Inmathematics, theLusternik–Schnirelmann theorem, akaLusternik–Schnirelmann–Borsuk theorem orLSB theorem, says as follows.

If thesphereSⁿ is covered byn + 1 closed sets, then one of these sets contains a pair (x, −x) of antipodal points.

This implies that theLusternik–Schnirelmann category of⁠${\displaystyle {\mathbb{R}\mathbb{P}}^n}$⁠ is at least⁠${\displaystyle n+2}$⁠ (and it is not difficult to show that⁠${\displaystyle n+2}$⁠ sets suffice, so the category is exactly⁠${\displaystyle n+2}$⁠).

It is named afterLazar Lyusternik andLev Schnirelmann, who published it in 1930.^[1][2][3]

The theorem can be proved by leveraging the relationship between${\displaystyle S^n}$ and${\displaystyle {\mathbb{R}\mathbb{P}}^n}$, and the Borsuk-Ulam theorem in higher dimensions.^[4]

Real Projective Space${\displaystyle {\mathbb{R}\mathbb{P}}^n}$ is defined as the quotient space of${\displaystyle S^n}$ by the antipodal map, where each point${\displaystyle x\in S^n}$ is identified with its antipodal point${\displaystyle -x\in S^n}$. In other words,${\displaystyle {\mathbb{R}\mathbb{P}}^n=S^n/\sim }$, where${\displaystyle x\sim -x}$.

A pair of antipodal points in${\displaystyle S^n}$ corresponds to a single point in${\displaystyle {\mathbb{R}\mathbb{P}}^n}$. The concept of "containing a pair of antipodal points" in a subset of${\displaystyle S^n}$ is equivalent to that subset having a non-empty intersection with the pre-image of some point in${\displaystyle {\mathbb{R}\mathbb{P}}^n}$ under the projection map${\displaystyle \pi :S^n\to {\mathbb{R}\mathbb{P}}^n}$.

TheBorsuk-Ulam theorem states that for any continuous map${\displaystyle f:S^n\to {\mathbb{R}}^n}$, there exists at least one pair of antipodal points${\displaystyle \{x,-x\}}$ in${\displaystyle S^n}$ such that${\displaystyle f(x)=f(-x)}$.

Consider${\displaystyle S^n}$ expressed as the union of${\displaystyle n+1}$ closed sets${\displaystyle A_1,A_2,\dots ,A_{n+1}}$.

Define distance functions${\displaystyle d_i(x)=\underset{y\in A_i}{inf}|x-y|}$ for${\displaystyle i=1,\dots ,n}$. These are continuous functions.

Construct a map${\displaystyle F:S^n\to {\mathbb{R}}^n}$ given by${\displaystyle F(x)=(d_1(x),d_2(x),\dots ,d_n(x))}$.

By the Borsuk-Ulam theorem, there exists a pair of antipodal points${\displaystyle \{x_0,-x_0\}}$ such that${\displaystyle F(x_0)=F(-x_0)}$, meaning${\displaystyle d_i(x_0)=d_i(-x_0)}$ for all${\displaystyle i=1,\dots ,n}$.

If any${\displaystyle d_i(x_0)=0}$, then${\displaystyle x_0\in A_i}$. Since${\displaystyle d_i(x_0)=d_i(-x_0)=0}$, then${\displaystyle -x_0\in A_i}$ as well (because${\displaystyle A_i}$ is closed). Thus,${\displaystyle A_i}$ contains a pair of antipodal points.

If${\displaystyle d_i(x_0)>0}$ for all${\displaystyle i=1,\dots ,n}$, then${\displaystyle x_0\notin A_i}$ for${\displaystyle i=1,\dots ,n}$. Since${\displaystyle S^n=\bigcup _{j=1}^{n+1}{A}_j}$, it must be that${\displaystyle x_0\in A_{n+1}}$. Similarly, since${\displaystyle d_i(-x_0)>0}$ for all${\displaystyle i=1,\dots ,n}$, then${\displaystyle -x_0\notin A_i}$ for${\displaystyle i=1,\dots ,n}$, implying${\displaystyle -x_0\in A_{n+1}}$. Therefore,${\displaystyle A_{n+1}}$ contains a pair of antipodal points.

There are several fixed-point theorems which come in three equivalent variants: analgebraic topology variant, a combinatorial variant and a set-covering variant. Each variant can be proved separately using totally different arguments, but each variant can also be reduced to the other variants in its row. Additionally, each result in the top row can be deduced from the one below it in the same column.^[5]

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