= Solution
For $\kappa<4$, each event $\{\gamma\text{ hits }[1/n,\infty)\}$ has probability zero by part (c). Every positive real point lies in one of these intervals. The countable union therefore has probability zero, proving \b[simultaneous avoidance of the entire positive real axis]:
$$
\boxed{\gamma([0,\infty))\cap(0,\infty)=\varnothing\quad\text{almost surely}.}
$$
For $\kappa>4$, every event $\{\gamma\text{ hits }[n,\infty)\}$ has probability one. Their countable intersection still has probability one. On that event the set of positive real points visited contains a point at least $n$ for every integer $n$, so \b[the visited positive boundary set is unbounded]:
$$
\boxed{\sup\bigl(\gamma([0,\infty))\cap(0,\infty)\bigr)=\infty
\quad\text{almost surely}.}
$$
This uses countably many <SLE boundary swallowing criteria>; no intersection over an uncountable family of probability-one events is needed.
Back to article page
