Topological Spaces and Continuity

Let $X$ be a set and write $\mathcal{P}(X)$ for its power set. A topology on $X$ is a family $\mathcal{T} \subseteq \mathcal{P}(X)$ such that

• (T1) $\varnothing \in \mathcal{T}$ and $X \in \mathcal{T}$;
• (T2) $\bigcup \mathcal{S} \in \mathcal{T}$ for every $\mathcal{S} \subseteq \mathcal{T}$;
• (T3) $U \cap V \in \mathcal{T}$ for all $U, V \in \mathcal{T}$.

A topological space is a pair $(X, \mathcal{T})$ consisting of a set $X$ and a topology $\mathcal{T}$ on it; the elements of $X$ are its points. When only one topology is in play we write $X$ for the pair; when several are, the topology is always named.

The members of $\mathcal{T}$ are the open sets of $(X,\mathcal{T})$. A subset $F \subseteq X$ is closed when its complement $X \setminus F$ is open, and clopen when it is both open and closed. By (T1) both $\varnothing$ and $X$ are clopen. Open and closed are not opposites: a set may be neither, and it may be both, so "not open" is never a synonym for "closed".

Finite intersections, and the empty one. Axiom (T3) is stated for two sets because that is all that is ever needed: if $n \ge 1$ and $U_1, \dots, U_n \in \mathcal{T}$, then $U_1 \cap \dots \cap U_n \in \mathcal{T}$ follows by iterating (T3) $n-1$ times. The intersection of the empty family of subsets of $X$ is $X$, since the condition "lies in every member of the empty family" is vacuous; that case is covered by (T1) rather than by (T3), which is one of the two reasons $X \in \mathcal{T}$ is demanded outright. The other is that (T2) alone gives only $\bigcup \varnothing = \varnothing$.

The closed sets, and the dual axiomatisation. Put $\mathcal{C} := \{\, X \setminus U : U \in \mathcal{T} \,\}$, the family of closed sets. Then

• (C1) $\varnothing \in \mathcal{C}$ and $X \in \mathcal{C}$;
• (C2) $\bigcap \mathcal{D} \in \mathcal{C}$ for every nonempty $\mathcal{D} \subseteq \mathcal{C}$;
• (C3) $C \cup D \in \mathcal{C}$ for all $C, D \in \mathcal{C}$.

Conversely, if a family $\mathcal{C} \subseteq \mathcal{P}(X)$ satisfies (C1), (C2) and (C3), then $\mathcal{T} := \{\, X \setminus C : C \in \mathcal{C} \,\}$ is a topology on $X$ whose closed sets are exactly the members of $\mathcal{C}$. So a topology may be presented equally well by its open sets or by its closed sets, and this library uses whichever is shorter.

Both directions are one computation, and it is carried out here rather than assumed. The map $U \mapsto X \setminus U$ is an involution of $\mathcal{P}(X)$, hence a bijection, and it carries the two De Morgan identities

$X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (X \setminus A_i) \quad (I \ne \varnothing), \qquad X \setminus \bigcap_{i \in I} A_i = \bigcup_{i \in I} (X \setminus A_i) \quad (I \ne \varnothing)$

term by term into each other. Given (T1)–(T3): $\varnothing = X \setminus X$ and $X = X \setminus \varnothing$ lie in $\mathcal{C}$, which is (C1); for nonempty $\mathcal{D} \subseteq \mathcal{C}$ the family $\mathcal{S} = \{\, X \setminus C : C \in \mathcal{D} \,\}$ lies in $\mathcal{T}$ and $\bigcap \mathcal{D} = X \setminus \bigcup \mathcal{S}$ is closed by (T2), which is (C2); and $C \cup D = X \setminus ((X \setminus C) \cap (X \setminus D))$ is closed by (T3), which is (C3). Given (C1)–(C3) the same three lines run backwards, the empty-union case $\bigcup \varnothing = \varnothing$ being supplied by (C1) rather than by (C2); and $\{\, X \setminus C : C \in \mathcal{C} \,\}$ has $\mathcal{C}$ as its family of complements because complementation is an involution.

