The second answer here shows that every $(0,\epsilon)$ has a $t$ with $B_t<0$ using law of iterated logarithm, where $B_t$ is a Brownian motion.
$$\liminf_{t\rightarrow 0}\frac{B_t}{\sqrt{2t\log\log(1/t)}}=-1$$
Similarly, $\limsup_{t\rightarrow 0}\frac{B_t}{\sqrt{2t\log\log(1/t)}}=1$ and so I guess I can say that every $(0,\epsilon)$ also has a $t$ with $B_t=0$. Here we've used time-inversion and I was wondering if we can use time-inversion and recurrence of a Brownian motion to prove the same? Thanks.
Similarly, with the Arcsin Law: the probability for the BM not to cancel between $a$ and $b$ is $$ \frac{2}{\pi}\arcsin\sqrt{\frac{a}{b}}. $$
Otherwise, $tB_{\frac{1}{t}}$ is also a BM.