 Solution

WriteThe Russo-Seymour-Welsh theorem and the Harris-FKG inequality give the standard one-arm extension estimate: there is $c>0$ such thatuniformly in $r$. Indeed, on the one-arm event to scale $r$, a fixed finite collection of open rectangle crossings in the annulus ${\Lambda }_{2r}\setminus {\Lambda }_r$, each having probability bounded below by RSW, joins that arm to $\partial {\Lambda }_{2r}$; FKG multiplies the lower bounds. Iteration shows that $\pi (ar)$ and $\pi (r)$ are comparable for every fixed $a>0$. This proves the estimate suggested in the hint.

Fix $x\in \partial {\Lambda }_n$ and choose $r=\lfloor n/4\rfloor$. If $0⟷x$, there is an open arm from $0$ to distance $r$ and another from $x$ to distance $r$. These are independent events because they use disjoint edge sets, so the extension estimate gives

For the reverse inequality, take one-arm events from $0$ and $x$ at scale comparable with $n$, in disjoint boxes. A fixed collection of open crossings of rectangles of bounded aspect ratio joins the two arms. The Russo-Seymour-Welsh theorem bounds the probability of every added crossing below uniformly in $n$ and in the position of $x$ along the four sides; the Harris-FKG inequality and the arm-extension estimate therefore giveThis is the usual RSW gluing lemma for two one-arm events. Enlarging the constants handles the finitely many small $n$, proving the claim with positive constants $c_1,c_2$.

Solved by gpt-5.6-sol high.