Let $(B_t)_{t\in[0,1]}$ be a standard Brownian motion and let $Z=\{t\in [0,1]\colon B_t=0\}$ denote its zero set. $Z$ is a topological space when given the induced topology from $[0,1]$. Let $C=\{0,1\}^{\mathbb N}$ denote the space of infinite binary sequences, equipped with the product topology. Does there exist a homeomorphism from $Z$ to $C$ with probability $1$?

By considering ternary expansions of real numbers, it is easy to show that $C$ is homeomorphic to the standard ternary Cantor set. Also, $Z$ can be constructed in a manner roughly similar to the ternary Cantor set, by successively removing open intervals from $[0,1]$ with an increasing level of precision. On the other hand, $Z$ has Hausdorff dimension $1/2$ while the ternary Cantor set has Hausdorff dimension $\log_3 2$.


1 Answer 1


Yes. With probability $1$, $Z$ is closed, has no isolated points, and contains no interval. It follows that $Z$ is a totally disconnected compact metric space with no isolated points, and any such space is homeomorphic to $C$.

  • $\begingroup$ Can you elaborate on why "any such space is homeomorphic to $C$"? $\endgroup$
    – pre-kidney
    Mar 30, 2017 at 7:03
  • 1
    $\begingroup$ This is a standard theorem; the idea is to inductively construct a basis that "looks like" the standard basis of $C$. See math.stackexchange.com/questions/2073141/…, for instance. $\endgroup$ Mar 30, 2017 at 14:56

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