Comparison of topologies. If $\mathcal{T}_1$ and $\mathcal{T}_2$ are topologies on the same set $X$ and $\mathcal{T}_1 \subseteq \mathcal{T}_2$, then $\mathcal{T}_1$ is coarser than $\mathcal{T}_2$ and $\mathcal{T}_2$ is finer than $\mathcal{T}_1$; the comparison is strict when the inclusion is. Inclusion is a partial order on the set of topologies on $X$ (Partial order and partially ordered set), and it is not a total one: two topologies on the same set may be incomparable.

• The topology is part of the data. Two different topologies on one set are two different topological spaces, exactly as two different metrics on one set are two different metric spaces. Every statement below that says "the space $X$" has a topology fixed in advance.

• Nothing is assumed beyond (T1)–(T3). In particular no separation property is built into the word space here: distinct points need not lie in disjoint open sets, singletons need not be closed, and both failures occur in the standard topologies defined on the next item. Where a separation property is used it is stated as a hypothesis.

• This library always writes coarser and finer. The synonyms smaller/larger are unambiguous, but weaker/stronger is used in both directions in the literature and is avoided here entirely.

• Why "arbitrary unions, finite intersections" and not the reverse. The asymmetry is not a convention: it is what makes the metric notion of open set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) an instance, and an arbitrary intersection of open sets genuinely need not be open. The failure is recorded on this page as a false statement, with witnesses in two different spaces.

Throughout, a topology is as in Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, and finite, at most countable and uncountable are as in Finite, countably infinite, countable, uncountable, so that "countable" always means "at most countable" and every finite set is countable. Let $X$ be a set. The six families below are topologies on $X$; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list.

1. Discrete topology. $\mathcal{T}_{\mathrm{disc}} := \mathcal{P}(X)$: every subset is open, hence every subset is closed, hence every subset is clopen.
2. Indiscrete topology. $\mathcal{T}_{\mathrm{ind}} := \{\varnothing, X\}$. Its closed sets are again $\varnothing$ and $X$.
3. Cofinite topology. $\mathcal{T}_{\mathrm{cof}} := \{\varnothing\} \cup \{\, U \subseteq X : X \setminus U \text{ is finite} \,\}$. Its closed sets are $X$ together with the finite subsets of $X$.
4. Cocountable topology. $\mathcal{T}_{\mathrm{coc}} := \{\varnothing\} \cup \{\, U \subseteq X : X \setminus U \text{ is at most countable} \,\}$. Its closed sets are $X$ together with the at most countable subsets of $X$.
5. Particular-point topology. Fix $p \in X$ and put $\mathcal{T}_p := \{\varnothing\} \cup \{\, U \subseteq X : p \in U \,\}$: the open sets are $\varnothing$ and the sets containing $p$. Its closed sets are $X$ together with the sets not containing $p$.
6. Sierpinski topology. On a two-point set $S = \{a, b\}$ with $a \ne b$, $\mathcal{T}_{\mathrm{Sier}} := \{\varnothing, \{b\}, S\}$. The pair $(S, \mathcal{T}_{\mathrm{Sier}})$ is Sierpinski space; $b$ is its open point and $a$ its closed point. This is exactly the particular-point topology of item 5 on a two-point set with particular point $b$, listed separately because it is quoted so often.

Two elementary facts about finite sets are used below, and both are proved here.

(i) A subset of a finite set is finite. Let $F \approx n$ with $n \in \mathbb{N}$ (Equinumerous sets, $A \approx B$ and $A \preceq B$, The natural numbers $\mathbb{N}$ (von Neumann)), witnessed by a bijection $\varphi : F \to n$, and let $B \subseteq F$. Then $\varphi$ restricts to a bijection of $B$ onto $\varphi[B] \subseteq n$ (Injection, surjection, bijection). Every element of the von Neumann natural $n$ is a natural number strictly smaller than $n$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$), so $\varphi[B]$ is a subset of $\mathbb{N}$ bounded above by $n$, hence finite by the sharper form of Every subset of an at most countable set is at most countable ("a subset $S \subseteq \mathbb{N}$ is finite if it is bounded above"). Since $\approx$ is symmetric and transitive, $B$ is finite.

