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Source: published paper, printed p. 220, Lemma 2.1.

Statement

For $L>2$, a $1/3$-separated set $P\subset\mathbb T_L^n$ satisfies

$$|P|\le(4\sqrt n\,L)^n.$$

The source denotes the period by $R$; the renamed parameter permits its use with $L=3R$ in the reconstructed main proof.

Full proof

The open radius-$1/6$ balls about the points of $P$ are disjoint. Since $1/6<L/2$, each is isometric to an ordinary Euclidean ball. A radius-$r$ ball has volume

$$v_n(r)=\frac{r^n\pi^{n/2}}{\Gamma(n/2+1)}.$$

Here $\Gamma(n/2+1)\le n^{n/2}$. For even $n=2m$, this follows from $m!\le m^m\le n^{n/2}$. For odd $n=2m+1$, the half-integer formula gives

$$\Gamma(n/2+1)=\frac{\sqrt\pi}{2} \prod_{j=1}^{m}\left(j+\frac12\right) \le\frac{\sqrt\pi}{2}(n/2)^m\le n^{m+1/2}.$$

The last inequality also holds for $m=0$, since $\pi<4$. Thus $v_n(1/6)\ge(\pi/(36n))^{n/2}$. Summing the disjoint volumes inside a torus of volume $L^n$ gives

$$|P|\le (36nL^2/\pi)^{n/2}<(4\sqrt n\,L)^n.$$

The same bound applies to every finite subset if finiteness was not initially assumed, and therefore rules out an infinite separated $P$.

Related proof pages. definitions.

Bears on. Problem 188.