TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member

Statement

Let $X$ be a topological space, let $I$ be a set and let $A_i \subseteq X$ be a connected subset of $X$ for every $i \in I$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets). Subsets carry 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). Then:

Common point. If there is $p \in X$ with $p \in A_i$ for every $i \in I$ , then $\bigcup_{i \in I} A_i$ is a connected subset of $X$ .
Common connected core. If $A \subseteq X$ is connected and $A \cap A_i \ne \varnothing$ for every $i \in I$ , then $A \cup \bigcup_{i \in I} A_i$ is a connected subset of $X$ .

No hypothesis of any kind is imposed on the index set: $I$ may be empty, finite or infinite, and no choice principle is used, since the point $p$ in claim 1 and the set $A$ in claim 2 are given rather than selected.

Facts & Assumptions

Given: A space $X$ , a set $I$ , connected subsets $A_i \subseteq X$ for $i \in I$ , and the two-point discrete space $\mathbf{2} = \{0,1\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A1]

A subset $S$ of a space is connected exactly when every continuous map $S \to \mathbf{2}$ is constant (For a topological space the following agree: no separation exists, the only clopen subsets are $\varnothing$ and $X$ , and every continuous map to the two-point discrete space is constant, claim 2, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A2]

The subspace topology is transitive: for $B \subseteq S \subseteq X$ the topology $B$ inherits from $S$ is the topology it inherits from $X$ ; and a restriction of a continuous map to a subspace is continuous (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, Continuity of a map of topological spaces at a point and globally).

[A3]

The empty space is connected, no separation of it existing (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

Proof

technique · direct

1.1

For claim 1 write $S := \bigcup_{i \in I} A_i$ and assume $p \in A_i$ for every $i \in I$ . If $I = \varnothing$ then $S = \varnothing$ , which is connected by [A3], so assume $I \ne \varnothing$ ; then $p \in S$ .

givenA3

1.2

Let $\chi : S \to \mathbf{2}$ be continuous. For each $i \in I$ the restriction $\chi|_{A_i} : A_i \to \mathbf{2}$ is continuous by [A2], the topology $A_i$ carries as a subspace of $S$ being the one it carries as a subspace of $X$ .

1.3

For claim 2 assume $A$ is connected and $A \cap A_i \ne \varnothing$ for every $i \in I$ . If $I = \varnothing$ the union is $A$ , which is connected by hypothesis, so assume $I \ne \varnothing$ ; then $A \ne \varnothing$ , and we fix $p \in A$ .

givenA3

2.1

Each $\chi|_{A_i}$ is constant by [A1], since $A_i$ is connected; and $p \in A_i$ , so that constant value is $\chi(p)$ . Hence $\chi(a) = \chi(p)$ for every $a \in A_i$ and every $i \in I$ .

step 1.2A1given

3.1

Every $s \in S$ lies in some $A_i$ , so $\chi(s) = \chi(p)$ by step 2.1; thus $\chi$ is constant. As $\chi$ was arbitrary, $S$ is connected by [A1]. This is claim 1.

step 1.1step 2.1A1

4.1

For each $i \in I$ the two sets $A$ and $A_i$ are connected and share a point of $A \cap A_i$ , so $A \cup A_i$ is connected by claim 1 applied to the two-member family $\{A, A_i\}$ .

step 3.1given

5.1

The family $\{\, A \cup A_i : i \in I \,\}$ consists of connected sets by step 4.1 and every member contains $p$ by step 1.3, so its union is connected by claim 1; and that union is $A \cup \bigcup_{i \in I} A_i$ , since every member contains $A$ and $I \ne \varnothing$ . This is claim 2.

step 3.1step 1.3step 4.1∎

Remarks

Why a common point and not merely pairwise intersection. Pairwise intersection does not give a common point, so it does not supply claim 1's hypothesis, and the failure is not exotic: three sets can meet pairwise with empty total intersection. Claim 2 is the form that covers that case, since it asks only that each member meet one fixed connected set — and for a nonempty pairwise-intersecting family one may take that fixed set to be any one member, so such a union is connected after all. Claim 1 is the special case in which the fixed set is a single point, a singleton being connected.
Chains are covered by iterating claim 2. If $A_0, A_1, A_2, \dots$ are connected and $A_n \cap A_{n+1} \ne \varnothing$ for every $n$ , then each partial union $A_0 \cup \dots \cup A_n$ is connected by induction using claim 2, and the total union is connected by claim 1 applied to the partial unions, all of which contain $A_0$ . The argument is written out where it is used rather than stated as a further clause here.
Nothing is assumed about openness or closedness of the members. The hypothesis is connectedness alone. This is what makes the theorem the workhorse for building components: an arbitrary union of connected sets through a fixed point is connected, and that is precisely what makes the component of a point well defined.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 50 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

Connected space (Wikipedia) (standard reference, not scraped)
The Stacks Project, Section 5.7: Connected components (standard reference, not scraped)