 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 simultaneous avoidance of the entire positive real axis: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 the visited positive boundary set is unbounded:This uses countably many SLE boundary swallowing criteria; no intersection over an uncountable family of probability-one events is needed.