CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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The punctured plane has connected complement in C but disconnected spherical complement

Statement refuted

A connected complex domain with connected complement in $C$ is simply connected.

Facts & Assumptions

Given: The punctured plane $Ω = C^{\times}$ .

[L1]

Assuming the Axiom of Choice, a plane domain is simply connected exactly when its spherical complement is connected (Assuming the Axiom of Choice, a plane domain is simply connected exactly when its spherical complement is connected).

[L2]

The fundamental group of $C^{\times}$ is $Z$ , detected by winding number (Winding number identifies the fundamental group of C times with the integers).

Counterexample

technique · direct

1.1

The complement of $Ω$ in $C$ is the singleton ${0}$ , hence connected. But the spherical complement is $\hat{C} ∖ Ω = {0, \infty},$ which is disconnected.

given

2.1

Assuming the Axiom of Choice, [L1] makes the disconnected spherical complement from step 1.1 enough to conclude that $Ω$ is not simply connected. Fact [L2] records the same failure independently as the nontrivial group $π_{1} (C^{\times}) ≅ Z$ .

step 1.1L1L2

3.1

Therefore connected complement in $C$ does not imply simple connectivity.

step 1.1step 2.1∎

Depends on

Used by

FALSE: a connected plane-domain complement in C already implies simple connectivity False statement

Dependency tree · two levels

9 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

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 4.3.1 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 3, §6 (standard reference, not scraped)