(ii) A union of two finite sets is finite. First, if $H$ is finite and $g$ is any object then $H \cup \{g\}$ is finite: if $g \in H$ there is nothing to prove, and otherwise a bijection $u : H \to k$ extends to a bijection $H \cup \{g\} \to k \cup \{k\} = \sigma(k)$ by setting $u(g) := k$, which is injective because $k \notin k$ (Every natural number is a transitive set and is not a member of itself). Now fix a finite set $F$ and argue by induction (The principle of mathematical induction) on $m \in \mathbb{N}$ over the statement "for every $G$ with $G \approx m$, the union $F \cup G$ is finite". At $m = 0$ we have $G = \varnothing$ and $F \cup G = F$. At $m = \sigma(j)$, a bijection $\psi : G \to \sigma(j)$ gives $g := \psi^{-1}(j)$ and $G' := G \setminus \{g\} \approx j$ (restrict $\psi$), so $F \cup G = (F \cup G') \cup \{g\}$ is finite by the induction hypothesis and the previous sentence.

Discharge of the topology axioms.

Discrete. Every subset of $X$ lies in $\mathcal{P}(X)$, so (T1), (T2) and (T3) hold with nothing to check.

Indiscrete. (T1) is the definition. For (T2), a subfamily of $\{\varnothing, X\}$ has union $\varnothing$ (if it is empty or $\{\varnothing\}$) or $X$ (otherwise). For (T3), $\varnothing \cap A = \varnothing$ and $X \cap X = X$.

Cofinite. (T1): $\varnothing$ is listed, and $X \setminus X = \varnothing$ is finite. (T2): let $\mathcal{S} \subseteq \mathcal{T}_{\mathrm{cof}}$. If every member is $\varnothing$ the union is $\varnothing$. Otherwise fix $U_0 \in \mathcal{S}$ with $U_0 \ne \varnothing$; then $X \setminus \bigcup \mathcal{S} \subseteq X \setminus U_0$, which is finite, so the left side is finite by (i). (T3): for nonempty $U, V$ with finite complements, $X \setminus (U \cap V) = (X \setminus U) \cup (X \setminus V)$ is finite by (ii); and if either of $U, V$ is empty so is $U \cap V$. The closed sets are the complements of the open ones, that is $X = X \setminus \varnothing$ together with the finite sets.

Cocountable. Identical to the cofinite case with "at most countable" in place of "finite": (i) is replaced by Every subset of an at most countable set is at most countable itself, and (ii) by the statement that a union of two at most countable sets is at most countable, which is the two-set instance of Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$ applied to the family $A_0 := U, A_1 := V, A_k := \varnothing$ for $k \ge 2$.

Particular point. (T1): $\varnothing$ is listed and $p \in X$. (T2): a subfamily whose members are all $\varnothing$ has union $\varnothing$; otherwise some member contains $p$, hence so does the union. (T3): if $U$ and $V$ both contain $p$ then so does $U \cap V$; and if either is $\varnothing$ then so is the intersection.

Sierpinski. The special case $X = \{a,b\}$, $p = b$ of the previous paragraph: the sets containing $b$ are $\{b\}$ and $S$, so $\mathcal{T}_b = \{\varnothing, \{b\}, S\} = \mathcal{T}_{\mathrm{Sier}}$.

• Two degenerate collapses. If $X$ is finite then the cofinite topology is the discrete one, since every subset then has finite complement by fact (i) above; if $X$ is at most countable the cocountable topology is discrete for the same reason. Both families are therefore interesting only on an infinite, respectively uncountable, set, and every statement made about them below names that hypothesis.

• Where the two extremes sit in the comparison order. The discrete topology is the finest and the indiscrete the coarsest topology on $X$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison): every topology is a subfamily of $\mathcal{P}(X)$ and contains $\varnothing$ and $X$. Every other topology on $X$ lies between them, and the cofinite topology is coarser than the cocountable one, because a finite set is at most countable.

