Subspaces, Products, and Quotients

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 $S\subseteq X$ carry the subspace topology ${\mathcal{T}}_S$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) and let $A\subseteq S$. Write $\bar{A}$ and $\mathrm{int}(A)$ for the closure and the interior of $A$ in $X$, and ${\mathrm{cl}}_S(A)$ and ${\mathrm{int}}_S(A)$ for those taken in the space $(S,{\mathcal{T}}_S)$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then:

1. Closure traces exactly. ${\mathrm{cl}}_S(A)=\bar{A}\cap S\mathrm{.}$
2. Interior traces only one way. $\mathrm{int}(A)\subseteq S$, so $\mathrm{int}(A)\cap S=\mathrm{int}(A)$, and $\mathrm{int}(A)\subseteq {\mathrm{int}}_S(A),$ an inclusion that may be strict.
3. Equality for an open subspace. If $S\in \mathcal{T}$ then ${\mathrm{int}}_S(A)=\mathrm{int}(A)$.
4. Density traces to open subspaces only. If $D\subseteq X$ is dense in $X$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and $S\in \mathcal{T}$, then $D\cap S$ is dense in $(S,{\mathcal{T}}_S)$. Without the hypothesis $S\in \mathcal{T}$ this fails.

Both failures are witnessed inside the proof, in Sierpinski space (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies): the unqualified forms of claims 2 and 3 and of claim 4 are false, and the counterexamples are two lines each rather than deferred.

• The same two failures occur in $\mathbb{R}$, and there they are the familiar ones. With the usual topology, $S=[0,1]$ and $A=[0,1]$ give ${\mathrm{int}}_S(A)=[0,1]$ while $\mathrm{int}(A)=(0,1)$; and $\mathbb{Q}$ is dense in $\mathbb{R}$ while its trace on the subspace of irrationals is empty, so a dense set need not trace to a dense set of a subspace that is not open. Sierpinski space is used in the proof only because it needs no real-number machinery.

• Why closure behaves better than interior. Claim 1 holds for every $S$, with no hypothesis, because the closed sets of a subspace are exactly the traces of the closed sets and tracing preserves the "smallest superset" that defines a closure. The interior is a largest subset, and tracing does not preserve that: a set can be open in $S$ without being the trace of any open set of $X$ that is contained in $A$, which is exactly what step 1.4 exhibits.

• Claim 4 is what makes "has a countable dense subset" behave the way it does. The property passes to open subspaces by claim 4, and it does not pass to arbitrary subspaces; the witness for the failure is worked on the companion page, where an uncountable discrete subspace is exhibited inside a space with a countable dense subset.

A property of topological spaces is a condition $P$ that is either true or false of each space, as in Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological; a topological property is one whose truth value is the same for homeomorphic spaces. Every subset of a space is regarded as a space by giving it the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Let $P$ be a property of topological spaces. Then $P$ is

• hereditary if, whenever a space $X$ has $P$, every subspace of $X$ has $P$;
• open-hereditary if, whenever $X$ has $P$, every subspace $S\subseteq X$ with $S$ open in $X$ has $P$;
• closed-hereditary if, whenever $X$ has $P$, every subspace $S\subseteq X$ with $S$ closed in $X$ has $P$.

A hereditary property is both open-hereditary and closed-hereditary, since the condition on $S$ is only a restriction of the range of subspaces quantified over. Neither of the two weaker notions implies the other, and neither implies heredity.

The definition is stable under the route by which a subspace is reached. If $S\subseteq T\subseteq X$ then the topology $S$ inherits from the subspace $T$ is the topology $S$ inherits from $X$, transitivity being discharged in Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace. So "every subspace of $X$" is unambiguous, and a hereditary property automatically passes from $X$ to a subspace of a subspace, with no separate induction.

Heredity is a statement about a property, not about a space. It quantifies over all spaces having $P$ and all their subspaces, so a single space whose subspaces all inherit $P$ says nothing; and a single space that has $P$ and has one subspace lacking $P$ refutes heredity outright. A space that lacks $P$ refutes nothing, however its subspaces behave. That asymmetry is why the failures are recorded here as counterexamples and the successes as theorems.

