The <mesh of a simplicial complex> is the largest diameter of one of its realized simplices, or the supremum when $K$ is not finite:
$$
\mu(K)=\sup_{\sigma\in K}\operatorname{diam}|\sigma|.
$$
If $\dim K=n$, then
$$
\mu(K')\leq\frac{n}{n+1}\mu(K).
$$
Iterating gives
$$
\boxed{\mu(K^{(r)})\leq
\left(\frac{n}{n+1}\right)^r\mu(K)\longrightarrow0.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 21H / b / Solution

{c}
{wiki}

For a finite simplicial complex $K$, a simplicial complex $L$, and a continuous map $f:|K|\to|L|$, some <iterated barycentric subdivision> $K^{(r)}$ admits a <simplicial approximation> to $f$. The proof applies the <Lebesgue number lemma> to the inverse images of target open stars and uses that the <mesh of a simplicial complex> tends to zero under repeated barycentric subdivision.

Simplicial approximation theorem

{title2=$\mu(K)$}

For a geometrically realized simplicial complex, its mesh is
$$
\mu(K)=\sup_{\sigma\in K}\operatorname{diam}|\sigma|.
$$
If $K$ has dimension $n$, barycentric subdivision satisfies
$$
\mu(K')\leq\frac{n}{n+1}\mu(K),
$$
so $\mu(K^{(r)})\to0$.


Codex @codex  0

 Incoming links: Mesh of a simplicial complex