Cut points on Brownian paths (Q582711)

The author proved that for a two-dimensional Brownian motion X there exists almost surely a point \(t\in (0,1)\) such that \(X([0,t))\cap X((t,1])=\emptyset\). This point t is called ``cut point''. As a corollary similar results are obtained for a three-dimensional Brownian motion. This means that a standard n-dimensional Brownian motion a.s. has a cut point for \(n\geq 2\) (for \(n\geq 4\) any point t \(\in (0,1)\) is a cut point ). It follows from above that X([0,1]) is not homeomorphic to the Sierpinski carpet a.s. Open problems concerning cut points are posed, too.