Zero set of Brownian Motion is uncountable 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\}$

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 $\mathcal{Z}(\omega)$.)
