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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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  • Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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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 (Xi,Ti)iI(X_i, \mathcal{T}_i)_{i \in I} be topological spaces and let P:=iIXiP := \prod_{i \in I} X_i carry the product topology, with projections πj\pi_j (The product set iIXi\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 PP making all of them continuous.
  2. Characteristic property. For every space ZZ and every function h:ZPh : Z \to P, h is continuous     πih is continuous for every iI.h \text{ is continuous } \iff \pi_i \circ h \text{ is continuous for every } i \in I . The functions πih\pi_i \circ h are the components of hh, and every family of functions hi:ZXih_i : Z \to X_i arises from exactly one hh, namely h(z)(i):=hi(z)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 XiX_i is nonempty then every πj\pi_j is surjective. For II 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 II 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 (Xi,Ti)iI(X_i,\mathcal{T}_i)_{i \in I}, the product P=iIXiP = \prod_{i \in I} X_i with the product topology and the projections πj(x)=xj\pi_j(x) = x_j, a space ZZ and a function h:ZPh : Z \to P, and an index jIj \in I.

[A1]

The product topology on PP is the initial topology of (πi)iI(\pi_i)_{i \in I}, and a basis for it is the family of boxes iUi\prod_i U_i with every UiU_i open and Ui=XiU_i = X_i for all but finitely many ii; a basis for the box topology is the family of all boxes iUi\prod_i U_i with every UiU_i open (The product set iIXi\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]

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

[L2]

If FF is a function with domain a natural number nn 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
1.1

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

A1L1
1.2

For a family of functions hi:ZXih_i : Z \to X_i the assignment h(z)(i):=hi(z)h(z)(i) := h_i(z) defines a function ZPZ \to P, since h(z)h(z) has domain II and h(z)(i)=hi(z)Xih(z)(i) = h_i(z) \in X_i; it satisfies πih=hi\pi_i \circ h = h_i, and any hh' with πih=hi\pi_i \circ h' = h_i for every ii satisfies h(z)(i)=hi(z)=h(z)(i)h'(z)(i) = h_i(z) = h(z)(i) for all zz and ii, hence h=hh' = h.

given
1.3

Let B=iUiB = \prod_i U_i be a box with every UiU_i open. If B=B = \varnothing then πj[B]=\pi_j[B] = \varnothing. If BB \ne \varnothing, fix bBb \in B; then πj[B]=Uj\pi_j[B] = U_j, since πj[B]Uj\pi_j[B] \subseteq U_j by definition, and for uUju \in U_j the function yy with yj:=uy_j := u and yi:=biy_i := b_i for iji \ne j lies in BB and has πj(y)=u\pi_j(y) = u.

A1choose
1.4

Assume every XiX_i is nonempty and II is a natural number nn. By [L2] applied to iXii \mapsto X_i there is a choice function gg for the family of values, and x(i):=g(Xi)x(i) := g(X_i) defines a point of PP; so PP \ne \varnothing.

L2
1.5

Assume every XiX_i is nonempty and II is arbitrary. By [L3] the product PP is nonempty.

L3
2.1

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 πj\pi_j of any open set of either topology is a union of sets each of which is \varnothing or an open UjXjU_j \subseteq X_j, hence open. This is claim 3.

step 1.3A1A2L4
2.2

Assume every XiX_i is nonempty and let tXjt \in X_j. By step 1.4 when II is a natural number, and by step 1.5 in general, there is a point pPp \in P; the function yy with yj:=ty_j := t and yi:=piy_i := p_i for iji \ne j then lies in PP and satisfies πj(y)=t\pi_j(y) = t. So πj\pi_j is surjective, which is claim 4.

step 1.4step 1.5
3.1

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

step 1.1step 1.2step 2.1step 2.2

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 RN\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 R2\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.

Depends on

Used by

…and 5 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources