Lusternik-Schnirelmann theorem

This article describes a theorem about spheres

Statement

If $S^{n}$ is the union of $n + 1$ closed subsets, then at least one of those subsets contains a pair of antipodal points.

Proof

Proof for two dimensions

The proof in the two-dimensional case follows from the Urysohn lemma, and the Borsuk-Ulam theorem in two dimensions. The idea is as follows:

Suppose $S^{2}$ is a union of closed subsets $F_{1}, F_{2}, F_{3}$ . Let $a$ denote the antipode map. Suppose that $F_{1} \cap a (F_{1})$ is empty and $F_{2} \cap a (F_{2})$ is empty. Then, by the Urysohn lemma, and the fact that $S^{2}$ is a normal space, construct a continuous function $g_{1} : S^{2} \to I$ that takes the value 0 on $F_{1}$ and 1 on $F_{2}$ . Analogously, construct a continuous function $g_{2} : S^{2} \to I$ that takes the value 0 on $- F_{1}$ and 1 on $- F_{2}$ . Then the function $f = (g_{1}, g_{2})$ is a continuous function from $S^{2}$ to $R^{2}$ .

By the Borsuk-Ulam theorem, there exists $x \in S^{2}$ such that $f (x) = f (a (x))$ . By assumption $x$ cannot be in $F_{1}$ because that would mean $g_{1} (a (x)) = g_{1} (x) = 0$ forcing $a (x) \in F_{1}$ . Similarly $x$ cannot be in $F_{2}$ . This forces $x$ to be in $F_{3}$ . Similarly, $- x$ must also be in $F_{3}$ , and we are done.