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

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Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space

Statement

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

Containment. $C (x) \subseteq Q (x)$ .
Closedness. $Q (x)$ is closed in $X$ .
Saturation. If $y \in Q (x)$ then $Q (y) = Q (x)$ ; consequently $C (y) \subseteq Q (x)$ for every $y \in Q (x)$ , and $Q (x) = ⋃ {C (y) : y \in Q (x)},$ so every quasicomponent is a union of components.
Partition. The quasicomponents are nonempty, pairwise disjoint, and cover $X$ .

No converse is asserted. Claim 1 is an inclusion and this theorem does not claim it is an equality; the question of when $C (x) = Q (x)$ is not settled on this page, and nothing here may be read as settling it.

Facts & Assumptions

Given: A topological space $X$ , a point $x \in X$ , and 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 largest connected subset of $X$ containing $x$ , and $Q (x)$ is the intersection of all clopen $K \subseteq X$ with $x \in K$ ; that family is nonempty, $X$ being clopen; every point lies in its own component (Connected components, quasicomponents, and totally disconnected spaces, The components of a space are its maximal connected subsets, they partition it, and each of them is closed, claims 1 and 2).

[A2]

A connected space has no clopen subset other than $\emptyset$ and the whole space; a subset $A \subseteq X$ is connected exactly when the only subsets of $A$ clopen in $(A, T_{A})$ are $\emptyset$ and $A$ (For a topological space the following agree: no separation exists, the only clopen subsets are $\emptyset$ and $X$ , and every continuous map to the two-point discrete space is constant, claims 1 and 2, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A3]

The traces $U \cap A$ of open sets are the open sets of $A$ , and the traces $F \cap A$ of closed sets are the closed sets of $A$ (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).

[A4]

A clopen set is closed; a nonempty intersection of closed sets is closed; the complement of a clopen set is clopen (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1A1A3

Let $K \subseteq X$ be clopen with $x \in K$ . Then $K \cap C (x)$ is both open and closed in the subspace $C (x)$ by [A3], and it contains $x$ , so it is nonempty.

1.2A1A4

Every clopen $K ∋ y$ also contains $x$ whenever $y \in Q (x)$ : otherwise $X ∖ K$ is a clopen set containing $x$ by [A4], so $Q (x) \subseteq X ∖ K$ by [A1], contradicting $y \in Q (x) \cap K$ .

1.3A1A4

$Q (x)$ is closed, being by [A1] the intersection of a nonempty family of clopen, hence closed, sets, and such an intersection is closed by [A4]; this is claim 2. And $x \in Q (x)$ , every member of that family containing $x$ .

2.1step 1.1A1A2

Since $C (x)$ is connected by [A1], its only clopen subsets are $\emptyset$ and $C (x)$ by [A2]; so step 1.1 forces $K \cap C (x) = C (x)$ , that is $C (x) \subseteq K$ .

2.2step 1.2A1

Let $y \in Q (x)$ . Every clopen $K$ with $x \in K$ contains $y$ , since $Q (x) \subseteq K$ by [A1], so $Q (y) \subseteq K$ ; hence $Q (y) \subseteq Q (x)$ . Conversely every clopen $K$ with $y \in K$ contains $x$ by step 1.2, so $Q (x) \subseteq K$ and therefore $Q (x) \subseteq Q (y)$ . Thus $Q (y) = Q (x)$ .

3.1step 2.1A1

As $K$ was an arbitrary clopen set containing $x$ , it follows that $C (x)$ is contained in the intersection of all of them, that is $C (x) \subseteq Q (x)$ ; this is claim 1.

4.1step 3.1step 2.2A1

So for $y \in Q (x)$ one has $C (y) \subseteq Q (y) = Q (x)$ by step 3.1 and step 2.2; and each such $y$ lies in $C (y)$ by [A1], so $Q (x) = ⋃ {C (y) : y \in Q (x)}$ . This is claim 3.

5.1step 1.3step 2.2∎

For claim 4: each $Q (x)$ is nonempty by step 1.3; if $z \in Q (x) \cap Q (y)$ then $Q (z) = Q (x)$ and $Q (z) = Q (y)$ by step 2.2, so $Q (x) = Q (y)$ , and hence two quasicomponents are equal or disjoint; and $x \in Q (x)$ by step 1.3, so they cover $X$ .

Remarks

Where the inclusion can be strict, and why the proof cannot be improved. Step 2.1 uses connectedness of $C (x)$ to promote "meets $K$ " to "is contained in $K$ ". Running the argument backwards would need every point of $Q (x)$ to be joined to $x$ by a connected set, and nothing in the definition of $Q$ provides one: $Q (x)$ records only that no clopen set separates the two points. That gap is real and not an artefact of this proof.
Both partitions are into closed sets, and they are nested. The components partition $X$ into closed sets (The components of a space are its maximal connected subsets, they partition it, and each of them is closed), the quasicomponents partition $X$ into closed sets by claims 2 and 4, and by claim 3 the second partition is coarser: every quasicomponent is a union of whole components.
Claim 3 is what makes the notion useful. A clopen set never cuts a component in half, so any argument that produces a clopen set separating two points has automatically shown that they lie in different components. That implication runs only in this direction, which is exactly claim 1.

Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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