TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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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.

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

Statement

Let $X$ be a topological space, let $A \subseteq X$ be a connected subset (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) and let $B$ satisfy

$A \subseteq B \subseteq \overline{A},$

the closure being taken in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then $B$ is a connected subset of $X$ , 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).

Taking $B = \overline{A}$ : the closure of a connected set is connected. Taking $B = A$ recovers the hypothesis, so the statement is a genuine interpolation between a connected set and its closure: every set squeezed between the two is connected, and one may stop anywhere.

Facts & Assumptions

Given: A space $X$ , a connected subset $A \subseteq X$ , and a set $B$ with $A \subseteq B \subseteq \overline{A}$ .

[A1]

A subset $S \subseteq X$ is disconnected exactly when $S = S_{1} \cup S_{2}$ with $S_{1}, S_{2}$ nonempty and separated in $X$ , that is $\overline{S_{1}} \cap S_{2} = \emptyset = S_{1} \cap \overline{S_{2}}$ ; equivalently $S$ is connected exactly when no such decomposition exists (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, 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]

Closure is monotone: if $P \subseteq Q$ then $\overline{P} \subseteq \overline{Q}$ , since $\overline{Q}$ is a closed set containing $P$ and $\overline{P}$ is the smallest such (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).

Proof

technique · direct

1.1

Suppose $B = B_{1} \cup B_{2}$ with $B_{1}$ and $B_{2}$ nonempty and separated in $X$ , so that $\overline{B_{1}} \cap B_{2} = \emptyset$ and $B_{1} \cap \overline{B_{2}} = \emptyset$ .

1.2

Put $A_{1} : = A \cap B_{1}$ and $A_{2} : = A \cap B_{2}$ ; then $A = A_{1} \cup A_{2}$ , because $A \subseteq B = B_{1} \cup B_{2}$ .

given

2.1

$A_{1}$ and $A_{2}$ are separated in $X$ : $\overline{A_{1}} \cap A_{2} \subseteq \overline{B_{1}} \cap B_{2} = \emptyset$ by [A2] and step 1.1, and symmetrically $A_{1} \cap \overline{A_{2}} \subseteq B_{1} \cap \overline{B_{2}} = \emptyset$ .

step 1.1step 1.2A2

3.1

Since $A$ is connected, [A1] and steps 1.2 and 2.1 forbid both $A_{1}$ and $A_{2}$ from being nonempty, so at least one is empty; the hypothesis of step 1.1 is symmetric in $B_{1}$ and $B_{2}$ , so after relabelling we may assume $A_{2} = A \cap B_{2} = \emptyset$ .

step 1.2step 2.1A1given

4.1

Then $A \subseteq B_{1}$ , since $A \subseteq B_{1} \cup B_{2}$ and $A$ meets $B_{2}$ in nothing by step 3.1; hence $\overline{A} \subseteq \overline{B_{1}}$ by [A2].

step 3.1A2given

5.1

Therefore $B_{2} \subseteq B \subseteq \overline{A} \subseteq \overline{B_{1}}$ , so $B_{2} \subseteq \overline{B_{1}} \cap B_{2} = \emptyset$ by step 1.1; that is $B_{2} = \emptyset$ , contradicting its nonemptiness in step 1.1.

step 1.1step 4.1given

6.1

So no decomposition as in step 1.1 exists, and $B$ is a connected subset of $X$ by [A1].

step 1.1step 5.1A1∎

Remarks

What fails without the upper bound $B \subseteq \overline{A}$ . The conclusion is false for an arbitrary superset of a connected set. In $R$ take $A = (0, 1)$ , which is connected, and $B = (0, 1) \cup {2}$ : the ambient open sets $(- 1, 1)$ and $(1, 3)$ meet $B$ in $(0, 1)$ and ${2}$ , two nonempty disjoint relatively open pieces covering $B$ , which is exactly the decomposition the proof rules out. The point $2$ lies outside $\overline{A} = [0, 1]$ , and that is what makes the separation available; in a general space lying outside the closure supplies only one half of a separation, so the hypothesis is stated as the inclusion $B \subseteq \overline{A}$ rather than as a condition on individual added points. The hypothesis is used only at step 5.1, and that is where it is needed: it forces $B_{2}$ to lie inside $\overline{A}$ , hence inside $\overline{B_{1}}$ , hence to be empty.
The interior of a connected set need not be connected. Nothing here transfers to interiors, and the two operations behave differently: closure adds points that cling to the set and cannot split it, whereas the interior may remove the very points holding two lumps together.
Where this is used. It is the second half of the standard method for building a connected set that is not path-connected: take a path-connected set, which is connected, and close it up. The closure is connected by this theorem regardless of how badly the added points behave, and that is what The graph of the piecewise-linear map oscillating between $0$ and $1$ on the intervals $[1 / (n + 2), 1 / (n + 1)]$ is path-connected, its closure adds the segment ${0} \times [0, 1]$ , and that closure is connected, is not path-connected because no path joins the segment to the graph, and is not locally connected exploits.

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

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

If AA is connected and A⊆B⊆A‾A⊆B⊆A then BB is connected; in particular the closure of a connected set is connected