False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

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FALSE: the closure of a path-connected subspace is path-connected

Statement

False claim: if $A$ is a path-connected subset of a topological space $X$ (Paths, path-connected spaces and path components, 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 $\overline{A}$ is path-connected (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

The corresponding statement for connectedness is true and is 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; the false claim above is that statement with "connected" replaced by "path-connected" throughout, and the replacement is not legitimate.

Witness. In $X = \mathbb{R}^2$ take $A = G$ , the graph of the zigzag function (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): $G$ is path-connected and $\overline{G}$ is not.

Facts & Assumptions

Given: The zigzag graph $G \subseteq \mathbb{R}^2$ and its closure $\overline{G}$ , with the subspace topology.

[A1]

A subset is path-connected when any two of its points are joined by a path with image in it; closure is taken in the ambient space (Paths, path-connected spaces and path components, 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, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L1]

$G$ is path-connected (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, claim 1).

[L2]

$\overline{G}$ is not path-connected (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, claim 4).

Refutation

technique · contradiction

1.1

Suppose, for contradiction, that the claim holds: the closure of every path-connected subset is path-connected.

assume-contra

2.1

$G$ is a path-connected subset of $\mathbb{R}^2$ by [L1], so the supposed claim applies to it.

step 1.1L1A1

3.1

It follows that $\overline{G}$ is path-connected.

step 2.1

4.1

This contradicts [L2]. So the claim is false.

step 3.1L2discharge-contradiction∎

Remarks

Exactly one word changes between the true statement and the false one. 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, with connectedness as in Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, holds in every topological space with no hypothesis at all, and the witness above shows that its path-connected analogue holds in none but the cases where some further hypothesis is present. The asymmetry has a cause: a point of $\overline{A} \setminus A$ is a limit of points of $A$ , which is enough to prevent a separation but not enough to produce a path, a path being a single continuous map defined on the whole unit interval.
The added set is as small as it can usefully be. By claim 2 of 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, $\overline{G} \setminus G$ is one segment. So a path-connected set can lose path-connectedness on adjoining a single closed segment, and no larger or more complicated addition is needed.
What the witness does not show. Nothing here says that the closure of a path-connected set is never path-connected; it usually is. The claim refuted is the universal one.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 102 results over 17 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)
Topologist's sine curve (Wikipedia) (standard reference, not scraped)
Keith Conrad, Spaces That Are Connected but Not Path-Connected (standard reference, not scraped)