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

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 subspace $A \subseteq X$ is disconnected exactly when $A = A_{1} \cup A_{2}$ with $A_{1}, A_{2}$ nonempty and separated in $X$ , which is the criterion this library already uses on the real line

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $A \subseteq X$ 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). Write $\overline{B}$ for the closure of $B$ in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then $A$ is a disconnected subset of $X$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) if and only if there are sets $A_{1}, A_{2}$ with

$A = A_{1} \cup A_{2}, A_{1} \neq \emptyset \neq A_{2}, \overline{A_{1}} \cap A_{2} = \emptyset = A_{1} \cap \overline{A_{2}} .$

Equivalently: $A$ is connected if and only if it admits no such decomposition. The two sets in such a decomposition are automatically disjoint, since $A_{1} \cap A_{2} \subseteq \overline{A_{1}} \cap A_{2} = \emptyset$ .

The displayed condition is the one Separated sets, disconnection, and connected subset of $R$ states for subsets of $R$ , with the closure of $R$ replaced by the closure of $X$ : there a disconnection of $E$ is a pair of nonempty separated sets whose union is $E$ , and $E$ is connected when none exists. That the two closures on $R$ are the same operation, and hence that the two definitions agree there, is proved later on this page; nothing in the present lemma asserts it.

Facts & Assumptions

Given: A topological space $(X, T)$ and a subset $A \subseteq X$ with the subspace topology $T_{A}$ .

[A1]

$A$ is a disconnected subset of $X$ exactly when the space $(A, T_{A})$ admits a separation, that is a pair $(W_{1}, W_{2})$ of sets open in $A$ , nonempty, disjoint, with $W_{1} \cup W_{2} = A$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, 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).

[A2]

For $B \subseteq A$ the closure of $B$ in the subspace $A$ is ${cl}_{A} (B) = \overline{B} \cap A$ (For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is ${\overline{A}}^{X} \cap S$ , while the interior only contains ${int}^{X} (A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ , claim 1).

[A3]

In any space a subset equals its own closure exactly when it is closed, and $B \subseteq \overline{B}$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[A4]

In any space a subset is closed exactly when its complement is open; two disjoint sets whose union is the whole space are each the complement of the other (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

Suppose $A$ is disconnected and fix a separation $(W_{1}, W_{2})$ of $(A, T_{A})$ as in [A1]; then $W_{1}$ and $W_{2}$ are nonempty subsets of $A$ with $W_{1} \cup W_{2} = A$ and $W_{1} \cap W_{2} = \emptyset$ .

1.2

Each of $W_{1}, W_{2}$ is closed in $A$ : being complementary in $A$ and both open in $A$ , each is the complement in $A$ of an open set.

A1A4

1.3

Conversely suppose $A = A_{1} \cup A_{2}$ with $A_{1}, A_{2}$ nonempty and $\overline{A_{1}} \cap A_{2} = \emptyset = A_{1} \cap \overline{A_{2}}$ ; then $A_{1} \cap A_{2} = \emptyset$ , since $A_{1} \cap A_{2} \subseteq \overline{A_{1}} \cap A_{2}$ by [A3].

2.1

In the situation of step 1.1, ${cl}_{A} (W_{1}) = W_{1}$ by step 1.2 and [A3], hence $\overline{W_{1}} \cap A = W_{1}$ by [A2]; symmetrically $\overline{W_{2}} \cap A = W_{2}$ .

step 1.1step 1.2A2A3

2.2

In the situation of step 1.3, $\overline{A_{1}} \cap A = (\overline{A_{1}} \cap A_{1}) \cup (\overline{A_{1}} \cap A_{2}) = A_{1} \cup \emptyset = A_{1}$ , using $A = A_{1} \cup A_{2}$ , the hypothesis $\overline{A_{1}} \cap A_{2} = \emptyset$ and $A_{1} \subseteq \overline{A_{1}}$ from [A3]; symmetrically $\overline{A_{2}} \cap A = A_{2}$ .

step 1.3A3

3.1

So in the situation of step 1.1 one has $\overline{W_{1}} \cap W_{2} \subseteq \overline{W_{1}} \cap A \cap W_{2} = W_{1} \cap W_{2} = \emptyset$ , because $W_{2} \subseteq A$ ; symmetrically $W_{1} \cap \overline{W_{2}} = \emptyset$ . Hence $A_{1} : = W_{1}$ and $A_{2} : = W_{2}$ are nonempty, have union $A$ , and are separated in $X$ .

step 1.1step 2.1

3.2

And in the situation of step 1.3 one has ${cl}_{A} (A_{1}) = \overline{A_{1}} \cap A = A_{1}$ by [A2] and step 2.2, so $A_{1}$ is closed in $A$ by [A3]; symmetrically $A_{2}$ is closed in $A$ .

step 1.3step 2.2A2A3

4.1

In the situation of step 1.3 the sets $A_{1}$ and $A_{2}$ are therefore disjoint, cover $A$ , and are each closed in $A$ by step 3.2, so each is the complement in $A$ of the other and hence open in $A$ by [A4]; being nonempty, $(A_{1}, A_{2})$ is a separation of $(A, T_{A})$ and $A$ is disconnected by [A1].

step 1.3step 3.2A1A4

5.1

Step 3.1 gives the forward implication and step 4.1 the backward one, so $A$ is disconnected exactly when the displayed decomposition exists; negating both sides gives the statement for connectedness.

step 3.1step 4.1∎

Remarks

Why the closures are taken in $X$ and the openness in $A$ . The two halves of the criterion live in different spaces on purpose. Relative openness is not visible from $X$ alone — a set open in $A$ need not be open in $X$ — whereas the closure operator of $A$ is computed from that of $X$ by [A2]. Trading the relatively open pieces for ambiently separated ones is exactly what makes the criterion usable when only $X$ is concretely known, which is the situation in every worked example on the companion page.
Separated is strictly stronger than disjoint, and that is what is needed. If the condition asked only for a partition into two nonempty disjoint pieces then every space with at least two points would be "disconnected". The two closure conditions are what forbid one piece from clinging to the other, and each of them is used once in the proof above.
The hypothesis $A_{i} \subseteq A$ is not imposed and is automatic. Both sets appear inside a union equal to $A$ , so each is contained in $A$ ; the statement is written without the redundant hypothesis so that it can be applied directly to a candidate pair.

Depends on

Used by

Dependency tree · two levels

19 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)
Separated sets (Wikipedia) (standard reference, not scraped)

A subspace A⊆XA⊆X is disconnected exactly when A=A1∪A2A=A1​∪A2​ with A1,A2A1​,A2​ nonempty and separated in XX, which is the criterion this library already uses on the real line