Only topological properties are worth asking about. Taking $S=X$ shows that a hereditary property holds of $X$ itself, and the subspace topology on $X$ is $\mathcal{T}$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, with $U\cap X=U$), so the definition is not vacuous at the top. But a condition that is not invariant under homeomorphism can be hereditary for uninteresting reasons, since a subspace is only determined up to the identification of its topology (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); every property named hereditary in this library is a topological property, and it is said so where it is proved.

• The three notions separate in practice. Metrizability and first countability are hereditary, and that is proved in the next item. "Has a countable dense subset" is open-hereditary, by claim 4 of For $A\subseteq S\subseteq X$ the closure of $A$ in $S$ is ${\bar{A}}^X\cap S$, while the interior only contains ${\mathrm{int}}^X(A)\cap S$, with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$, and is not hereditary; the witness is worked on the companion page, where an uncountable discrete subspace is exhibited inside a space that has a countable dense subset.

• What is deliberately not settled here. Whether the separation properties beyond the Hausdorff condition of Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not are hereditary is a question about axioms that are not available at this point in the reading order, and no claim about them is made on this page.

• Products have their own word. A property preserved by arbitrary products is usually called productive, and the same three-way refinement (finite products, countable products, arbitrary products) applies to it. No item on this page uses that word, because the productive theorems it would organise are not available at this point in the reading order.

Both of the following properties of topological spaces are hereditary (Hereditary, open-hereditary and closed-hereditary properties of topological spaces).

1. Metrizability (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). If $\mathcal{T}$ is induced by a metric $d$ on $X$ and $S\subseteq X$, then the subspace topology ${\mathcal{T}}_S$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) is induced by the subspace metric $d_S=d\upharpoonright (S\times S)$ (Isometry, isometric embedding, and the subspace metric on a subset). So a subspace of a metrizable space is metrizable, and a metric inducing its topology is available explicitly and not merely asserted to exist.
2. First countability (First countable space: a countable neighbourhood base at every point). If every point of $X$ has an at most countable neighbourhood base and $S\subseteq X$, then every point of $S$ has an at most countable neighbourhood base in $(S,{\mathcal{T}}_S)$, namely the family of traces on $S$ of the members of a base at that point in $X$.

Claim 1 is a corollary in the strict sense: the identification of the subspace topology with the metric topology of the subspace metric is discharged inside Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, and nothing is reproved here.

• No choice principle is spent. The metric $d_S$ is a restriction, and the neighbourhood base ${\mathcal{B}}_x^S$ is the image of a given family under an explicit map, so the enumeration of step 3.1 is produced from a given enumeration rather than selected. The only selections in the proof are the single metric of step 1.1 and the single family of step 1.2, each of which exists by hypothesis for the one space under consideration.

• Neither converse holds, and neither is claimed. A subspace of a non-metrizable space may perfectly well be metrizable, every one-point subspace being so; heredity is a statement in one direction only.

• The metric is not canonical, and the topology is. Claim 1 produces a metric on $S$, the restriction of the one chosen on $X$; a different metric on $X$ inducing the same topology restricts to a different metric on $S$ inducing the same subspace topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). What is hereditary is the existence of a metric, which is a property of the topology alone.

Two constructions are defined here, one for maps into spaces and one for maps out of spaces. Every further construction on this page is an instance of one of them.

Initial topology. Let $X$ be a set, let $I$ be an index set, let $(Y_i,{\mathcal{T}}_i)$ be a topological space for each $i\in I$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and let $f_i:X\to Y_i$ be a function for each $i$. The initial topology on $X$ induced by the family $(f_i{)}_{i\in I}$ is

${\mathcal{T}}^{\mathrm{i}\mathrm{n}}:=⟨\{f_i^{-1}[V]:i\in I,V\in {\mathcal{T}}_i\}⟩,$

the topology generated by the preimages of the open sets of the $Y_i$ (Basis and subbasis for a topology, and the topology generated by a family of sets).

This is well posed, and the obligation is discharged by the item cited. For an arbitrary family $\mathcal{S}$ of subsets of $X$, the family $⟨\mathcal{S}⟩$ is a topology on $X$, contains $\mathcal{S}$, and is contained in every topology on $X$ containing $\mathcal{S}$ (Basis and subbasis for a topology, and the topology generated by a family of sets); no further verification is needed here. By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis a basis for ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$ is the family of intersections of finitely many sets $f_i^{-1}[V]$, the intersection of none being $X$.

