CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-24

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One member of every three-set closed cover of $S^{2}$ contains an antipodal pair

Statement

If three closed subsets $A_{1}, A_{2}, A_{3}$ cover $S^{2}$ , then one of them contains a pair of antipodal points: there are $i \in {1, 2, 3}$ and $x \in S^{2}$ with $x, - x \in A_{i}$ .

Facts & Assumptions

Given: Closed subsets $A_{1}, A_{2}, A_{3} \subseteq S^{2}$ with $S^{2} = A_{1} \cup A_{2} \cup A_{3}$ .

[L1]

For every continuous map $f : S^{2} \to R^{2}$ , there is an $x \in S^{2}$ with $f (x) = f (- x)$ (Borsuk–Ulam theorem in dimension two).

[L2]

If $A$ is a nonempty subset of a metric space, then $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so $x \mapsto d (x, A)$ is continuous ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

[L3]

For every closed subset $C$ of a metric space there is a continuous real-valued function with zero set $C$ ; for nonempty $C$ one may use $d (x, C)$ , and for $C = \emptyset$ one may use the constant function $1$ (In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal).

[F1]

The sphere $S^{2}$ carries the Euclidean subspace metric (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

Proof

technique · direct

1.1givenF1L2L3

For $i = 1, 2$ , define $δ_{i} (x) = d (x, A_{i})$ when $A_{i} \neq \emptyset$ , and define $δ_{i} (x) = 1$ when $A_{i} = \emptyset$ . By [L2] and [L3], each $δ_{i}$ is continuous and its zero set is exactly $A_{i}$ .

2.1step 1.1L1choose

Apply [L1] to $δ = (δ_{1}, δ_{2}) : S^{2} \to R^{2}$ . There is $x \in S^{2}$ such that $δ_{1} (x) = δ_{1} (- x)$ and $δ_{2} (x) = δ_{2} (- x)$ .

3.1step 1.1step 2.1L3given∎

If either common value is zero, then $x$ and $- x$ both lie in the corresponding $A_{i}$ . If both common values are positive, neither point lies in $A_{1} \cup A_{2}$ , so the covering hypothesis puts both in $A_{3}$ . In every case one cover member contains the antipodal pair.

Depends on

Used by

Four closed sets can cover S² without any one containing an antipodal pair Counterexample

Dependency tree · two levels

45 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

Allen Hatcher, Algebraic Topology, Corollary 1.11 (standard reference, not scraped)

One member of every three-set closed cover of S2S2 contains an antipodal pair