TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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The complement of a compact plane set has exactly one unbounded connected component

Statement

Let $K \subseteq C$ be compact. Then $C ∖ K$ has exactly one unbounded connected component $U_{\infty}$ , and every other component is bounded. Moreover, whenever $R > 0$ satisfies $K \subseteq {z : ∣ z ∣ \leq R}$ , the exterior ${z : ∣ z ∣ > R}$ is contained in $U_{\infty}$ .

The empty set is covered: $C ∖ \emptyset = C$ has the single component $C$ , which is unbounded.

Facts & Assumptions

Given: A compact set $K \subseteq C$ ; 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.

[L1]

For $c \in C$ and $R > 0$ , the set ${z : ∣ z - c ∣ > R}$ is path-connected and a connected subset of $C$ (The exterior of a closed disc in the plane is path-connected).

[L2]

The connected component $C (x)$ is the union of all connected subsets containing $x$ (Connected components, quasicomponents, and totally disconnected spaces).

[L3]

$C (x)$ contains every connected subset containing $x$ ; distinct components are disjoint; every point lies in its own component, and the components cover the space (The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

[L4]

A union of connected subsets with a common point is connected (A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member).

[L5]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

[L6]

A subset $A$ of a metric space is bounded when $A = \emptyset$ or $A \subseteq B (x_{0}, r)$ for some $x_{0}$ and some real $r > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

[L7]

A set is closed exactly when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L8]

For $n \geq 1$ , $R^{n}$ is polygonally connected and connected ( $R^{n}$ is polygonally connected, connected, locally path-connected and locally connected).

[L9]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

technique · direct

1.1

By [L5] the set $K$ is closed and bounded, so by [L6] there are $x_{0}$ and $s > 0$ with $K \subseteq B (x_{0}, s)$ , and then $K \subseteq {z : ∣ z ∣ \leq R}$ for $R = ∣ x_{0} ∣ + s > 0$ ; if $K = \emptyset$ any $R > 0$ serves. By [L7] the complement $C ∖ K$ is open.

givenL5L6L7

1.2

Fix any $R > 0$ with $K \subseteq {z : ∣ z ∣ \leq R}$ and put $E_{R} = {z : ∣ z ∣ > R}$ . Then $E_{R} \subseteq C ∖ K$ , and $E_{R}$ is a connected subset of $C$ by [L1]; it is nonempty, since $2 R \in E_{R}$ , and unbounded, since for every real $r > 0$ and every $x_{0}$ the number $R + ∣ x_{0} ∣ + r$ has modulus exceeding $R$ and lies outside $B (x_{0}, r)$ , so no ball of [L6] contains $E_{R}$ .

givenL1L6L9

2.1

Let $z_{0} \in E_{R}$ and let $U_{\infty} : = C (z_{0})$ be its component in $C ∖ K$ . The set $E_{R}$ is a connected subset of $C ∖ K$ containing $z_{0}$ , so $E_{R} \subseteq U_{\infty}$ by [L2] and [L3]; hence $U_{\infty}$ is unbounded by step 1.2 and [L6]. This holds for every admissible $R$ , which is the final clause of the statement.

step 1.1step 1.2L2L3L6

3.1

Let $C$ be a component of $C ∖ K$ with $C \neq U_{\infty}$ . By [L3] the two are disjoint, so $C \cap E_{R} = \emptyset$ by step 2.1, that is $C \subseteq {z : ∣ z ∣ \leq R} \subseteq B (0, R + 1)$ ; hence $C$ is bounded by [L6].

step 2.1L3L6

4.1

Steps 2.1 and 3.1 give exactly one unbounded component, namely $U_{\infty}$ , with every other component bounded. When $K = \emptyset$ the complement is $C$ , which is connected by [L8], so by [L3] and [L4] it is its own single component and that component is $U_{\infty}$ .

step 2.1step 3.1L3L4L8∎

Depends on

Used by

Dependency tree · two levels

55 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)
J. Lebl, Complex Analysis, Ch. 4 §4.1 (standard reference, not scraped)