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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31
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The germ neighborhoods form a Hausdorff, second-countable Riemann-surface atlas

Statement

Let R(ξ0,Ω) be the germ space of a complete analytic function. Then the sets N(f,U) of The germ space of a complete analytic function form a basis for a topology on R(ξ0,Ω). With that topology, the maps

ϕf,U:N(f,U)→U,ϕf,U([f]z)=z,

form a holomorphic atlas. The resulting space is Hausdorff and second countable.

Facts & Assumptions

Given: The germ space R(ξ0,Ω) and its subsets N(f,U).

[L1]

The germ space, its basic candidate sets N(f,U), and the projection p([f]z)=z are those of The germ space of a complete analytic function.

[L2]

A family is a basis exactly when it covers the set and every point of an intersection of two members lies in a third member inside that intersection (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis).

[L3]

If two holomorphic functions agree on a set with an accumulation point in a complex domain, then they agree on that whole domain (Identity theorem for holomorphic functions).

[L4]

Hausdorff means that distinct points admit disjoint open neighbourhoods, and second countable means that the topology has a countable basis (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Second countability: an at most countable basis for the topology).

Proof

technique · direct
1.1L1

Every point of R(ξ0,Ω) is, by [L1], a germ [f]z coming from some function element (f,U), and then [f]z∈N(f,U). So the family {N(f,U)} covers the germ space.

1.2L1L2

Suppose ξ=[f]z lies in N(f,U)∩N(g,V). Then ξ=[g]z as well, so [L1] gives equality of the germs of f and g at z. Hence there is a disc D centered at z with D⊆U∩V and f=g on D. For each w∈D this implies [f]w=[g]w, so N(f,D)=N(g,D)⊆N(f,U)∩N(g,V). Thus [L2] makes the family {N(f,U)} a basis for a topology on the germ space.

1.3L3L4cases

The space is Hausdorff. If [f]z and [g]w have z≠w, choose disjoint discs Dz∋z and Dw∋w; then N(f,Dz) and N(g,Dw) are disjoint basis neighbourhoods. If z=w but [f]z≠[g]z, choose discs Df⊆U and Dg⊆V centered at z so small that Df∩Dg is connected. If N(f,Df) and N(g,Dg) met, then f and g would agree as germs at some point of Df∩Dg, and [L3] would force f=g on that connected overlap, hence near z, contradiction. So distinct germs have disjoint neighbourhoods, exactly as [L4] requires.

1.4L1algebra

To prove second countability, let D be the countable family of rational open discs contained in Ω. For every finite chain D0,…,Dm of discs in D with a0∈D0 and Dj−1∩Dj≠∅, at most one branch of the complete analytic function is determined on Dm by continuing the initial germ successively across that chain. So the basis sets arising from such rational-disc chains form a countable family.

2.1step 1.2L3algebra

On each basis element, ϕf,U is bijective with inverse z↦[f]z. If N(f,U)∩N(g,V)≠∅, then step 1.2 gives a disc D⊆U∩V on which f=g, so on ϕf,U(N(f,D))=D the transition map ϕg,V∘ϕf,U−1:D→D is the identity. Therefore the charts are holomorphically compatible.

3.1L1L4L5step 1.4construct∎

Let [f]z∈N(f,U). By [L5], there is a polygonal path in Ω from a0 to z. Cover its compact image by finitely many rational discs from D that lie inside the function-element neighborhoods of one continuation chain to [f]z, and choose them in the order encountered along the path, with the last disc contained in U and containing z. The resulting rational-disc chain determines the same terminal branch on that last disc, so it produces a countable-basis neighbourhood of [f]z contained in N(f,U). Thus the topology has a countable basis, and [L4] makes the germ space second countable.

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