• No choice principle is needed for any of the six, despite the citation. The only appeal above that carries a choice hypothesis is Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$, whose statement assumes $\mathrm{AC}_\omega$, and it is used for a union of two sets only, padded with copies of $\varnothing$. That instance is provable in ZF alone, by interleaving two given enumerations, exactly as The irrationals are uncountable records for the union of the rationals and the irrationals; the general theorem is cited because it is the form in which this library states the union result, not because the strength is needed. Nothing about the cocountable topology depends on countable choice.

• The Sierpinski point that is open is a genuine choice of labelling. Both $\{\varnothing,\{b\},S\}$ and $\{\varnothing,\{a\},S\}$ are topologies, and they are carried to each other by the transposition of $a$ and $b$; this library fixes the first and always names the open point.

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

A family $\mathcal{B} \subseteq \mathcal{T}$ is a basis for $\mathcal{T}$ if every open set is a union of members of $\mathcal{B}$: for every $U \in \mathcal{T}$ there is $\mathcal{B}_U \subseteq \mathcal{B}$ with $U = \bigcup \mathcal{B}_U$. Equivalently, and this is the form used in proofs,

$\text{for every } U \in \mathcal{T} \text{ and every } x \in U \text{ there is } B \in \mathcal{B} \text{ with } x \in B \subseteq U .$

The two forms say the same thing. If every open $U$ is such a union and $x \in U$, then $x$ lies in one of the sets united, which is a member of $\mathcal{B}$ inside $U$. Conversely, if the displayed condition holds then $U = \bigcup \{\, B \in \mathcal{B} : B \subseteq U \,\}$, since each such $B$ is contained in $U$ and each $x \in U$ lies in one of them. Note that $\bigcup \varnothing = \varnothing$, so the empty open set is covered by the empty subfamily and needs no member of $\mathcal{B}$. The members of a basis are called basic open sets.

The topology generated by a family. Let $\mathcal{S} \subseteq \mathcal{P}(X)$ be any family of subsets of $X$. Then

$\langle \mathcal{S} \rangle := \bigcap \{\, \mathcal{T}' : \mathcal{T}' \text{ is a topology on } X \text{ with } \mathcal{S} \subseteq \mathcal{T}' \,\}$

is a topology on $X$, it contains $\mathcal{S}$, and it is contained in every topology on $X$ that contains $\mathcal{S}$. It is called the topology generated by $\mathcal{S}$, and $\mathcal{S}$ is a subbasis for a topology $\mathcal{T}$ when $\mathcal{T} = \langle \mathcal{S} \rangle$.

This is well posed, and the obligation is discharged here. The collection being intersected is nonempty, because $\mathcal{P}(X)$ is a topology on $X$ containing $\mathcal{S}$; so the intersection is an intersection of a nonempty family of subsets of $\mathcal{P}(X)$ and is a set. It is a topology: $\varnothing$ and $X$ lie in every topology on $X$, hence in the intersection, which is (T1); if $\mathcal{S}' \subseteq \langle \mathcal{S} \rangle$ then $\mathcal{S}'$ is a subfamily of each $\mathcal{T}'$ in the collection, so $\bigcup \mathcal{S}'$ lies in each $\mathcal{T}'$ and hence in the intersection, which is (T2); and the same argument with $U \cap V$ gives (T3). It contains $\mathcal{S}$ because every $\mathcal{T}'$ in the collection does, and it is contained in each such $\mathcal{T}'$ because an intersection is contained in each of its members. So $\langle \mathcal{S} \rangle$ is the coarsest topology on $X$ containing $\mathcal{S}$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and in particular it is uniquely determined by $\mathcal{S}$.

Neither notion is intrinsic to the family alone. "$\mathcal{B}$ is a basis for $\mathcal{T}$" and "$\mathcal{S}$ is a subbasis for $\mathcal{T}$" are relations between a family and a topology, not properties of the family. The question of which families are a basis for some topology, and how the topology generated by a subbasis is computed from it, is settled by the next item.

• Every topology is a basis for itself, so a basis always exists; the point of a basis is to be smaller and more explicit than $\mathcal{T}$, and the point of a subbasis is to be smaller still at the cost of one round of finite intersections.

