Theorem 13.27.4126E2D85CA62

Proof. Step 1: a decreasing family of symmetric relations. Set $C=1+M$ and $m(p)=L(p)/C$, so that $0<m(p)<1$. Put $a_1=1$, $a_{k+1}=na_ k+1$, and choose $0<\varepsilon _1<1$. Recursively choose $\varepsilon _{k+1}>0$ so small that, with $r_ k=4^{a_{k+1}}\varepsilon _{k+1}$,

Let $\mathcal U_0=\mathcal M_+(\theta )^2$ and, for $k\geq 1$, set

These symmetric sets contain the diagonal and decrease with $k$, and their intersection is the diagonal by proposition 13.22.4. We claim that

where composition of relations means joining successive pairs. For $k=0$ this is immediate. For $k\geq 1$, suppose $(p_{i-1},p_ i)\in \mathcal U_{k+1}$ for $i=1,2,3$, and put $m_*=\min _{0\leq i\leq 3}m(p_ i)$. The reciprocal differences along this chain have total less than $3\varepsilon _{k+1}<1$, so $1/m(p_ i)>1/m_*-1$. Hence $m(p_ i)\leq 4m_*$ for every $i$: if $m_*\leq 3/4$, this follows from $1/m_*-1\geq 1/(4m_*)$, and otherwise from $m(p_ i)<1$. Each successive comparison is therefore less than

All four models have lower volume at least $Cm_*$. Applying (13.22.6) with $\delta =Cm_*$ twice, first with middle point $p_2$ and then with middle point $p_1$, yields

Here $(a_{k+1}-1)/n=a_ k$, $a_{k+1}\geq a_ k$ and $m_*<1$. The endpoint reciprocal difference is less than $3\varepsilon _{k+1}<\varepsilon _ k$, proving (13.27.6).

Step 2: the chain metric. Define

where the infimum runs over finite chains $p=p_0,p_1,\ldots ,p_ N=q$ and integers $k_ i\geq 0$ with $(p_{i-1},p_ i)\in \mathcal U_{k_ i}$; we call $\sum _ i2^{-k_ i}$ the cost of the chain. A single edge in $\mathcal U_0$ gives $\mathfrak d_\theta \leq 1$. Reversing and concatenating chains gives symmetry and the triangle inequality, while diagonal edges with arbitrarily large indices give $\mathfrak d_\theta (p,p)=0$.

We next prove the inclusions

The second inclusion follows by using a single edge. For the first, we prove by induction on the number of edges that every chain of cost less than $2^{-k}$ has its endpoints in $\mathcal U_{k-1}$, simultaneously for all $k\geq 1$. For one edge this follows because the $\mathcal U_ k$ decrease. For a chain of cost $A<2^{-k}$, choose $l\geq k$ with

Split the chain at the first edge at which the cumulative cost reaches $2^{-(l+1)}$. The subchains before and after this edge both have cost less than $2^{-(l+1)}$, so by induction their endpoint pairs lie in $\mathcal U_ l$; an empty subchain is interpreted as a diagonal pair. Every edge cost is a power of two smaller than $2^{-l}$, so the middle edge lies in $\mathcal U_{l+1}\subset \mathcal U_ l$. By (13.27.6), the endpoints of the whole chain form a pair in $\mathcal U_{l-1}\subset \mathcal U_{k-1}$. This completes the induction and proves (13.27.8). Since the $\mathcal U_ k$ intersect in the diagonal, $\mathfrak d_\theta $ separates points, so it is a metric.

Step 3: convergence and completeness. By (13.27.8), $\mathfrak d_\theta (p_ j,p)\to 0$ if and only if $(p_ j,p)$ eventually lies in every $\mathcal U_ k$. This implies $d_\theta (p_ j,p)\to 0$. Conversely, if $d_\theta (p_ j,p)\to 0$, then proposition 13.27.2 gives $L(p_ j)\to L(p)>0$, so for each fixed $k$ both inequalities defining $\mathcal U_ k$ hold eventually. This proves (13.27.5).

The same inclusions show that a $\mathfrak d_\theta $-Cauchy sequence is $d_\theta $-Cauchy and has Cauchy reciprocal lower volumes. These reciprocals are bounded, so a tail of the sequence lies in some $\mathcal M_\delta (\theta )$. By theorem 13.22.7, the sequence converges there for $d_\theta $, and hence for $\mathfrak d_\theta $. This proves completeness. Conversely, a $d_\theta $-Cauchy sequence with Cauchy reciprocal lower volumes has a common positive lower-volume bound on a tail, and then both inequalities defining each $\mathcal U_ k$ hold for all pairs of sufficiently late terms. This proves the Cauchy criterion.

On $\mathcal M_\delta (\theta )$, (13.27.3) controls the reciprocal differences uniformly in terms of $d_\theta $, while $m(p)\geq \delta /C$ controls the second threshold in $\mathcal U_ k$. Thus $\mathfrak d_\theta $ and $d_\theta $ define the same uniform structure there. Continuity of the lower volume makes $\mathcal M_\delta (\theta )$ closed for $\mathfrak d_\theta $, so the restriction of $\mathfrak d_\theta $ to it is complete. $\square$