Final topology. Let $X$ be a set, let $(Z_i,{\mathcal{S}}_i)$ be a topological space for each $i\in I$, and let $g_i:Z_i\to X$ be a function for each $i$. The final topology on $X$ induced by the family $(g_i{)}_{i\in I}$ is

${\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}:=\{U\subseteq X:g_i^{-1}[U]\in {\mathcal{S}}_i\text{ for every }i\in I\}\mathrm{.}$

This is a topology, and the verification is carried out here rather than assumed. Preimage commutes with the three set operations named in the axioms: $g_i^{-1}[\mathrm{\varnothing }]=\mathrm{\varnothing }$ and $g_i^{-1}[X]=Z_i$, which gives (T1); $g_i^{-1}[\bigcup \mathcal{U}]=\bigcup \{g_i^{-1}[U]:U\in \mathcal{U}\}$ for every family $\mathcal{U}\subseteq {\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}$, which with (T2) in $Z_i$ gives (T2); and $g_i^{-1}[U\cap V]=g_i^{-1}[U]\cap g_i^{-1}[V]$, which with (T3) in $Z_i$ gives (T3) (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Two degenerate cases, stated because they are used. For $I=\mathrm{\varnothing }$ the initial topology is $⟨\mathrm{\varnothing }⟩=\{\mathrm{\varnothing },X\}$, the indiscrete topology, and the final topology is $\mathcal{P}(X)$, the discrete topology, the defining condition being vacuous (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Basis and subbasis for a topology, and the topology generated by a family of sets). For a family consisting of a single map the definitions read the same way with the index dropped.

The subspace topology is the model initial topology. Let $(X,\mathcal{T})$ be a space, let $S\subseteq X$ and let $\iota :S\to X$ be the inclusion. Then ${\iota }^{-1}[U]=U\cap S$ for every $U\subseteq X$, so the generating family for the initial topology of the one-element family $(\iota )$ is exactly ${\mathcal{T}}_S=\{U\cap S:U\in \mathcal{T}\}$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). That family is already a topology, so generating adds nothing and the initial topology of $\iota$ is the subspace topology. Everything Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace proves about it is therefore quoted here and not reproved: its closed sets are the traces of the closed sets, a basis of $X$ traces to a basis of $S$, the inclusion is continuous (Continuity of a map of topological spaces at a point and globally, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\bar{A})\subseteq \overline{f(A)}$), ${\mathcal{T}}_S$ is the coarsest topology making $\iota$ continuous, and a map $g:Z\to S$ is continuous if and only if $\iota \circ g$ is.

Terminology. A family $(f_i)$ as above is the defining family of the initial topology, and likewise for the final one; both topologies depend on the whole family and not on any one member. The characteristic properties that make these two constructions worth naming are the subject of the next item.

• The direction of the arrows is what distinguishes the two. An initial topology is put on the source of its defining maps and is the coarsest making them continuous; a final topology is put on the target and is the finest making them continuous. Both facts are proved in the next item rather than built into the definitions above.

• Why the initial topology is generated and the final one is not. The preimages $f_i^{-1}[V]$ need not be closed under unions or finite intersections, so a topology has to be generated from them; the final family is closed under both operations already, because preimage commutes with both, so it is a topology as it stands. The asymmetry is a fact about preimages and not a choice of presentation.

• Both constructions are determined by a universal property, so they are unique once that property is stated. That is the content of the next item, and it is what lets the product, the coproduct and the quotient be treated as three instances of two constructions rather than as three separate theories.

Let $X$ be a set and let $I$ be an index set.

