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In section 7.4.1 of Durrett's Probability:Theory and Examples, to show that the set of zeros of Brownian motion is uncountable, firstly it shows that there are no isolated points in the set below. $$ \mathcal{Z} = \{t:B_t(w)=0\}$$ It assumes that if $u\in \mathcal{Z}(w)$ is isolated on the left, then it is, with probability one, a decreasing limit of points in $\mathcal{Z}(w)$. How can we conclude that it is a decreasing limit of points?

I think the fact that $P_x (T_0 \circ \theta_{R_t}>0\text{ for some rational }t)=0$ is used to conclude that it is decreasing limit of points, but I can't figure it out. What is the logic here?

cf) $ T_0 = \inf\{u>0:B_u=0\}, R_t = \inf\{u>t :B_u=0\}$

Dongri
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The first step shows (using the strong Markov property) that each point in $\{T_q:q\in \mathbb{Q}_{\ge 0}\}$ is not isolated from the right w.p.1. Next, consider a point $z\in \mathcal{Z}(\omega)$ isolated from the left, i.e., $z=T_q(\omega)$ for some $q\in\mathbb{Q}_{\ge 0}$. By the previous argument, it is not isolated from the right (a.s.). (You can construct a decreasing sequence using the fact that for each $\epsilon>0$, $(T_q(\omega),T_q(\omega)+\epsilon)$ contains a point in $\mathcal{Z}(\omega)$.)

  • Regrading the last sentence, how does $(T_q(w), T_q(w)+\epsilon)$ contains a point in $\mathcal{Z} (w)$? – Dongri May 29 '21 at 07:11
  • @Dongri It follows from the first step, i.e. once $B$ hits $0$, it changes its sign infinitely many times near that point. –  May 29 '21 at 09:13
  • Ah ok I understood, thank you!!! – Dongri May 30 '21 at 02:28