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For $q\ge 1$, finite random-cluster measures with all boundary graph vertices identified have a weak infinite-volume limit of a sequence. On a regular lattice it is invariant under translations and lattice reflections and has positive association of the random-cluster model. Its conditional single-edge open probability lies between $p/(p+q(1-p))$ and $p$. Connection events can be approximated by increasing finite-volume path events, so their positive-association inequalities pass to the limit of a sequence.

Write $\varphi ={\varphi }_{p,q}^1$ and $C_n=\varphi (0\leftrightarrow e_n)$. We use these standard facts about the infinite-volume wired random-cluster measure for $q\ge 1$: it is translation invariant and invariant under coordinate reflections/permutations; it has positive association of the random-cluster model; and every conditional edge-open probability is at leastThe last bound follows because an edge either leaves its component count unchanged on opening, giving probability $p$, or merges two components, giving $p/(p+q(1-p))$. These are standard infinite-volume properties of the limit obtained from the finite wired laws by weak convergence of probability measures. Positive association of random variables applies to connection events by increasing approximation with finite-path events.

Opening the $n$ edges of the straight segment gives $C_n\ge r^n$, so its logarithm is finite. The events $A=\{0\leftrightarrow e_m\}$ and $B=\{e_m\leftrightarrow e_{m+n}\}$ are increasing, and their intersection implies connection to $e_{m+n}$. ThusThe sequence $a_n=-\log C_n$ is a subadditive sequence, with $0\le a_n\le -n\log r$. Part (a) proves the random-cluster inverse correlation length exists and is finite:Since $a_n/n\ge \alpha$ for every $n$, exponentiating yields the requested bound

For $\omega \in \{0,1{\}}^E$, write $o(\omega )=|\eta (\omega )|$ and $c(\omega )=|E|-o(\omega )$. The random-cluster model isFor $0\le p\le 1$ and $q\ge 1$, its positive association of the random-cluster model states that increasing functions $f,g$ satisfy ${\varphi }_{p,q}(fg)\ge {\varphi }_{p,q}(f){\varphi }_{p,q}(g)$.

For the two parallel edges $e_1,e_2$, the four unnormalized weights for $00,10,01,11$ are respectivelyFor the increasing events $A=\{e_1\text{ open}\}$ and $B=\{e_2\text{ open}\}$, positive association is equivalent to $w_{00}{w}_{11}\ge w_{10}{w}_{01}$, which reduces to $q\ge 1$. It therefore fails whenever $p,q\in (0,1)$.

At $p=1/2$, all factors involving the number of open edges are equal, so the weight is proportional to $q^{k(\omega )}$. The smallest possible component count is one. Dividing numerator and denominator by $q$ and sending $q\downarrow 0$ leaves equal weight precisely on connected spanning subgraphs, proving the stated uniform connected-subgraph limit of the random-cluster model.

Finally let $p,q\downarrow 0$ with $q/p\to 0$, and let $N=|V|$. Fix a spanning tree $\tau$. The ratio of the weight of $\omega$ to that of $\tau$ iswhere $d(\omega )=o(\omega )+k(\omega )-N\ge 0$. Equality $d=0$ means that the open graph is a forest. The ratio tends to zero unless $d=0$ and $k=1$, which means precisely that $\omega$ is a spanning tree. All spanning trees have equal weight, so the limiting law is the uniform spanning-tree limit of the random-cluster model:where $\mathcal{T}$ is the set of spanning trees of $G$.