• Two extremes of the generated topology. $\langle \varnothing \rangle$ is the indiscrete topology, since $\{\varnothing, X\}$ is a topology containing the empty family and is contained in every topology. At the other end, $\langle \mathcal{P}(X) \rangle = \mathcal{P}(X)$ is the discrete topology.

• Generation is monotone and idempotent. If $\mathcal{S}_1 \subseteq \mathcal{S}_2$ then every topology containing $\mathcal{S}_2$ contains $\mathcal{S}_1$, so $\langle \mathcal{S}_1 \rangle \subseteq \langle \mathcal{S}_2 \rangle$; and $\langle \langle \mathcal{S} \rangle \rangle = \langle \mathcal{S} \rangle$ because $\langle \mathcal{S} \rangle$ is itself a topology containing $\mathcal{S}$. Both are used silently below.

Let $X$ be a set, and for $\mathcal{B} \subseteq \mathcal{P}(X)$ write

$\mathcal{T}_{\mathcal{B}} := \{\, U \subseteq X : \text{for every } x \in U \text{ there is } B \in \mathcal{B} \text{ with } x \in B \subseteq U \,\} .$

1. $\mathcal{B}$ is a basis for some topology on $X$ (Basis and subbasis for a topology, and the topology generated by a family of sets) if and only if

   • (B1) $\bigcup \mathcal{B} = X$, and
   • (B2) for all $B_1, B_2 \in \mathcal{B}$ and every $x \in B_1 \cap B_2$ there is $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$.

   When (B1) and (B2) hold, that topology is unique: it is $\mathcal{T}_{\mathcal{B}}$, which is also exactly the family of all unions of subfamilies of $\mathcal{B}$.

2. Let $\mathcal{S} \subseteq \mathcal{P}(X)$ be an arbitrary family and let $\mathcal{B}_{\mathcal{S}}$ be the family of intersections of finitely many members of $\mathcal{S}$. Then $\mathcal{B}_{\mathcal{S}}$ satisfies (B1) and (B2), and $\mathcal{T}_{\mathcal{B}_{\mathcal{S}}} = \langle \mathcal{S} \rangle$, the topology generated by $\mathcal{S}$. So the finite intersections of any subbasis form a basis for the topology it generates.

The nullary intersection: this library takes the empty intersection to be $X$. In claim 2 the phrase "finitely many" includes none, and the intersection of the empty subfamily of $\mathcal{S}$ is $X$, because the defining condition "lies in every member of the empty family" holds of every point of $X$. Hence $X \in \mathcal{B}_{\mathcal{S}}$ for every $\mathcal{S}$, including $\mathcal{S} = \varnothing$, and no covering hypothesis is imposed on a subbasis. The competing convention takes only nonempty finite intersections and compensates by requiring $\bigcup \mathcal{S} = X$; under it claim 2 holds verbatim once that hypothesis is added, and the two conventions differ only in which of the two devices supplies (B1). The choice made here is recorded again among this page's conventions, and it is the reason $\langle \varnothing \rangle$ comes out as the indiscrete topology $\{\varnothing, X\}$ rather than being undefined.

• What (B1) and (B2) are not. (B2) does not say that $\mathcal{B}$ is closed under intersections; it says only that the intersection of two members is a union of members. The basis of open intervals of $\mathbb{R}$ satisfies (B2) outright, since an intersection of two open intervals is an open interval or empty, whereas the basis of half-open intervals of the Sorgenfrey line uses the same closure property; a basis of open balls in a metric space uses the weaker form in an essential way.

• The subbasis clause is what makes generation computable. The definition of $\langle \mathcal{S} \rangle$ as an intersection of topologies says nothing about what its members look like; claim 2 says they are exactly the unions of finite intersections of members of $\mathcal{S}$, which is how every generated topology in this library is actually described.

• A family may be a basis for at most one topology, but it is a subbasis for at most one as well, and the two roles differ: $\mathcal{S}$ is a subbasis for $\langle \mathcal{S} \rangle$ and a basis for it exactly when $\mathcal{S}$ already satisfies (B1) and (B2).

