TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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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.

The components of a space are its maximal connected subsets, they partition it, and each of them is closed

Statement

Let $X$ be a topological space and let $C (x)$ be the connected component of $x \in X$ (Connected components, quasicomponents, and totally disconnected spaces). Then:

Maximality. $C (x)$ is connected, contains $x$ , and contains every connected subset of $X$ that contains $x$ . So the components are exactly the maximal connected subsets of $X$ : a connected $A \subseteq X$ is a component if and only if no connected subset of $X$ properly contains $A$ — except in the empty space, where $\emptyset$ is vacuously maximal and yet is not a component, there being no points; for nonempty $X$ no exception is needed, since $\emptyset$ is properly contained in a connected singleton and so is never maximal.
Partition. For $x, y \in X$ , either $C (x) = C (y)$ or $C (x) \cap C (y) = \emptyset$ ; every point lies in its own component; and $X = ⋃_{x \in X} C (x)$ . So the components are nonempty, pairwise disjoint, and cover $X$ .
Closedness. Every component is closed in $X$ .

Components need not be open, and no clause above says they are. Openness of the components is a genuine extra hypothesis on $X$ , taken up later on this page under the name local connectedness.

Facts & Assumptions

Given: A topological space $X$ , with subsets carrying 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).

[A1]

$C (x)$ is the union of all connected subsets of $X$ containing $x$ ; it is connected, contains $x$ , and contains every connected set through $x$ (Connected components, quasicomponents, and totally disconnected spaces, 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, claim 1).

[A2]

A union of connected subsets with a point in common is connected (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, claim 1).

[A3]

If $A$ is connected and $A \subseteq B \subseteq \overline{A}$ then $B$ is connected (If $A$ is connected and $A \subseteq B \subseteq \overline{A}$ then $B$ is connected; in particular the closure of a connected set is connected).

[A4]

$A \subseteq \overline{A}$ , and $A$ is closed exactly when $A = \overline{A}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[A5]

A singleton 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

Claim 1 is [A1]: $C (x)$ is connected, contains $x$ by [A5] since ${x}$ is one of the sets united, and contains every connected $A ∋ x$ because such an $A$ is one of the sets united.

A1A5

2.1

A connected $A \subseteq X$ with $A \neq \emptyset$ satisfies $A \subseteq C (a)$ for every $a \in A$ , by step 1.1; so $A$ is maximal among connected subsets exactly when $A = C (a)$ , and every component is nonempty.

A1A5

2.2

Suppose $z \in C (x) \cap C (y)$ . Then $C (x) \cup C (y)$ is connected by [A2], the two sets being connected by [A1] and sharing $z$ .

step 1.1A1A2

2.3

For claim 3, apply [A3] with $A = C (x)$ and $B = \overline{C (x)}$ , the hypothesis $A \subseteq B \subseteq \overline{A}$ holding by [A4]; so $\overline{C (x)}$ is connected, and it contains $x$ , hence $\overline{C (x)} \subseteq C (x)$ by step 1.1.

step 1.1A3A4

3.1

That union contains $x$ , so it is contained in $C (x)$ by step 1.1, whence $C (y) \subseteq C (x)$ ; it also contains $y$ , so symmetrically $C (x) \subseteq C (y)$ , and therefore $C (x) = C (y)$ .

step 1.1step 2.2

4.1

So for any $x, y$ either $C (x) \cap C (y) = \emptyset$ or $C (x) = C (y)$ by step 3.1; and $x \in C (x)$ by step 1.1, so $X = ⋃_{x \in X} C (x)$ and every component is nonempty by step 2.1. This is claim 2.

step 1.1step 2.1step 3.1

5.1

With $C (x) \subseteq \overline{C (x)}$ from [A4] this gives $C (x) = \overline{C (x)}$ , so $C (x)$ is closed by [A4].

step 2.3A4∎

Remarks

The exception for $\emptyset$ in claim 1 is not a quibble. The empty set is connected under the convention of Connected components, quasicomponents, and totally disconnected spaces, and it is contained in every set, so "maximal connected subset" has to be read as "maximal among the nonempty connected subsets" for the identification with components to be exact. Step 2.1 is where that is pinned down.
Closed but not open is the typical case. Claim 3 uses only that the closure of a connected set is connected, which is available in every space. There is no matching argument for openness, because the union of the connected sets through a point carries no information about neighbourhoods; that is what local connectedness supplies, and it is a genuine extra hypothesis rather than a missing step.
A component of a subspace is computed in that subspace. For $S \subseteq X$ the components of $S$ are the maximal connected subsets of the space $S$ , and claim 3 then says each is closed in $S$ , not in $X$ . Closedness in $X$ follows only when $S$ itself is closed in $X$ .

Depends on

Used by

Dependency tree · two levels

16 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

Connected space (Wikipedia) (standard reference, not scraped)
The Stacks Project, Lemma 5.7.3 (standard reference, not scraped)