{Spec}R$ and $Z=\mathrm{Spec}(R/(r))$. The idempotent $e$ determines the disjoint clopen sets $Z_A=D_Z(e)$ and $Z_B=D_Z(1-e)$. Each is closed in $T$. Define the open subsetsThey cover $T$ and satisfy $T_A\cap T_B=T\setminus Z=D(r)$. The global section $r$ defines morphisms of schemes $T_A\to A$ and $T_B\to B$, both sending the coordinate $t$ to $r$. On their intersection $r$ is a unit, so both maps land in the common open subscheme $A\cap B=\mathrm{Spec}\mathbb{Z}[t,t^{-1}]$, where they agree. They therefore glue uniquely:For an entirely explicit description, choose any lift $a\in R$ of $e$. Then $a(1-a)\in (r)$ andThe equality $a+(1-a)=1$ proves that these open subsets cover. The inclusion $D(a(1-a))\subseteq D(r)$ proves that their intersection is exactly $D(r)$. Their description as complements of $Z_A,Z_B$ proves independence of the chosen lift. In particular, no lift of $e$ to an idempotent of $R$ is required.

Label the two charts of the affine line with doubled origin by $A,B$, and use $t$ for their common coordinate. Let $u:T=\mathrm{Spec}R\to X$ be an $R$-valued element of the functor of points. Its composite with $\varphi$ corresponds to a unique $r\in R$ under the affine-target adjunction for schemes:The closed subscheme $Z=V(r)=\mathrm{Spec}(R/(r))$ maps into the disjoint union of the two zero sections,The inverse image $Z_A$ of $A_0$ is a clopen set in $Z$. By the idempotent–clopen correspondence, it determines a unique idempotent $e\in R/(r)$, equal to $1$ on $Z_A$ and $0$ on the inverse image $Z_B$ of $B_0$. HenceOver $\mathbb{Z}$, the two “origins” are entire zero sections, rather than single closed points; this construction uses their full scheme-theoretic fibers.

For an idempotent $e\in R$, define the principal open subset $D(e)\subseteq \mathrm{Spec}R$. Since $e(1-e)=0$, a prime ideal contains at least one of $e,1-e$, and it cannot contain both. ThereforeThus $D(e)$ is a clopen set.

Conversely, let $C$ be a clopen set in $T=\mathrm{Spec}R$. The structure sheaf glues the section $1$ on $C$ and the section $0$ on its complement to a unique global section $e\in \Gamma (T,{\mathcal{O}}_T)=R$. Checking on these two open subsets gives $e^2=e$, and $D(e)=C$. Conversely, the restriction of an idempotent $e$ to $D(e)$ is $1$ because $e$ is a unit there, and its restriction to $D(1-e)$ is $0$. Thus the two constructions are inverse, including for the zero ring:

Starting from $(r,e)$, the construction in (iii) preserves the composite to the affine line and sends $Z_A$ to the $A$ zero section and $Z_B$ to the $B$ zero section. Part (ii) therefore recovers $(r,e)$.

Conversely, start with $u:T\to X$ and construct $(r,e)$. The only points of $X$ outside $A$ lie on $B_0$, and the only points outside $B$ lie on $A_0$. Consequently $u^{-1}(A)=T\setminus Z_B=T_A$ and $u^{-1}(B)=T\setminus Z_A=T_B$. On each of these open subsets, $u$ is the unique morphism of schemes to the appropriate affine line with coordinate $r$. Thus part (iii) recovers $u$.

Finally, a ring homomorphism $R\to R^{\prime }$ sends $r$ to $r^{\prime }$ and induces $R/(r)\to R^{\prime }/(r^{\prime })$, sending $e$ to an idempotent $e^{\prime }$. Inverse images of the clopen sets $Z_A,Z_B$ are precisely those determined by $e^{\prime }$. The constructions therefore commute with this base change. We have a natural bijectionso $F$ is a functor represented by a scheme, represented by the affine line with doubled origin. The representing scheme is not required to be an affine scheme.