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $x \in X$.

A set $N \subseteq X$ is a neighbourhood of $x$ if there is an open $U \in \mathcal{T}$ with $x \in U \subseteq N$. The family of all neighbourhoods of $x$ is written $\mathcal{N}(x)$ and called the neighbourhood filter at $x$. A neighbourhood that is itself open is an open neighbourhood.

Convention, and it is a live fork: in this library a neighbourhood need not be open. The competing convention, used by Munkres among others, defines a neighbourhood of $x$ to be an open set containing $x$. Both are in current use; this library follows the one above and writes "open neighbourhood" in full whenever openness is wanted, so that no statement here depends on which convention a reader brings.

A family $\mathcal{B}_x \subseteq \mathcal{N}(x)$ is a neighbourhood base at $x$ if every neighbourhood of $x$ contains a member of $\mathcal{B}_x$: for every $N \in \mathcal{N}(x)$ there is $B \in \mathcal{B}_x$ with $B \subseteq N$.

Four immediate consequences, established here because they are used constantly.

1. $X$ is a neighbourhood of each of its points, since $x \in X \subseteq X$ and $X$ is open by (T1); so $\mathcal{N}(x) \ne \varnothing$ and every point has at least one neighbourhood base, namely $\mathcal{N}(x)$ itself.
2. A superset of a neighbourhood of $x$ is a neighbourhood of $x$: if $x \in U \subseteq N \subseteq M$ with $U$ open then $x \in U \subseteq M$.
3. The intersection of two neighbourhoods of $x$ is a neighbourhood of $x$: if $x \in U \subseteq N$ and $x \in V \subseteq M$ with $U, V$ open, then $U \cap V$ is open by (T3) and $x \in U \cap V \subseteq N \cap M$. By iteration the same holds for any intersection of finitely many neighbourhoods of $x$.
4. A set is open exactly when it is a neighbourhood of each of its points. If $U$ is open and $x \in U$ then $x \in U \subseteq U$. Conversely, if $N$ is a neighbourhood of each of its points, choose for each $x \in N$ an open $U_x$ with $x \in U_x \subseteq N$; then $N = \bigcup_{x \in N} U_x$ is open by (T2). No choice principle is involved: $U_x$ may be taken to be the union of all open subsets of $N$ containing $x$, which is determined by $x$ and $N$, and is open by (T2).

Basic sets give neighbourhood bases. If $\mathcal{B}$ is a basis for $\mathcal{T}$ (Basis and subbasis for a topology, and the topology generated by a family of sets) then $\mathcal{B}_x := \{\, B \in \mathcal{B} : x \in B \,\}$ is a neighbourhood base at $x$ consisting of open sets. Each such $B$ is open and contains $x$, hence is a neighbourhood of $x$; and if $N \in \mathcal{N}(x)$, fix open $U$ with $x \in U \subseteq N$ and then $B \in \mathcal{B}$ with $x \in B \subseteq U$, which gives $B \in \mathcal{B}_x$ with $B \subseteq N$. A member of $\mathcal{B}_x$ is called a basic neighbourhood of $x$.

• The convention costs nothing and buys one thing. Every statement of the form "for every neighbourhood $N$ of $x$ ... " whose predicate is preserved when $N$ is enlarged is equivalent to the statement with $N$ restricted to open neighbourhoods: every neighbourhood contains an open one, and the predicate then passes to the larger set. Eventual-membership and the usual local-existence tests have this form; an arbitrary predicate need not. What the wider notion buys is that $\mathcal{N}(x)$ is a filter on $X$ in the sense of Filter on a set: consequence 1 is (F1), consequence 3 is (F3), consequence 2 is (F4), and (F2) holds because $x \in U \subseteq \varnothing$ is impossible, so $\varnothing$ is a neighbourhood of no point. Under the narrower convention the family of open sets containing $x$ fails (F4) as soon as some non-open set contains an open set around $x$, so the name "neighbourhood filter" would not be available.

• A neighbourhood base is not required to be closed under intersection, and the bases used below usually are not; what is required is only that its members be cofinal downwards among neighbourhoods.

