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Coordinate balls form a basis of a topological manifold

Statement

Let $M$ be a topological $n$ -manifold and let $p \in M$ . For every open neighbourhood $O$ of $p$ there is a chart $(U, φ)$ of $M$ at $p$ and an open Euclidean ball $B (c, r) \subseteq R^{n}$ such that

$p \in φ^{- 1} [B (c, r)] \subseteq O, \overline{B (c, r)} \subseteq φ (U) .$

Consequently the sets $φ^{- 1} [B (c, r)]$ of this form constitute a basis of the topology of $M$ . Their closures in $M$ are compact.

Facts & Assumptions

Given: A topological $n$ -manifold $M$ , a point $p \in M$ , and an open neighbourhood $O$ of $p$ .

[F1]

Every point of a topological manifold has a chart $(U, φ)$ onto an open subset of $R^{n}$ (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces, Manifold charts, coordinate domains, and coordinate functions).

[L1]

For $n \geq 1$ , Euclidean closed balls are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact).

[A1]

If $W \subseteq R^{n}$ is open and $x \in W$ , then there exists $r > 0$ with $\overline{B (x, r)} \subseteq W$ ; this is the usual metric-ball shrinking property of Euclidean open sets.

[A2]

Homeomorphisms preserve openness, closures inside their domains, and compactness of subsets.

Proof

technique · direct

1.1

By [F1] choose a chart $(U, φ)$ of $M$ at $p$ . Since $O$ is open and $p \in U \cap O$ , replacing $U$ by $U \cap O$ and $φ$ by its restriction still gives a chart at $p$ whose image is the open set $φ (U \cap O)$ in $R^{n}$ . So we may assume from the start that $U \subseteq O$ .

givenF1choose

2.1

Put $c : = φ (p)$ . Because $φ (U)$ is open, [A1] gives $r > 0$ with $\overline{B (c, r)} \subseteq φ (U)$ . Then $p \in φ^{- 1} [B (c, r)] \subseteq U \subseteq O$ . This gives the required coordinate ball inside $O$ .

A1step 1.1choose

3.1

The closure of $φ^{- 1} [B (c, r)]$ in $M$ is contained in $φ^{- 1} [\overline{B (c, r)}]$ , and [A2] identifies $φ^{- 1} [\overline{B (c, r)}]$ with the homeomorphic image of the Euclidean closed ball $\overline{B (c, r)}$ . For $n \geq 1$ this set is compact by [L1], hence its homeomorphic image is compact by [A2] and the smaller closure in $M$ is compact as a closed subset of a compact set. When $n = 0$ , the chart image is the one-point space $R^{0}$ , so the same conclusion is immediate.

L1A2step 2.1

4.1

Since step 2.1 works for every point $p$ and every open neighbourhood $O$ of $p$ , the coordinate balls form a basis of the topology of $M$ .

step 2.1∎

Depends on

Used by

Topological manifolds are locally compact and locally path connected Proposition
Topological manifolds are sigma-compact Theorem

Dependency tree · two levels

11 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, §2.1 (standard reference, not scraped)