LemmaStatement: AI-adaptedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A countable coordinate-bump map embeds a manifold in countable Euclidean data

Statement

Assume ${A C}_{ω}$ , and let $M^{n}$ be a smooth manifold. Then there are countably many smooth coordinate balls $(U_{j}, x_{j})$ , open coordinate balls $V_{j} ⋐ U_{j}$ covering $M$ , and smooth bump functions $ϕ_{j}$ supported in $U_{j}$ and equal to $1$ on $V_{j}$ such that the globally defined smooth blocks $B_{j} (p) : = {\begin{cases} (ϕ_{j} (p), ϕ_{j} (p) x_{j}^{1} (p), \dots, ϕ_{j} (p) x_{j}^{n} (p)), & p \in U_{j}, \\ 0, & p \notin U_{j} \end{cases} \in R^{n + 1} .$ separates points and tangent vectors: if $p \neq q$ , then $B_{j} (p) \neq B_{j} (q)$ for some $j$ , and for each $p \in M$ there is an index $j$ with $p \in V_{j}$ such that the last $n$ coordinates of $B_{j}$ give the chart coordinates on a neighbourhood of $p$ .

Facts & Assumptions

Given: A smooth $n$ -manifold $M$ and ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The underlying topological manifold is second countable (Smooth manifolds and their smooth charts).

[L1]

At every point of a smooth manifold there is a smooth chart onto an open Euclidean set (Smooth manifolds and their smooth charts). Euclidean balls can be nested inside that set, and the inner coordinate ball then has compact closure inside the outer one by the compact-closure argument of Coordinate balls form a basis of a topological manifold.

[L2]

Under ${A C}_{ω}$ , an open cover of a second countable space has a countable subcover (Assuming countable choice, every second countable space is Lindelöf).

[L3]

For compact $K \subseteq W$ with $W$ open in a smooth manifold, a smooth function supported in $W$ equals $1$ on a neighbourhood of $K$ (A manifold bump for a compact set inside an open set).

Proof

technique · direct

1.1givenL1L2L3choose

Let $V$ consist of all smooth coordinate balls $V$ for which there exists a smooth coordinate ball $(U, x)$ with $\overline{V}$ compact and contained in $U$ . This is an open cover of $M$ : at a point $p$ , take a smooth chart $(W, x)$ and concentric Euclidean balls $B (x (p), r) \subset B (x (p), R)$ whose outer closed ball lies in $x (W)$ . Their inverse images give the required $V$ and $U$ by [L1]. By [L2], choose a countable subcover $(V_{j})$ . For each $j$ , use countable choice to select a witnessing smooth coordinate ball $(U_{j}, x_{j})$ , and then apply [L3] and countable choice to select a smooth $ϕ_{j} : M \to [0, 1]$ supported in $U_{j}$ and equal to $1$ on an open neighbourhood of $\overline{V_{j}}$ . Thus $V_{j} ⋐ U_{j}$ and $ϕ_{j} = 1$ on $V_{j}$ . If $M$ is empty, take the family to be empty.

2.1step 1.1construct

Define the coordinate blocks $B_{j}$ by the piecewise formula in the statement. The chart-coordinate formula is smooth on $U_{j}$ . Its coordinate products vanish on a neighbourhood of each point outside $U_{j}$ , because $supp ϕ_{j} \subseteq U_{j}$ is closed. Hence extension by zero is smooth on all of $M$ .

3.1step 1.1step 2.1algebra

If $p \neq q$ , choose $j$ with $p \in V_{j}$ . If $B_{j} (p) = B_{j} (q)$ , then $ϕ_{j} (p) = 1$ , hence $ϕ_{j} (q) = 1$ , so both points lie in $U_{j}$ . Equality of the last $n$ coordinates of $B_{j}$ then gives $x_{j} (p) = x_{j} (q)$ , contradicting the injectivity of the chart map. Therefore some block separates $p$ and $q$ .

4.1step 1.1algebra∎

Fix $p \in M$ and choose $j$ with $p \in V_{j}$ . On $V_{j}$ one has $ϕ_{j} \equiv 1$ , so the last $n$ coordinates of $B_{j}$ are exactly the chart coordinates $x_{j}^{1}, \dots, x_{j}^{n}$ . Their differential is an isomorphism at $p$ , so this single block already detects every nonzero tangent vector at $p$ . Thus the family $(B_{j})$ separates tangent vectors as claimed.

Depends on

Used by

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Dependency tree · two levels

23 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

Marco Gualtieri, Topology I: Smooth Manifolds, Part 11 (standard reference, not scraped)
John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 6 (standard reference, not scraped)