LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The exterior of a closed disc in the plane is path-connected

Statement

Let $c \in C$ . Then:

for every real $R \geq 0$ , the open exterior $E_{>} : = {z \in C : ∣ z - c ∣ > R}$ is path-connected, and therefore a connected subset of $C$ ; taking $R = 0$ , the punctured plane $C ∖ {c}$ is path-connected;
for every real $R > 0$ , the closed exterior $E_{\geq} : = {z \in C : ∣ z - c ∣ \geq R}$ is path-connected, and therefore a connected subset of $C$ .

Facts & Assumptions

Given: A point $c \in C$ and a real $R$ , with $R \geq 0$ in clause 1 and $R > 0$ in clause 2; the plane is read as $R^{2}$ with its Euclidean metric through $C = R [x] / (x^{2} + 1)$ as the Euclidean plane and as a normed real algebra: what the identification preserves. Write $E$ for whichever of the two sets is under discussion.

[L1]

For $n \geq 2$ the unit sphere $S^{n - 1} \subseteq R^{n}$ is path-connected and connected (For $n \geq 2$ , the sphere $S^{n - 1}$ is path-connected and connected).

[L2]

For $n \geq 1$ the map $ρ (x) = x / ∥ x ∥_{2}$ from $R^{n} ∖ {0}$ to $S^{n - 1}$ is continuous (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ ).

[L3]

For $n \geq 1$ , $S^{n - 1} = S_{2} (0, 1) = {x \in R^{n} : ∥ x ∥_{2} = 1}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L4]

A subset is path-connected when any two of its points are joined by a continuous map from $[0, 1]$ whose image lies in it (Paths, path-connected spaces and path components).

[L5]

A path-connected subset of a topological space is a connected subset (Every path-connected space is connected, and every path component lies inside a component).

[L6]

A composite of continuous maps is continuous, and a function whose restrictions to the members of a finite closed cover are continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L7]

$B (x, r) = {y : d (x, y) < r}$ (Open ball, closed ball and sphere in a metric space).

[L8]

$∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex $z, w$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1givenL2L3L7

Let $z, w \in E$ and put $ρ = \max (∣ z - c ∣, ∣ w - c ∣)$ . In clause 1 this gives $ρ \geq ∣ z - c ∣ > R \geq 0$ and $ρ \geq ∣ w - c ∣ > R$ , and in clause 2 it gives $ρ \geq ∣ z - c ∣ \geq R > 0$ and $ρ \geq ∣ w - c ∣ \geq R$ ; in both cases $ρ > 0$ and $z \neq c$ , $w \neq c$ , so $u = (z - c) / ∣ z - c ∣$ and $v = (w - c) / ∣ w - c ∣$ lie on the unit circle $S^{1}$ by [L2] and [L3].

1.2L1L4

By [L1] with $n = 2$ there is a continuous $σ : [0, 1] \to S^{1}$ with $σ (0) = u$ and $σ (1) = v$ .

2.1step 1.1L6L7L8

The map $μ (s) = c + ((1 - s) ∣ z - c ∣ + s ρ) u$ is continuous on $[0, 1]$ by [L6], joins $z$ to $c + ρ u$ , and satisfies $∣ μ (s) - c ∣ = (1 - s) ∣ z - c ∣ + s ρ$ by [L8] and $∣ u ∣ = 1$ ; that value lies between $∣ z - c ∣$ and $ρ$ , so it exceeds $R$ in clause 1 and is at least $R$ in clause 2, and $μ$ has image in $E$ . The same formula with $w$ and $v$ gives a continuous $ν : [0, 1] \to E$ joining $w$ to $c + ρ v$ .

2.2step 1.1step 1.2L6L7L8

The map $s \mapsto c + ρ σ (s)$ is continuous on $[0, 1]$ by [L6], joins $c + ρ u$ to $c + ρ v$ , and has $∣ c + ρ σ (s) - c ∣ = ρ$ by [L8], which exceeds $R$ in clause 1 and is at least $R$ in clause 2, so its image lies in $E$ .

3.1step 2.1step 2.2L4L5L6∎

Concatenating $μ$ , the path of step 2.2 and the reversal of $ν$ , each on a closed subinterval of $[0, 1]$ and agreeing at the two shared endpoints, gives by [L6] a continuous map $[0, 1] \to E$ from $z$ to $w$ . Since $z, w \in E$ were arbitrary, $E$ is path-connected by [L4], hence a connected subset of $C$ by [L5]; the argument was run for both clauses at once, and at $R = 0$ clause 1 reads $C ∖ {c}$ .

Depends on

Used by

A connected plane domain that is not homologically simply connected Counterexample
A nonvanishing holomorphic function on a domain with no holomorphic logarithm Counterexample
A genus-two compact surface gives a cocompact Fuchsian group Example
Every cycle in a round annulus has one period, that of the central circle Example
First Cousin gluing on the pseudoconvex domain ℂ Example
Any two points admit a minimizing geodesic False statement
A circle traversed k times has winding number k inside and 0 outside Theorem
The complement of a compact plane set has exactly one unbounded connected component Theorem

Dependency tree · two levels

44 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

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §2.1 (standard reference, not scraped)