# A proof that Sperner's Lemma implies KKM?

Wikepedia KKM says that Brouwer fixed-point theorem, Sperner's lemma, and Knaster–Kuratowski–Mazurkiewicz lemma are equivalent.

In "A course of topological combinatorics", I find a proof that Sperner implies Brouwer. I am wondering is there a proof that Sperner implies KKM?

Consider the $$k$$th barycentric division, which is a triangulation, $$T^k$$ of $$\Delta_n:=\{(x_1,...,x_{n+1}): x_i\ge 0, \sum x_i=1\}$$.
We color each vertex $$v$$ of $$T_k$$ by minimum $$i$$ such that $$v_i\neq 0$$ and $$v\in A_i$$, where $$A_1,...,A_n$$ are the closed sets in KKM lemma. Then this coloring is a legal Sperner coloring and we know that there exists a simplex $$\sigma_k$$ in $$T^k$$ whose vertices are rainbow (in $$n$$ colors).
As the volume of $$\sigma_k$$ tends to 0 as $$k$$ goes to infinity, we know that there exists $$z\in \Delta_n$$ that is a common accumulation point of all the vertices in an infinite subsequence of $$\sigma_k$$ (say $$\sigma_{k_\ell}$$). As $$A_i$$ are closed and the vertices of $$\sigma_k$$ are rainbow, we know that $$z\in\bigcap_i A_i$$. (For the vertices $$v(k_\ell,i)$$, which are the vertices in $$\sigma_{k_\ell}$$ and are colored by $$i$$, by definition we have $$v(k_\ell,i)\in A_i$$, and we have $$v=\lim_\ell v(k_\ell,i)$$ is also in $$A_i$$ as $$A_i$$ are closed.)
• "As the volume of σk tends to 0 as k goes to infinity, we know that there exists z∈Δn that is a common accumulation point of all the vertices in an infinite subsequence of σk" I do not understand why. In theory, if you have a black box that returns a rainbow-simplex for each $k$, it can return a rainbow simplex in a different region each time, so that the sequence does not converge. How do you know that there is a converging sub-sequence? Aug 18, 2022 at 17:13
• That is Bolzano–Weierstrass theorem for $\mathbb{R}^n$ as $\Delta_n$ is a compact set so that each infinite sequence in it has a convergent subsequence. Aug 21, 2022 at 9:30