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 $\mathbb {R} ^{2}$.

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.