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $A \subseteq X$ and let $x \in X$. Neighbourhoods are as in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, so a neighbourhood need not be open.

• The interior of $A$ is $\operatorname{int}(A) := \bigcup \{\, U \in \mathcal{T} : U \subseteq A \,\}$.
• The closure of $A$ is $\overline{A} := \bigcap \{\, F \subseteq X : F \text{ closed and } A \subseteq F \,\}$.
• The exterior of $A$ is $\operatorname{ext}(A) := \operatorname{int}(X \setminus A)$.
• The boundary of $A$ is $\partial A := \overline{A} \setminus \operatorname{int}(A)$.
• $x$ is a limit point (accumulation point) of $A$ if every neighbourhood $N$ of $x$ satisfies $N \cap (A \setminus \{x\}) \ne \varnothing$. The set of limit points of $A$ is the derived set $A'$.
• $x$ is an isolated point of $A$ if $x \in A$ and some neighbourhood $N$ of $x$ satisfies $N \cap A = \{x\}$.

Both operators are well posed, and the two names are justified rather than asserted. The interior is a union of open sets, hence open by (T2), it is contained in $A$, and it contains every open $U \subseteq A$: so $\operatorname{int}(A)$ is the largest open subset of $A$. The family being intersected in the definition of $\overline{A}$ is nonempty, since $X$ is closed and contains $A$, so the intersection is a set; it is closed by (C2) of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, it contains $A$, and it is contained in every closed $F \supseteq A$: so $\overline{A}$ is the smallest closed superset of $A$. In particular

$\operatorname{int}(A) \subseteq A \subseteq \overline{A},$

and $A$ is open if and only if $A = \operatorname{int}(A)$, and closed if and only if $A = \overline{A}$, in each case because one inclusion is automatic and the other says exactly that $A$ belongs to the family in question.

Interior and closure are exchanged by complementation. For every $A \subseteq X$,

$X \setminus \operatorname{int}(A) = \overline{X \setminus A}, \qquad X \setminus \overline{A} = \operatorname{int}(X \setminus A) = \operatorname{ext}(A).$

Indeed $U \mapsto X \setminus U$ is a bijection between the open subsets of $A$ and the closed supersets of $X \setminus A$, and it turns unions into intersections (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); applying complementation to the first identity gives the second. So $\operatorname{ext}(A) = X \setminus \overline{A}$, and every statement about interiors has a mirror statement about closures.

The pointwise description of the interior. $x \in \operatorname{int}(A)$ if and only if $A$ is a neighbourhood of $x$. If $x \in \operatorname{int}(A)$ then $\operatorname{int}(A)$ is an open set with $x \in \operatorname{int}(A) \subseteq A$; conversely a neighbourhood relation $x \in U \subseteq A$ with $U$ open puts $x$ in the union defining $\operatorname{int}(A)$. The corresponding description of the closure is proved as the next item, because it is the statement that does the work in every later proof.

$X$ is the disjoint union of the three regions. Since $\operatorname{int}(A) \subseteq \overline{A}$, the three sets $\operatorname{int}(A)$, $\partial A = \overline{A} \setminus \operatorname{int}(A)$ and $\operatorname{ext}(A) = X \setminus \overline{A}$ are pairwise disjoint and their union is $X$. This is recorded again, with the identities for interior and closure of unions and intersections, in the lemma two items below.

• The boundary is symmetric in $A$ and its complement: $\partial A = \overline{A} \cap \overline{X \setminus A}$, because $\overline{A} \setminus \operatorname{int}(A) = \overline{A} \cap (X \setminus \operatorname{int}(A)) = \overline{A} \cap \overline{X \setminus A}$ by the complementation identity above. Hence $\partial A = \partial(X \setminus A)$, and $\partial A$ is closed, being an intersection of two closed sets.

• A limit point of $A$ need not lie in $A$, and a point of $A$ need not be a limit point of $A$. The points of $A$ that are not limit points of $A$ are exactly its isolated points, directly from the two definitions. The relation $\overline{A} = A \cup A'$ is a theorem, proved next, not a restatement.

