False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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.

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FALSE: every connected topological space is path-connected

Statement

False claim: every connected topological space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) is path-connected (Paths, path-connected spaces and path components).

The implication holds in the other direction — every path-connected space is connected (Every path-connected space is connected, and every path component lies inside a component) — and it is that true statement which the false one attempts to reverse.

Witness. The closure $\overline{G}$ of the zigzag graph (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), a subspace of $R^{2}$ (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), is connected and is not path-connected.

Facts & Assumptions

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

[A1]

A space is path-connected when any two of its points are joined by a path, and connected when it admits no separation (Paths, path-connected spaces and path components, 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).

[L1]

$\overline{G}$ is 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 3).

[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: every connected space is path-connected.

assume-contra

1.2

$\overline{G}$ is a topological space, being a subspace of $R^{2}$ , and it is connected by [L1].

L1A1

2.1

Applying the supposed claim to $\overline{G}$ gives that $\overline{G}$ is path-connected.

step 1.1step 1.2

3.1

This contradicts [L2], which says $\overline{G}$ is not path-connected. So the claim is false.

step 2.1L2discharge-contradiction∎

Remarks

What survives is the converse, and only the converse. Every path-connected space is connected, and every path component lies inside a component is a theorem: path-connectedness implies connectedness, always. The false claim is its reversal, and $\overline{G}$ shows that no amount of connectedness alone produces a path.
Where the failure sits in the witness. 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 the space $\overline{G}$ is the graph $G$ together with the segment ${0} \times [0, 1]$ . The graph on its own is path-connected; adjoining the segment keeps the space connected, because the closure of a connected set is connected, and destroys path-connectedness, because a path reaching the segment from $G$ would have to take the values $0$ and $1$ in its second coordinate arbitrarily late.
A hypothesis that does repair it. A connected space that is also locally path-connected is path-connected (A connected, locally path-connected space is path-connected, because its path components are open), and $\overline{G}$ fails that extra hypothesis at every point of the segment.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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)
Topologist's sine curve (Wikipedia) (standard reference, not scraped)
Keith Conrad, Spaces That Are Connected but Not Path-Connected (standard reference, not scraped)