[proofplan] By induction on $r$, it suffices to prove the one-step bound $\mu(K') \leq \frac{n}{n+1}\, \mu(K)$, where $n = \dim K$. An edge of $K'$ connects the barycenters $\hat{\sigma}$ and $\hat{\tau}$ of two simplices of $K$ that form a strict face relation $\sigma \subsetneq \tau$. We bound $|\hat{\sigma} - \hat{\tau}|$ by computing $\hat{\tau} - \hat{\sigma}$ in terms of the vertices of $\tau$, recognising that the difference has the form $\frac{d}{d+1}(c - \hat{\sigma})$ where $d = \dim \tau$ and $c$ is the barycenter of the complementary vertex set; then estimating $|c - \hat{\sigma}| \leq \operatorname{diam}(\tau) \leq \mu(K)$, the factor $\frac{d}{d+1} \leq \frac{n}{n+1}$ finishes the bound. Iterating gives the $r$-step bound, and $\left(\frac{n}{n+1}\right)^r \to 0$ completes the limit statement. [/proofplan] [step:Reduce to the one-step bound by induction on $r$] We show: if $\mu(K') \leq \frac{n}{n+1}\, \mu(K)$ whenever $\dim K = n$, then $\mu(K^{(r)}) \leq \left(\frac{n}{n+1}\right)^r \mu(K)$ for all $r \geq 0$. The dimension of a subdivision equals the dimension of the original complex: an $r$-fold subdivision of an $n$-dimensional complex remains $n$-dimensional, since the simplices of $K'$ are indexed by chains of faces in $K$ of length at most $n + 1$ (the longest chain being $\text{vertex} \subsetneq \text{edge} \subsetneq \cdots \subsetneq \text{full } n\text{-simplex}$), giving simplices of dimension $\leq n$ in $K'$; and the top-dimensional simplices of $K$ give chains of length $n + 1$, realising dimension $n$. Hence $\dim K^{(r)} = n$ for all $r \geq 0$. By induction on $r$: the base case $r = 0$ is $\mu(K) \leq \mu(K)$. For the inductive step, assuming $\mu(K^{(r)}) \leq \left(\frac{n}{n+1}\right)^r \mu(K)$, apply the one-step bound to the $n$-dimensional complex $K^{(r)}$: \begin{align*} \mu(K^{(r+1)}) = \mu((K^{(r)})') \leq \frac{n}{n+1}\, \mu(K^{(r)}) \leq \frac{n}{n+1} \cdot \left(\frac{n}{n+1}\right)^r \mu(K) = \left(\frac{n}{n+1}\right)^{r+1} \mu(K). \end{align*} It therefore suffices to prove the one-step bound. [/step] [step:Identify the edges of $K'$ as barycenters of comparable simplices] An edge of $K'$ is a $1$-simplex $\langle \hat{\sigma}, \hat{\tau} \rangle$ corresponding to a strict chain of faces $\sigma \subsetneq \tau$ in $K$ of length $2$. (Reversing the chain gives the same edge.) The mesh of $K'$ is \begin{align*} \mu(K') = \max\left\{|\hat{\sigma} - \hat{\tau}| : \sigma \subsetneq \tau,\ \sigma, \tau \in K \right\}. \end{align*} Since $K$ is finite (simplicial complexes in the convention of this course are finite), the maximum is attained. Fix a pair $\sigma \subsetneq \tau$ achieving this maximum; write $\tau = \langle a_0, \ldots, a_d \rangle$ with $d = \dim \tau \leq n$, and WLOG $\sigma = \langle a_0, \ldots, a_m \rangle$ for some $0 \leq m < d$ (after relabelling the vertices of $\tau$). The barycenters are \begin{align*} \hat{\sigma} = \frac{1}{m+1}\sum_{i=0}^m a_i, \qquad \hat{\tau} = \frac{1}{d+1}\sum_{i=0}^d a_i. \end{align*} [/step] [step:Express $\hat{\tau} - \hat{\sigma}$ as a scaled difference of two points of $\tau$] We compute \begin{align*} \hat{\tau} - \hat{\sigma} = \frac{1}{d+1}\sum_{i=0}^d a_i - \frac{1}{m+1}\sum_{i=0}^m a_i = \frac{1}{d+1}\sum_{i=m+1}^d a_i + \left(\frac{1}{d+1} - \frac{1}{m+1}\right)\sum_{i=0}^m a_i. \end{align*} Let $c = \frac{1}{d - m}\sum_{i=m+1}^d a_i$ be the barycenter of the complementary face $\langle a_{m+1}, \ldots, a_d \rangle$. Then $\sum_{i=m+1}^d a_i = (d - m) c$, and the identity becomes \begin{align*} \hat{\tau} - \hat{\sigma} = \frac{d - m}{d+1}\, c + \left(\frac{1}{d+1} - \frac{1}{m+1}\right)(m+1)\, \hat{\sigma} = \frac{d - m}{d+1}\, c + \left(\frac{m+1}{d+1} - 1\right) \hat{\sigma} = \frac{d - m}{d+1}\left(c - \hat{\sigma}\right). \end{align*} Taking norms, \begin{align*} |\hat{\tau} - \hat{\sigma}| = \frac{d - m}{d + 1}\, |c - \hat{\sigma}|. \end{align*} [guided] The goal of this step is to rewrite $\hat{\tau} - \hat{\sigma}$ as a scalar multiple of a difference of points $c - \hat{\sigma}$ both lying inside $\tau$, so that the diameter of $\tau$ provides an a priori bound on $|c - \hat{\sigma}|$. Split the sum defining $\hat{\tau}$ into the terms coming from $\sigma$ (indices $0, \ldots, m$) and the remaining vertices (indices $m+1, \ldots, d$): \begin{align*} \hat{\tau} = \frac{1}{d+1}\left(\sum_{i=0}^m a_i + \sum_{i=m+1}^d a_i\right) = \frac{m+1}{d+1}\, \hat{\sigma} + \frac{d - m}{d+1}\, c, \end{align*} where $c := \frac{1}{d-m}\sum_{i=m+1}^d a_i$ is the barycenter of the "complementary" face — the face spanned by the vertices of $\tau$ that are *not* in $\sigma$. This expression shows $\hat{\tau}$ is a convex combination of $\hat{\sigma}$ and $c$: the weight of $\hat{\sigma}$ is $\frac{m+1}{d+1}$, the weight of $c$ is $\frac{d-m}{d+1}$, and the two weights sum to $1$. Subtracting $\hat{\sigma}$ from both sides: \begin{align*} \hat{\tau} - \hat{\sigma} = \left(\frac{m+1}{d+1} - 1\right)\hat{\sigma} + \frac{d-m}{d+1}\, c = -\frac{d-m}{d+1}\hat{\sigma} + \frac{d-m}{d+1}\, c = \frac{d - m}{d+1}(c - \hat{\sigma}). \end{align*} So $|\hat{\tau} - \hat{\sigma}| = \frac{d-m}{d+1}|c - \hat{\sigma}|$. The factor $\frac{d-m}{d+1}$ is the *contraction ratio* for this particular pair; since $1 \leq d - m \leq d$, this factor is at most $\frac{d}{d+1}$. Why does this matter? Because $\hat{\sigma}$ and $c$ are both barycenters of faces of $\tau$, hence both points of the convex set $\tau$. So $|c - \hat{\sigma}| \leq \operatorname{diam}(\tau)$, and the diameter of $\tau$ is bounded by the longest edge of $\tau$, which is at most $\mu(K)$. [/guided] [/step] [step:Bound $|c - \hat{\sigma}|$ by $\mu(K)$ using convexity and the diameter of $\tau$] Since $\hat{\sigma} = \frac{1}{m+1}\sum_{i=0}^m a_i$ is a convex combination of the vertices $a_0, \ldots, a_m$ of $\tau$, and $c = \frac{1}{d-m}\sum_{i=m+1}^d a_i$ is a convex combination of the vertices $a_{m+1}, \ldots, a_d$ of $\tau$, both $\hat{\sigma}$ and $c$ lie in the convex hull of the vertex set $\{a_0, \ldots, a_d\}$, which is $\tau$. Hence \begin{align*} |c - \hat{\sigma}| \leq \operatorname{diam}(\tau). \end{align*} We claim $\operatorname{diam}(\tau) = \max_{0 \leq i < j \leq d} |a_i - a_j|$. Indeed, for any two points $p = \sum u_i a_i$, $q = \sum v_i a_i$ of $\tau$ (each expressed in barycentric coordinates), the triangle inequality gives \begin{align*} |p - q| = \left|\sum_i u_i a_i - \sum_j v_j a_j\right| = \left|\sum_i u_i \sum_j v_j (a_i - a_j)\right| \leq \sum_{i,j} u_i v_j |a_i - a_j| \leq \max_{i,j}|a_i - a_j|, \end{align*} where the middle step uses $\sum v_j = 1$ (so $a_i = \sum_j v_j a_i$) and $\sum u_i v_j = 1$ (so the convex-combination bound applies). The maximum is attained on a pair of vertices $a_i, a_j$, and each such pair is an edge of $\tau$ (in the combinatorial sense, $\langle a_i, a_j \rangle$ is a face of $\tau$, hence an edge of $K$). Therefore $\operatorname{diam}(\tau) \leq \mu(K)$, giving \begin{align*} |c - \hat{\sigma}| \leq \mu(K). \end{align*} [guided] We want to bound $|c - \hat{\sigma}|$ by $\mu(K)$. Since both $c$ and $\hat{\sigma}$ are barycenters of faces of $\tau$, they are both convex combinations of vertices of $\tau$, hence both lie in the convex hull of $\{a_0, \ldots, a_d\}$, which equals the simplex $\tau$ itself. So $|c - \hat{\sigma}| \leq \operatorname{diam}(\tau)$. Now we must bound $\operatorname{diam}(\tau)$ by the longest edge of $\tau$. This is the standard fact that the diameter of a simplex is attained on vertices. The reason: for any $p = \sum u_i a_i$ and $q = \sum v_j a_j$ in $\tau$ (with $u_i, v_j \geq 0$, $\sum u_i = \sum v_j = 1$), \begin{align*} p - q = \sum_i u_i a_i - \sum_j v_j a_j = \sum_i u_i \left(\sum_j v_j\right) a_i - \sum_j v_j \left(\sum_i u_i\right) a_j = \sum_{i,j} u_i v_j (a_i - a_j), \end{align*} using $\sum_j v_j = \sum_i u_i = 1$ to insert the artificial factors. Taking norms and using the triangle inequality, since each $u_i v_j \geq 0$ and $\sum_{i,j} u_i v_j = 1$: \begin{align*} |p - q| \leq \sum_{i,j} u_i v_j |a_i - a_j| \leq \max_{i,j}|a_i - a_j|. \end{align*} Thus $\operatorname{diam}(\tau) \leq \max_{i,j}|a_i - a_j|$, with equality attained when $p$ and $q$ are the two vertices achieving the maximum. Each pair $\{a_i, a_j\}$ spans a $1$-simplex of $\tau$, which is an edge of $\tau$ and hence an edge of $K$ (since $\tau \in K$ and $K$ is closed under faces). Every edge of $\tau$ therefore appears in the maximum defining $\mu(K)$, giving $\operatorname{diam}(\tau) \leq \mu(K)$. [/guided] [/step] [step:Combine the estimates and conclude the one-step bound] From Steps 3 and 4, \begin{align*} |\hat{\tau} - \hat{\sigma}| = \frac{d - m}{d + 1}\, |c - \hat{\sigma}| \leq \frac{d - m}{d + 1}\, \mu(K). \end{align*} The factor $\frac{d - m}{d + 1}$ is bounded by $\frac{d}{d+1}$ (since $m \geq 0$), which in turn satisfies $\frac{d}{d+1} \leq \frac{n}{n+1}$ because $d \leq n$ and $t \mapsto \frac{t}{t+1}$ is increasing on $[0, \infty)$. Hence \begin{align*} |\hat{\tau} - \hat{\sigma}| \leq \frac{n}{n+1}\, \mu(K). \end{align*} Since $\langle \hat{\sigma}, \hat{\tau} \rangle$ was an arbitrary edge of $K'$, and the maximum in the mesh is attained, \begin{align*} \mu(K') \leq \frac{n}{n + 1}\, \mu(K). \end{align*} Combined with Step 1, this proves \begin{align*} \mu(K^{(r)}) \leq \left(\frac{n}{n+1}\right)^r \mu(K) \end{align*} for all $r \geq 0$. Since $\frac{n}{n+1} < 1$ (for $n \geq 1$; the case $n = 0$ is trivial as $K$ has no edges and $\mu(K) = 0$), the sequence $\left(\frac{n}{n+1}\right)^r$ tends to $0$ as $r \to \infty$, so $\mu(K^{(r)}) \to 0$. This completes the proof. [/step]