PropositionStatement: Literature-sourcedProof: AI-generatedprecheck 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

1.1F2L1A1

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.

2.1F1step 1.1choose

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 $m (C)$ .

3.1L2step 2.1

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.

4.1step 1.1step 3.1∎

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

Depends on

Used by

Closed embedded submanifolds of complete Riemannian manifolds are complete Corollary
Compact Riemannian manifolds are geodesically complete Corollary
Sufficiently short geodesic segments are uniquely minimizing Corollary
Null-cobordant closed manifolds Definition
Pontryagin numbers of a closed oriented manifold Definition
Stiefel-Whitney numbers of a closed manifold Definition
The suspension foliation of a representation of the fundamental group Definition
Canonical divisors on hyperelliptic curves Example
Nonsingular affine and projective curves as Riemann surfaces Example
A Lie subalgebra need not integrate to a closed subgroup False statement
The exponential map is always defined on all of TM False statement
A finite endpoint of a maximal unit-speed geodesic produces a Cauchy curve Lemma
Boundary of a compact 1-manifold has even cardinality Lemma
Radial geodesics from one point reach every point under global exponential domain Lemma
Uniform short-geodesic scale on a compact surface Lemma
Compactness up to breaking of Morse trajectory spaces Theorem
Existence and uniqueness of maximal connected integral manifolds Theorem
Length minimizers are constant-speed geodesics up to reparametrization Theorem

Dependency tree · two levels

20 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

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