Initial. Let $(Y_i,{\mathcal{T}}_i)$ be spaces and $f_i:X\to Y_i$ functions, and give $X$ the initial topology ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$ of the family $(f_i{)}_{i\in I}$ (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology). Then:

1. Every $f_i$ is continuous for ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$, and ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$ is the coarsest topology on $X$ with that property: every topology on $X$ making all the $f_i$ continuous contains ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$.
2. Characteristic property. For every space $Z$ and every function $h:Z\to X$, $h\text{ is continuous }⟺f_i\circ h\text{ is continuous for every }i\in I\mathrm{.}$

Final. Let $(Z_i,{\mathcal{S}}_i)$ be spaces and $g_i:Z_i\to X$ functions, and give $X$ the final topology ${\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}$ of the family $(g_i{)}_{i\in I}$. Then:

3. Every $g_i$ is continuous for ${\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}$, and ${\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}$ is the finest topology on $X$ with that property: every topology on $X$ making all the $g_i$ continuous is contained in ${\mathcal{T}}^{\mathrm{f}\mathrm{i}\mathrm{n}}$.
4. Characteristic property. For every space $W$ and every function $k:X\to W$, $k\text{ is continuous }⟺k\circ g_i\text{ is continuous for every }i\in I\mathrm{.}$

Claims 2 and 4 determine their topologies: a topology on $X$ satisfying claim 2 for every $Z$ and $h$ must equal ${\mathcal{T}}^{\mathrm{i}\mathrm{n}}$, and likewise for claim 4, by the argument recorded in the remarks.

• The characteristic property pins the topology down. Suppose two topologies ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ on $X$ both satisfy claim 2 for every space $Z$ and every function $h$. Apply claim 2 for ${\mathcal{T}}_1$ to $Z=(X,{\mathcal{T}}_2)$ and $h=\mathrm{i}\mathrm{d}$: the composites $f_i$ are continuous on $(X,{\mathcal{T}}_2)$ by claim 2 for ${\mathcal{T}}_2$ applied to the identity of $(X,{\mathcal{T}}_2)$, so the identity $(X,{\mathcal{T}}_2)\to (X,{\mathcal{T}}_1)$ is continuous, that is ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$. Exchanging the roles gives equality. The same argument with the arrows reversed does claim 4.

• Only continuity of the composites is tested, never their openness. Claim 2 says nothing about whether $h$ is open or closed, and claim 4 says nothing about $k$; the constructions below acquire such properties one at a time and each is proved where it is used.

• The one-element family is not a degenerate case but the main one. The subspace topology is the initial topology of a single inclusion and the quotient topology is the final topology of a single surjection (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology), so claims 2 and 4 with $I$ a one-element set already carry the characteristic properties of both.

The product set. Let $I$ be a set and let $X_i$ be a set for each $i\in I$. The product is

${\prod }_{i\in I}{X}_i:=\{x:x\text{ is a function with domain }I\text{ and }x(i)\in X_i\text{ for every }i\in I\},$

and we write $x_i:=x(i)$, the $i$-th coordinate of $x$. Two elements of the product are equal exactly when they agree at every index, functions being equal when they have the same domain and the same values. For $j\in I$ the $j$-th projection is

${\pi }_j:{\prod }_{i\in I}{X}_i\to X_j,{\pi }_j(x):=x_j\mathrm{.}$

Notation for a finite product. For $I=n$ a natural number, which is the set $\{0,1,\dots ,n-1\}$ of its predecessors, an element of ${\prod }_{k<n}{X}_k$ is a function on $n$ and we write it $(x_0,\dots ,x_{n-1})$. In particular $I=2$ gives the binary product, written $X\times Y$ for ${\prod }_{i<2}{X}_i$ with $X_0=X$ and $X_1=Y$, whose elements are written $(u,v)$ for the function $0\mapsto u$, $1\mapsto v$. This is the only meaning the symbol $X\times Y$ carries on this page.

Two facts about when the product is nonempty, stated because they are used and because they cost something. If some $X_{i_0}$ is empty then the product is empty, since no function can take a value in $X_{i_0}$. Conversely, suppose every $X_i$ is nonempty.

• For $I=n$ a natural number, the product is nonempty, and this is a theorem of ZF: Every natural-number-indexed list of nonempty sets has a choice function on its family of values applied to the function $i\mapsto X_i$ on $n$ supplies a choice function $g$ for the family of values, and $x(i):=g(X_i)$ defines a member of ${\prod }_{k<n}{X}_k$.
• For an arbitrary $I$ the assertion "${\prod }_{i\in I}{X}_i\mathrm{\ne }\mathrm{\varnothing }$ whenever every $X_i$ is nonempty" is the Axiom of Choice: it is the formulation recorded in The Axiom of Choice, and the choice function of Choice function is exactly a point of the product of a family by itself. Every use of it below is flagged at the step that spends it.

