PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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Components of a topological manifold are open and at most countable

Statement

Let $M$ be a topological manifold. Then every connected component of $M$ is open. Moreover the set of connected components of $M$ is at most countable.

Facts & Assumptions

Given: A topological manifold $M$ .

[F1]

A topological manifold is second countable (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces).

[F2]

A topological manifold is locally path connected (Topological manifolds are locally compact and locally path connected).

[L1]

In a locally connected space, the connected components of every open set are open; applied to the open set $M$ itself, this makes components of $M$ open (A space is locally connected exactly when every component of every open subspace is open; in that case the components of the space itself are clopen).

[L2]

Connected components partition the space and are pairwise disjoint (The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

[A1]

Every path-connected open neighbourhood is connected, so local path connectedness implies local connectedness.

Proof

technique · direct

1.1

By [F2] every point of $M$ has a path-connected open neighbourhood; by [A1] such a neighbourhood is connected, so $M$ is locally connected. Therefore [L1] applied to the open set $M$ shows that every connected component of $M$ is open.

F2L1A1

2.1

Because $M$ is second countable by [F1], choose a countable basis [F1, step 1.1, choose] $B = (B_{n})_{n \in N}$ . Let $C$ be the set of connected components of $M$ . By step 1.1 each $C \in C$ is a nonempty open set, so there exists at least one index $n$ with $B_{n} \subseteq C$ and $B_{n} \neq \emptyset$ . Choose the least such index and call it $m (C)$ .

F1step 1.1choose

3.1

If $C, D \in C$ and $m (C) = m (D) = n$ , then $B_{n} \subseteq C \cap D$ . [L2, step 2.1] Since components are disjoint by [L2], this forces $C = D$ . Thus $C \mapsto m (C)$ is injective from $C$ into $N$ , so $C$ is at most countable.

L2step 2.1

4.1

Step 1.1 proves openness of components and step 3.1 proves that there are at most countably many of them.

step 1.1step 3.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Rob van der Vorst, Introduction to differentiable manifolds, §1 (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds, §2.2 (standard reference, not scraped)