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DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27
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The disjoint union (coproduct) ⨆iXi with the final topology of the canonical injections: a set is open exactly when each of its traces is

Definition

The underlying set. Let I be a set and let Xi be a set for each i∈I. The disjoint union is

⨆i∈IXi  :=  ⋃i∈I(Xi×{i}),

whose elements are the pairs (x,i) with i∈I and x∈Xi. For j∈I the j-th canonical injection is

κj:Xj→⨆i∈IXi,κj(x):=(x,j).

The construction is what makes the word "disjoint" honest. Each κj is injective (Injection, surjection, bijection), since (x,j)=(x′,j) forces x=x′; the images κj[Xj]=Xj×{j} are pairwise disjoint, since the second coordinate determines j; and their union is the whole set. So no assumption that the Xi are disjoint as sets is needed, and none is made: the tag i separates the copies even when Xi=Xi′ for i≠i′.

The trace of a subset. For U⊆⨆iXi and j∈I write

Uj  :=  κj−1[U]  =  { x∈Xj:(x,j)∈U }⊆Xj,

the trace of U on the j-th summand. A subset is determined by its family of traces, since U=⋃iκi[Ui].

The topology. Now let each Xi carry a topology Ti (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). The disjoint union topology (also coproduct topology, or topological sum) on ⨆iXi is the final topology of the family (κi)i∈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), that is

T⊔  :=  { U⊆⨆iXi  :  Ui∈Ti for every i∈I }:

a set is open exactly when each of its traces is open. That this is a topology is discharged in 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, where the final topology of any family is verified to satisfy (T1), (T2) and (T3); nothing further is needed here.

Closed sets, dually. F⊆⨆iXi is closed exactly when every trace Fi is closed in Xi. Indeed the trace operation commutes with complementation, κi−1[ ⨆jXj∖F ]=Xi∖Fi, so F is closed if and only if the complement is open if and only if every Xi∖Fi is open.

Each summand sits inside as a clopen subspace. The set κj[Xj]=Xj×{j} has traces Xj at j and ∅ elsewhere, both open and both closed, so it is clopen in the union. Its 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) is carried across by κj from Tj, and κj is an embedding (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological); both statements are proved in the next item rather than assumed here.

Degenerate cases. For I=∅ the disjoint union is the empty set with its only topology. For I a one-element set the map κ is a bijection carrying T to T⊔, so the construction returns the one summand up to homeomorphism and changes nothing.

Remarks

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources