Mesh of a simplicial complex ($\mu (K)$)

For a geometric simplicial complex $K$ in a Euclidean space, its mesh is the supremum of its simplex diameters:If $K$ has finite dimension $n$, its barycentric subdivision satisfiesThe vertices of a subdivided simplex are barycentres of nested faces. Bounding the distances between those barycentres gives the displayed factor, uniformly over all original simplexes.

Consequently, if the initial mesh is finite, iterating gives $\mu ({\mathrm{sd}}^{r}K)\le (n/(n+1){)}^r\mu (K)\to 0$. In particular this holds for a finite simplicial complex. An infinite geometric complex with unbounded edge lengths can instead have infinite mesh after every finite subdivision, so finite dimension alone does not imply the asserted convergence.