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TheoremStatement: AI-adaptedProof: 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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A connected, locally path-connected space is path-connected, because its path components are open

Statement

Let X be a locally path-connected topological space (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point). Then:

  1. Path components are open, hence clopen, hence unions of them are clopen.
  2. Components and path components agree: P(x)=C(x) for every xX (Paths, path-connected spaces and path components, Connected components, quasicomponents, and totally disconnected spaces).
  3. If X is moreover connected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) then X is path-connected.

Claim 3 is the statement in the title; claims 1 and 2 are what carry it, and both are worth having on their own. Local path-connectedness alone does not make a space path-connected — a two-point discrete space is locally path-connected and is not path-connected — so the connectedness hypothesis in claim 3 is not removable.

Facts & Assumptions

[A1]

For every xX and every open Ux there is an open path-connected V with xVU; in particular, taking U=X, an open path-connected Vx exists (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A2]

The path components partition X; yP(x) exactly when a path in X joins x to y; a path-connected subset of X containing x is contained in P(x), since each of its points is joined to x inside it and hence in 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).

[A3]

A union of open sets is open, and a set is open when each of its points has an open set around it inside it (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A4]

A space is connected exactly when its only clopen subsets are and the whole space; a subset A is connected exactly when the only subsets of A clopen in A are and A; the traces of open and of closed sets are the open and the closed sets of a subspace (For a topological space the following agree: no separation exists, the only clopen subsets are and X, and every continuous map to the two-point discrete space is constant, claims 1 and 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).

[A5]

C(x) is the largest connected subset of X containing x; a path-connected space, and a path-connected subset, is connected (Connected components, quasicomponents, and totally disconnected spaces, Every path-connected space is connected, and every path component lies inside a component, claim 2).

Proof

technique · direct
1.1

Let xX and let yP(x). By [A1] there is an open path-connected V with yVX, and VP(y)=P(x) by [A2], the set V being path-connected and containing y, which lies in P(x).

A1A2
1.2

For claim 2, P(x)C(x) by [A5], P(x) being a path-connected subset containing x, hence connected, and C(x) the largest such.

A2A5
2.1

So every point of P(x) has an open set around it inside P(x), whence P(x) is open in X by [A3].

step 1.1A3
3.1

P(x) is also closed: its complement is the union of the remaining path components, which partition X by [A2], and each of them is open by step 2.1; so the complement is open by [A3]. Hence every path component is clopen, and so is any union of them, being a union of open sets with complement a union of open sets. This is claim 1.

step 2.1A2A3
4.1

Conversely P(x)C(x) is clopen in the subspace C(x) by step 3.1 and [A4], being the trace on C(x) of a clopen subset of X, and it is nonempty, containing x; since C(x) is connected, [A4] forces P(x)C(x)=C(x), that is C(x)P(x).

step 3.1A4A5
5.1

Claim 2 follows from steps 1.2 and 4.1.

step 1.2step 4.1
6.1

For claim 3 assume X is connected. If X= it is path-connected by [A2], having no pair of points to join. Otherwise fix xX; then P(x) is clopen by step 3.1 and nonempty, so P(x)=X by [A4], which says exactly that every point of X is joined to x by a path, and hence any two points are joined to each other. So X is path-connected.

step 3.1A2A4

Remarks

  • Why the argument is about path components and not about paths. The hypothesis gives small open path-connected sets, and the only use made of them is that they cannot straddle two path components. That turns a local statement into the global partition of claim 1 with no construction of a long path anywhere; the path joining two given points is produced only at the very end, by the definition of P(x).

  • Claim 2 is why local path-connectedness is the right hypothesis in practice. Under it the two partitions of X coincide, so "connected" and "path-connected" become interchangeable for subspaces that are open, and every connectedness computation can be done with paths.

  • What fails without local path-connectedness. Claim 1 is exactly where the hypothesis is spent: without it a path component need not be open, its complement need not be open, and the clopen argument collapses. A connected space whose path components are not open, and which is therefore connected and not path-connected, is constructed later on this page.

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