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Proposition 19.9.6E6B4502ECB3B

Proposition 19.9.6.E6B4502ECB3B For every such algebra,

\[ \limsup _{m\to \infty }m^{-n}\dim W_ m \leq \operatorname {vol}_{\mathbb R^ n}\Delta (W_\bullet ). \]

In particular, if $\Delta (W_\bullet )$ has empty interior, then $\dim W_ m=o(m^ n)$.

Proof. The empty case is immediate. Otherwise, let $S_ m=\{ \nu (s):0\neq s\in W_ m\} $. By lemma 19.9.2, $S_ m$ has $\dim W_ m$ elements, and $m^{-1}S_ m\subseteq \Delta (W_\bullet )$. Attach to each $a\in S_ m$ the half-open cube $m^{-1}(a+[0,1)^ n)$. These cubes are disjoint, their total volume is $m^{-n}\dim W_ m$, and their union is contained in

\[ \Delta (W_\bullet )+[0,m^{-1}]^ n. \]

These compact neighborhoods decrease to $\Delta (W_\bullet )$ as $m\to \infty $, so their volumes converge to its volume. This proves the inequality. A compact convex set has positive $n$-dimensional volume exactly when it has nonempty interior, which gives the last statement. $\square$

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