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

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The zigzag curve and its closure worked out: the components, the path components, and the points at which local connectedness fails

Example

Let $G$ be the graph of the zigzag function and $\overline{G} = G \cup (\{0\} \times [0,1])$ its closure in $\mathbb{R}^2$ (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 2). Write $\Sigma := \{0\} \times [0,1]$ for the added segment. Then, in the space $\overline{G}$ with its 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):

One component. $\overline{G}$ is connected, so it has exactly one component, namely $\overline{G}$ itself, and exactly one quasicomponent, also $\overline{G}$ (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, Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space).
Two path components, namely $G$ and $\Sigma$ (Paths, path-connected spaces and path components).
$G$ is open in $\overline{G}$ and $\Sigma$ is closed in it; neither is clopen, since $\overline{G}$ is connected.
Local connectedness holds at every point of $G$ and fails at every point of $\Sigma$ (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point). So the set of points at which $\overline{G}$ fails to be locally connected is exactly $\Sigma$ .

Claim 2 is the sharp form of "not path-connected": the failure is not that some pair of points is unjoined but that the space splits into exactly two path classes, one of which is the whole added segment.

Facts & Assumptions

Given: $G$ , $\Sigma = \{0\} \times [0,1]$ and $\overline{G} = G \cup \Sigma$ as subspaces of $\mathbb{R}^2$ .

[L1]

$G$ is path-connected, connected and locally connected; $\overline{G} = G \cup \Sigma$ ; $\overline{G}$ is connected; no path in $\overline{G}$ joins a point of $\Sigma$ to a point of $G$ ; and $\overline{G}$ is not locally connected at any point of $\Sigma$ (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, claims 1, 2, 3, 4, 5).

[A1]

The component of a point is the largest connected set containing it, the components partition the space, and every quasicomponent contains the component of each of its points (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, Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space).

[A2]

The path components partition the space and each is path-connected; a path-connected subset containing $x$ lies inside the path component of $x$ (Paths, path-connected spaces and path components).

[A3]

A map into $\mathbb{R}^2$ is continuous exactly when both components are, constant maps and the inclusion of an interval are continuous, and a continuous image of a connected space is connected (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Continuity of a map of topological spaces at a point and globally, A continuous image of a connected space is connected, and connectedness is a topological property, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A4]

A product of closed subsets of $\mathbb{R}$ is closed in $\mathbb{R}^2$ ; for $A \subseteq S$ the closure of $A$ in the subspace $S$ is $\overline{A} \cap S$ ; a set is closed exactly when it equals its closure (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, 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, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ , The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[A5]

$X$ is locally connected at $x$ when every open $U \ni x$ contains an open connected $V$ with $x \in V \subseteq U$ ; if $S$ is open in $X$ then a subset of $S$ is open in $S$ exactly when it is open in $X$ (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point, 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).

Verification

technique · direct

1.1

$\overline{G}$ is connected by [L1], so the largest connected subset containing any of its points is $\overline{G}$ itself; by [A1] it is therefore the unique component, and by [A1] the unique quasicomponent contains it and is contained in the space, hence equals it. This is claim 1.

L1A1

1.2

$\Sigma = \{0\} \times [0,1]$ is path-connected: for $(0,s), (0,u) \in \Sigma$ the map $t \mapsto (0, s + t(u-s))$ is continuous into $\mathbb{R}^2$ by [A3] and takes values in $\Sigma$ , since $s + t(u-s)$ lies between $s$ and $u$ , hence in $[0,1]$ .

2.1

$\Sigma$ is closed in $\mathbb{R}^2$ by [A4], being a product of two closed subsets of $\mathbb{R}$ ; hence $\Sigma = \overline{\Sigma} \cap \overline{G}$ is closed in $\overline{G}$ by [A4], and $G = \overline{G} \setminus \Sigma$ is open in $\overline{G}$ . This is claim 3, the "not clopen" half following from claim 1, a clopen proper nonempty subset being impossible in a connected space.

step 1.1A4

2.2

$G$ and $\Sigma$ are path-connected subsets of $\overline{G}$ by [L1] and step 1.2, so each lies inside a single path component by [A2]; and no path joins a point of one to a point of the other by [L1], so the two lie in different path components. Since $G \cup \Sigma = \overline{G}$ , the path components are exactly $G$ and $\Sigma$ . This is claim 2.

step 1.2L1A2

3.1

Local connectedness holds at every point of $G$ : let $p \in G$ and let $U$ be open in $\overline{G}$ with $p \in U$ . Then $U \cap G$ is open in $\overline{G}$ by step 2.1, hence open in $G$ by [A5], $G$ being open; $G$ is locally connected by [L1], so there is $V$ open in $G$ and connected with $p \in V \subseteq U \cap G$ ; and $V$ is open in $\overline{G}$ by [A5].

step 2.1L1A5

4.1

Local connectedness fails at every point of $\Sigma$ by [L1]; with step 3.1 this shows the failure set is exactly $\Sigma$ , which is claim 4.

step 3.1L1A5∎

Remarks

The two path components have different topological characters. $\Sigma$ is homeomorphic to a closed bounded interval and $G$ to a half-open one, and only the first is closed in $\overline{G}$ . So the partition into path components is not a partition into clopen pieces, which is exactly what A connected, locally path-connected space is path-connected, because its path components are open would supply if the space were locally path-connected — and it is not, by claim 4.
Why $\overline{G}$ has one component and two path components at once. Connectedness is destroyed only by a clopen splitting, and adjoining $\Sigma$ to $G$ creates none: every neighbourhood of a point of $\Sigma$ meets $G$ by 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 1, and 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. Path-connectedness is destroyed by a single continuity failure along one map, and that is enough.
The quasicomponent carries no extra information here. Since the space is connected, its unique quasicomponent equals its unique component, so this witness says nothing about the strictness of the inclusion $C(x) \subseteq Q(x)$ ; a separate space is needed for that.

Depends on

Used by

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Sources

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