= Solution
It is enough by <Scaling invariance of SLE> and reflection in the imaginary axis to treat $x=1$. Take $r=1+\epsilon$. The drift
$$
b(z)=\frac{\kappa-4}{2}-\frac2{1+e^z}
$$
tends to $(\kappa-8)/2<0$ as $z\to-\infty$. Choose $L<0$ and $a<0$ such that $b(z)\leq a$ for $z\leq L$. Before $\widetilde Z$ reaches $L$,
$$
\widetilde Z_t
\leq\log\epsilon+\sqrt\kappa W_t+at.
$$
The <infinite-horizon crossing probability for Brownian motion with negative drift> shows that this process has a finite running maximum almost surely, so
$$
\mathbb P_{\log\epsilon}(T_L<\infty)\longrightarrow0
\qquad(\epsilon\downarrow0).
$$
Order preservation for the <Loewner chain>[Loewner flow] gives $\tau_1\leq\tau_{1+\epsilon}$. On $\{\tau_1<\tau_{1+\epsilon}\}$, one has $V_t^1\to0$ while $V_t^{1+\epsilon}$ remains positive, so $\widetilde Z_t\to+\infty$ and in particular $T_L<\infty$. Hence
$$
\mathbb P(\tau_1<\tau_{1+\epsilon})\longrightarrow0,
\qquad
\mathbb P(\tau_1=\tau_{1+\epsilon})\longrightarrow1.
$$
If the trace itself hit the fixed boundary point $1$, then $1$ would be the right endpoint of the swallowed interval and $\tau_1<\tau_{1+\epsilon}$ for every $\epsilon>0$. Its probability is therefore zero. Scaling and reflection prove that <SLE does not hit a fixed nonzero boundary point> for every $x\in\mathbb R\setminus\{0\}$.
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