TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice

Statement

Let $(X_{i}, T_{i})_{i \in I}$ be topological spaces and let $P : = \prod_{i \in I} X_{i}$ carry the product topology, with projections $π_{j}$ (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:

The projections are continuous, and the product topology is the coarsest topology on $P$ making all of them continuous.
Characteristic property. For every space $Z$ and every function $h : Z \to P$ , $h is continuous ⟺ π_{i} \circ h is continuous for every i \in I .$ The functions $π_{i} \circ h$ are the components of $h$ , and every family of functions $h_{i} : Z \to X_{i}$ arises from exactly one $h$ , namely $h (z) (i) : = h_{i} (z)$ .
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.
Surjectivity. If every $X_{i}$ is nonempty then every $π_{j}$ 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.

Facts & Assumptions

Given: Topological spaces $(X_{i}, T_{i})_{i \in I}$ , the product $P = \prod_{i \in I} X_{i}$ with the product topology and the projections $π_{j} (x) = x_{j}$ , a space $Z$ and a function $h : Z \to P$ , and an index $j \in I$ .

[A1]

The product topology on $P$ is the initial topology of $(π_{i})_{i \in I}$ , and a basis for it is the family of boxes $\prod_{i} U_{i}$ with every $U_{i}$ open and $U_{i} = X_{i}$ for all but finitely many $i$ ; a basis for the box topology is the family of all boxes $\prod_{i} U_{i}$ with every $U_{i}$ open (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, Basis and subbasis for a topology, and the topology generated by a family of sets).

[A2]

$f$ is an open map when $f [U]$ is open in the target for every open $U$ in the source (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L1]

For a topology given as an initial topology of a family $(f_{i})$ : each $f_{i}$ is continuous, the topology is the coarsest with that property, and a map $h$ into it is continuous exactly when every $f_{i} \circ h$ is (Characteristic properties: a map into a space with the initial topology is continuous iff every composite with the defining family is, a map out of a space with the final topology is continuous iff every composite with the defining family is, and the two topologies are respectively the coarsest and the finest making that family continuous, claims 1 and 2; 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, Continuity of a map of topological spaces at a point and globally).

[L2]

If $F$ is a function with domain a natural number $n$ whose values are nonempty sets, then the family of its values has a choice function (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Choice function).

[L3]

If every member of a family of sets is nonempty then the product of the family is nonempty; this is the Axiom of Choice (The Axiom of Choice, Choice function).

[L4]

The image of a union is the union of the images, and an arbitrary union of open sets is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1A1L1

By [A1] the product topology is an initial topology, so [L1] gives claim 1 and claim 2 at once, the defining family being $(π_{i})_{i \in I}$ .

1.2given

For a family of functions $h_{i} : Z \to X_{i}$ the assignment $h (z) (i) : = h_{i} (z)$ defines a function $Z \to P$ , since $h (z)$ has domain $I$ and $h (z) (i) = h_{i} (z) \in X_{i}$ ; it satisfies $π_{i} \circ h = h_{i}$ , and any $h^{'}$ with $π_{i} \circ h^{'} = h_{i}$ for every $i$ satisfies $h^{'} (z) (i) = h_{i} (z) = h (z) (i)$ for all $z$ and $i$ , hence $h^{'} = h$ .

1.3A1choose

Let $B = \prod_{i} U_{i}$ be a box with every $U_{i}$ open. If $B = \emptyset$ then $π_{j} [B] = \emptyset$ . If $B \neq \emptyset$ , fix $b \in B$ ; then $π_{j} [B] = U_{j}$ , since $π_{j} [B] \subseteq U_{j}$ by definition, and for $u \in U_{j}$ the function $y$ with $y_{j} : = u$ and $y_{i} : = b_{i}$ for $i \neq j$ lies in $B$ and has $π_{j} (y) = u$ .

1.4L2

Assume every $X_{i}$ is nonempty and $I$ is a natural number $n$ . By [L2] applied to $i \mapsto X_{i}$ there is a choice function $g$ for the family of values, and $x (i) : = g (X_{i})$ defines a point of $P$ ; so $P \neq \emptyset$ .

1.5L3

Assume every $X_{i}$ is nonempty and $I$ is arbitrary. By [L3] the product $P$ is nonempty.

2.1step 1.3A1A2L4

Both the box topology and the product topology have a basis consisting of boxes, by [A1], and the image of a union of basic sets is the union of their images; so by step 1.3 the image under $π_{j}$ of any open set of either topology is a union of sets each of which is $\emptyset$ or an open $U_{j} \subseteq X_{j}$ , hence open. This is claim 3.

2.2step 1.4step 1.5

Assume every $X_{i}$ is nonempty and let $t \in X_{j}$ . By step 1.4 when $I$ is a natural number, and by step 1.5 in general, there is a point $p \in P$ ; the function $y$ with $y_{j} : = t$ and $y_{i} : = p_{i}$ for $i \neq j$ then lies in $P$ and satisfies $π_{j} (y) = t$ . So $π_{j}$ is surjective, which is claim 4.

3.1step 1.1step 1.2step 2.1step 2.2∎

Step 1.1 gives claims 1 and 2, step 1.2 gives the bijection between maps into $P$ and families of component maps, step 2.1 gives claim 3 and step 2.2 gives claim 4.

Remarks

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 $R^{N}$ 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 $R^{2}$ , 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.

A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources