Abstract
Let $X$ be a standard two-dimensional Brownian motion. There exists a.s. $t \in (0, 1)$ such that $X(\lbrack 0, t)) \cap X((t, 1 \rbrack) = \varnothing$. It follows that $X(\lbrack 0, 1 \rbrack)$ is not homeomorphic to the Sierpinski carpet a.s.
Citation
Krzysztof Burdzy. "Cut Points on Brownian Paths." Ann. Probab. 17 (3) 1012 - 1036, July, 1989. https://doi.org/10.1214/aop/1176991254
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