The box topology. Now let each $X_i$ carry a topology ${\mathcal{T}}_i$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Put

$\mathcal{R}:=\{{\prod }_{i\in I}{U}_i:U_i\in {\mathcal{T}}_i\text{ for every }i\in I\},$

the family of boxes. $\mathcal{R}$ is a basis for a topology (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis): it contains ${\prod }_i{X}_i$, so it covers the product, and it is closed under binary intersections, since

$\left({\prod }_i{U}_i\right)\cap \left({\prod }_i{V}_i\right)={\prod }_i(U_i\cap V_i)$

and each $U_i\cap V_i$ is open by (T3). The topology it generates is the box topology ${\mathcal{T}}^{\mathrm{\square }}$, and $\mathcal{R}$ is a basis for it (Basis and subbasis for a topology, and the topology generated by a family of sets).

The product topology. The product topology ${\mathcal{T}}^{\mathrm{\Pi }}$ on ${\prod }_i{X}_i$ is the initial topology of the family of projections $({\pi }_i{)}_{i\in I}$ (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology): the topology generated by the subbasis

$\mathcal{G}:=\{{\pi }_i^{-1}[U]:i\in I,U\in {\mathcal{T}}_i\},{\pi }_i^{-1}[U]={\prod }_{j\in I}{W}_j\text{ with }{W}_i=U\text{ and }{W}_j=X_j\text{ for }j\mathrm{\ne }i\mathrm{.}$

By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections of members of $\mathcal{G}$ form a basis for ${\mathcal{T}}^{\mathrm{\Pi }}$, and those finite intersections are exactly the boxes with all but finitely many factors unrestricted:

${\mathcal{R}}^{\mathrm{\Pi }}=\{{\prod }_{i\in I}{U}_i:U_i\in {\mathcal{T}}_i\text{ for every }i,\text{ and }{U}_i=X_i\text{ for all but finitely many }i\}\mathrm{.}$

Indeed the intersection of ${\pi }_{i_1}^{-1}[U_1],\dots ,{\pi }_{i_n}^{-1}[U_n]$ is the box whose factor at $i$ is the intersection of those $U_m$ with $i_m=i$ and is $X_i$ when no $i_m$ equals $i$; and the intersection of no members is the whole product, the box with every factor $X_i$. Conversely a box with $U_i=X_i$ off a finite set is such an intersection. Members of ${\mathcal{R}}^{\mathrm{\Pi }}$ are called basic product-open sets, and members of $\mathcal{R}$ boxes. So ${\mathcal{R}}^{\mathrm{\Pi }}\subseteq \mathcal{R}$, with equality when $I$ is a natural number.

The empty product. For $I=\mathrm{\varnothing }$ there is exactly one function with domain $\mathrm{\varnothing }$, the empty function, so ${\prod }_{i\in \mathrm{\varnothing }}{X}_i$ is a one-point set. A one-point set carries exactly one topology, namely $\{\mathrm{\varnothing },\{\mathrm{\varnothing }\}\}$, since a topology must contain the empty set and the whole set and there is nothing else to contain (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); so the box topology and the product topology agree there, and both equal the discrete topology and the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which coincide on a one-point set. There are no projections to speak of, and the initial topology of the empty family is indeed the indiscrete one (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology).

Convention. Unless the box topology is named explicitly, ${\prod }_i{X}_i$ always carries the product topology in this library. That is not a matter of taste: the product topology is the one with the characteristic property of the next item, and the box topology has no such property.

• Where the two topologies actually differ. The box topology is finer than the product topology by construction, since ${\mathcal{R}}^{\mathrm{\Pi }}\subseteq \mathcal{R}$. They agree whenever $I$ is finite; and, assuming the Axiom of Choice, for a family of nonempty spaces they differ for infinite $I$ as soon as infinitely many factors have a nonempty proper open subset. Nonemptiness is not decoration: if one factor is empty then the product is empty and carries exactly one topology, so the two agree however the other factors are chosen. Both statements are proved two items below, with that hypothesis, and the failure is recorded on this page as a false statement.

