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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 $\log_3 2$.

pre-kidney
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1 Answers1

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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 $C$.

Eric Wofsey
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  • Can you elaborate on why "any such space is homeomorphic to $C$"? – pre-kidney Mar 30 '17 at 07:03
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    This is a standard theorem; the idea is to inductively construct a basis that "looks like" the standard basis of $C$. See http://math.stackexchange.com/questions/2073141/compact-hausdorff-spaces-w-o-isolated-points-and-having-countable-bases-consisti, for instance. – Eric Wofsey Mar 30 '17 at 14:56