• These are the metric notions when the topology is a metric topology. For a metric space the definitions of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space are stated with balls in place of neighbourhoods, and the balls around $x$ are a neighbourhood base at $x$; the identification is carried out where metrizable spaces are defined, later on this page, and it is what allows metric examples to be quoted here without reproof.

Let $(X, \mathcal{T})$ be a topological space, let $\mathcal{B}$ be a basis for $\mathcal{T}$ (Basis and subbasis for a topology, and the topology generated by a family of sets), let $A \subseteq X$ and let $x \in X$. Closure, derived set and limit points are as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space. Then:

1. The following four conditions are equivalent.
   • (a) $x \in \overline{A}$;
   • (b) $N \cap A \ne \varnothing$ for every neighbourhood $N$ of $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open);
   • (c) $U \cap A \ne \varnothing$ for every open $U$ with $x \in U$;
   • (d) $B \cap A \ne \varnothing$ for every $B \in \mathcal{B}$ with $x \in B$.
2. $\overline{A}$ is closed, contains $A$, and is contained in every closed $F \subseteq X$ with $A \subseteq F$; so it is the smallest closed superset of $A$, and $A$ is closed if and only if $A = \overline{A}$.
3. $\overline{A} = A \cup A'$.

Claim 2 is recorded here for reference and is discharged in Interior, closure, boundary, exterior, derived set and isolated point in a topological space, where it is what makes the definition of $\overline{A}$ well posed; claims 1 and 3 are proved below. Claim 1 is the form in which the closure is used everywhere afterwards, and clause (d) is what makes a closure computable from a basis rather than from all open sets.

• Clause (d) is relative to a basis and clause (c) is not. Different bases for one topology give different families of test sets in (d), and the theorem says all of them detect the same closure. This is why a closure in a metric space may be computed with balls alone, and a closure in $\mathbb{R}$ with bounded open intervals alone.

• The empty set and the whole space. $\overline{\varnothing} = \varnothing$, since $\varnothing$ is closed, and $\overline{X} = X$; claim 1 reads correctly in both cases, no neighbourhood meeting $\varnothing$ and every neighbourhood meeting $X$.

• Nothing here assumes that singletons are closed. In the indiscrete topology on a set with at least two points, $\overline{\{x\}} = X$ for every $x$, since the only neighbourhood of any point is $X$; claim 3 then says $A' = X \setminus \{x\}$ is contained in $\overline{\{x\}}$, which it is.

Let $(X, \mathcal{T})$ be a topological space, with interior, closure, boundary and exterior as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space. Then:

1. Monotonicity. $A \subseteq B$ implies $\operatorname{int}(A) \subseteq \operatorname{int}(B)$ and $\overline{A} \subseteq \overline{B}$.
2. The two identities. For all $A, B \subseteq X$, $\operatorname{int}(A \cap B) = \operatorname{int}(A) \cap \operatorname{int}(B), \qquad \overline{A \cup B} = \overline{A} \cup \overline{B} .$
3. The two reverse combinations are inclusions only. For all $A, B \subseteq X$, $\operatorname{int}(A) \cup \operatorname{int}(B) \subseteq \operatorname{int}(A \cup B), \qquad \overline{A \cap B} \subseteq \overline{A} \cap \overline{B},$ and both inclusions are strict for $A = \{p\}$ and $B = X \setminus \{p\}$ in the cofinite topology on an infinite set $X$ with $p \in X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
4. Both identities of claim 2 fail for infinite families. In the same space, with $I := X \setminus \{p\}$, $\bigcup_{x \in I} \overline{\{x\}} = I \subsetneq X = \overline{\bigcup_{x \in I} \{x\}}, \qquad \operatorname{int}\Big(\bigcap_{x \in I} (X \setminus \{x\})\Big) = \varnothing \subsetneq \{p\} = \bigcap_{x \in I} \operatorname{int}(X \setminus \{x\}) .$
5. Trichotomy of position. For every $A \subseteq X$ the three sets $\operatorname{int}(A)$, $\partial A$ and $\operatorname{ext}(A)$ are pairwise disjoint and their union is $X$.