CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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The comb space is path-connected and fails to be locally connected at every point of the limit tooth strictly above the base, so path-connectedness does not imply local connectedness

Statement refuted

Refuted: that a path-connected space is locally connected. Neither Every path-connected space is connected, and every path component lies inside a component nor any statement on the page it belongs to asserts this, and it is false.

Witness, the comb space. Writing $\iota$ for the canonical natural so that $1/(n+1)$ means $1/\iota(n+1)$ with $n \in \mathbb{N}$ and $\mathbb{N}$ containing $0$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), put

$T \;:=\; \{0\} \cup \Bigl\{\tfrac{1}{n+1} : n \in \mathbb{N}\Bigr\} \subseteq [0,1], \qquad C \;:=\; \bigl([0,1] \times \{0\}\bigr) \;\cup\; \bigl(T \times [0,1]\bigr)$

as a subspace of $\mathbb{R}^2$ with the product topology (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, 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 $C$ is path-connected, hence connected, and $C$ is not locally connected at any point $(0,t)$ with $t > 0$ .

The base $[0,1] \times \{0\}$ is the spine of the comb, the sets $\{c\} \times [0,1]$ for $c \in T$ are its teeth, and the failure occurs along the limit tooth $\{0\} \times [0,1]$ above the base.

Facts & Assumptions

Given: $\mathbb{R}^2$ with the product topology and the comb $C$ above, with $\pi_0$ the first projection.

[A1]

The sets $(c,d) \times (e,g)$ form a basis of $\mathbb{R}^2$ ; a map into $\mathbb{R}^2$ is continuous exactly when both components are; the projections are continuous; an affine map of $\mathbb{R}$ into $\mathbb{R}$ is continuous (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, 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, 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, Continuity of a map of topological spaces at a point and globally, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[A2]

Being joined by a path is an equivalence relation on a space, so joins compose; a space is path-connected when any two of its points are joined; path-connected implies connected (Paths, path-connected spaces and path components, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Every path-connected space is connected, and every path component lies inside a component, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A3]

A subset of $\mathbb{R}$ is a connected subset exactly when it is order-convex, and a continuous image of a connected space is connected (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$ ", A continuous image of a connected space is connected, and connectedness is a topological property, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[A4]

$X$ is locally connected at $x$ when every open $U \ni x$ contains an open connected $V$ with $x \in V \subseteq U$ ; a neighbourhood need not be open (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, Connected components, quasicomponents, and totally disconnected spaces).

[A5]

For every real $\varepsilon > 0$ there is a natural $k \ge 1$ with $1/k < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[A6]

A point lies in the closure of a set exactly when every basic open set containing it meets the set (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, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Counterexample

technique · direct

1.1

Every point of $C$ is joined by a path in $C$ to the origin $(0,0)$ . For $(x,0)$ on the base, $t \mapsto ((1-t)x, 0)$ is continuous by [A1] and stays in the base; for $(c,u)$ on a tooth with $c \in T$ , the map $t \mapsto (c, (1-t)u)$ is continuous by [A1], stays in $\{c\} \times [0,1] \subseteq C$ and joins $(c,u)$ to $(c,0)$ , which lies on the base.

A1A2

1.2

For $t > 0$ and $\eta := t/2 > 0$ put $U_t := C \cap \bigl(\mathbb{R} \times (t - \eta,\ t + \eta)\bigr)$ , an open subset of $C$ containing $(0,t)$ ; every point of $U_t$ has second coordinate $> t - \eta = t/2 > 0$ , so $U_t$ contains no point of the base and $\pi_0[U_t] \subseteq T$ .

1.3

No subset of $T$ with two distinct points is order-convex: for $p < q$ in $T$ there is a real strictly between them and outside $T$ , namely the midpoint of $1/(k+2)$ and $1/(k+1)$ for a suitable $k$ , since between consecutive members of $T$ there is no member of $T$ and every element of $T$ other than $0$ is some $1/(k+1)$ .

A3A5

2.1

By step 1.1 and [A2] any two points of $C$ are joined to each other through $(0,0)$ , so $C$ is path-connected and hence connected.

step 1.1A2

2.2

Let $t > 0$ and suppose $V$ is open in $C$ , connected, with $(0,t) \in V \subseteq U_t$ . Then $\pi_0[V]$ is a connected subset of $\mathbb{R}$ by [A1] and [A3], hence order-convex, and $\pi_0[V] \subseteq T$ by step 1.2; so $\pi_0[V]$ has at most one point by step 1.3, and containing $0$ it equals $\{0\}$ . Hence $V \subseteq \{0\} \times [0,1]$ .

step 1.2step 1.3A1A3

3.1

But $V$ is open in $C$ and contains $(0,t)$ , so by [A1] there is $\eta' > 0$ with $C \cap \bigl((-\eta',\eta') \times (t-\eta', t+\eta')\bigr) \subseteq V$ ; by [A5] there is $k \ge 1$ with $1/k < \eta'$ , and with $n := k-1 \in \mathbb{N}$ the point $(1/(n+1), t)$ lies in that trace, hence in $V$ , while its first coordinate is not $0$ . This contradicts step 2.2.

step 2.2A1A5A6

4.1

So no such $V$ exists and $C$ fails to be locally connected at $(0,t)$ for every $t > 0$ , by [A4], while being path-connected and connected by step 2.1.

step 2.1step 3.1A4∎

Remarks

The comb and the zigzag closure fail in different ways, which is why both are on this page. The zigzag closure is connected and not path-connected; the comb is path-connected and still not locally connected. So local connectedness is not implied even by the strongest of the three global conditions, and the three properties are genuinely independent.
Where the hypothesis $t > 0$ is used. At $(0,0)$ the comb is locally connected: small neighbourhoods of the origin contain a piece of the base together with the bottoms of nearby teeth, and that set is path-connected by the argument of step 1.1. The failure needs the point to sit strictly above the base, so that the small neighbourhood $U_t$ of step 1.2 misses the base entirely and the teeth become disconnected from one another.
Deleting the base is what makes it a counterexample and not a curiosity. Without the base the teeth are disjoint clopen segments and the space is disconnected; with the base they are welded at the bottom, so a path may always descend, travel and climb. Local connectedness fails precisely because that detour is not available inside a small neighbourhood high above the base.

Depends on

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Sources

Comb space (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)