• The product set is a set of functions, and that is not a technicality. The factors are indexed by an arbitrary set, so there is no "list" to write down; writing $x=(x_i{)}_{i\in I}$ is notation for the function $x$. The finite case recovers the familiar tuple, and the identification of ${\prod }_{k<n}\mathbb{R}$ with the ${\mathbb{R}}^n$ of ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it is literal, that item defining ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$.

• The projections carry no hypothesis. They are defined for every product, including the empty one and products with an empty factor; what does need a hypothesis is their surjectivity, which is the point at which choice enters and which is stated separately in the next item.

Let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be topological spaces and let $P:={\prod }_{i\in I}{X}_i$ carry the product topology, with projections ${\pi }_j$ (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then:

1. The projections are continuous, and the product topology is the coarsest topology on $P$ making all of them continuous.
2. Characteristic property. For every space $Z$ and every function $h:Z\to P$, $h\text{ is continuous }⟺{\pi }_i\circ h\text{ is continuous for every }i\in I\mathrm{.}$ The functions ${\pi }_i\circ h$ are the components of $h$, and every family of functions $h_i:Z\to X_i$ arises from exactly one $h$, namely $h(z)(i):=h_i(z)$.
3. The projections are open maps (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), for the product topology and for the box topology alike. They need not be closed; that failure is recorded on this page as a false statement.
4. Surjectivity. If every $X_i$ is nonempty then every ${\pi }_j$ is surjective. For $I$ a natural number this is a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values); for an arbitrary $I$ it is the Axiom of Choice (The Axiom of Choice), and this is the only place in the item where a choice principle is used.

• Exactly where choice is spent, and where it is not. Openness of the projections (claim 3) is choice free: step 1.3 uses a single point of the box in question, which is given by the assumption that the box is nonempty, and builds the required preimage from it by changing one coordinate. Surjectivity (claim 4) is different, because there the point has to be produced from nothing but nonemptiness of the factors, and for an infinite index set that is the Axiom of Choice itself.

• The characteristic property is what makes the product topology the right one. The box topology has no analogue of claim 2: a map into a box-topologised product may have all components continuous and fail to be continuous, and the companion page exhibits the diagonal of ${\mathbb{R}}^{\mathbb{N}}$ doing exactly that.

• Openness does not survive to closedness. A projection is always open and is in general not closed, and the standard witness, the hyperbola in ${\mathbb{R}}^2$, is worked in the false statement on this page. There is no asymmetry of taste here: images of open boxes are computed coordinatewise, while a closed set of the product need not be a union of closed boxes at all.

Let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be topological spaces, let $P:={\prod }_{i\in I}{X}_i$, and let ${\mathcal{T}}^{\mathrm{\Pi }}$ and ${\mathcal{T}}^{\mathrm{\square }}$ be the product and the box topology on $P$ (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then:

1. ${\mathcal{T}}^{\mathrm{\Pi }}\subseteq {\mathcal{T}}^{\mathrm{\square }}$: the box topology is finer than the product topology (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).
2. If $I$ is a natural number then ${\mathcal{T}}^{\mathrm{\Pi }}={\mathcal{T}}^{\mathrm{\square }}$; this is a theorem of ZF.
3. Assume the Axiom of Choice. Suppose every $X_i$ is nonempty and let $J:=\{i\in I:X_i\text{ has an open subset }U\text{ with }\mathrm{\varnothing }\mathrm{\ne }U\mathrm{\ne }{X}_i\}\mathrm{.}$ If $J$ is not finite then ${\mathcal{T}}^{\mathrm{\Pi }}⊊{\mathcal{T}}^{\mathrm{\square }}$: the inclusion of claim 1 is strict.

Claim 3 spends the Axiom of Choice twice (The Axiom of Choice), once to produce a point of $P$ and once to select an open set together with a point of it in each factor indexed by $J$; both uses are flagged at the steps that make them. The hypothesis is stated in terms of open subsets rather than as "infinitely many factors are non-trivial", because a factor may have more than one point and still have no open set other than $\mathrm{\varnothing }$ and itself, as the indiscrete topology shows (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and for such a factor the conclusion fails.