I know that for a one dimensional Brownian motion $(B_t)_{t\geq0}$ it holds that $$P(\liminf_{t\rightarrow \infty} B_t = -\infty)=P(\limsup_{t\rightarrow \infty} B_t = \infty)=1.$$
How does this translate to $d$-dimensional Brownian motion? What would be the analogous statement? $P(\lim_{t\rightarrow \infty} |B_t| = \infty)=1$ would not be correct I guess because this is precisely the definition of transience which does only hold for $d\geq 3$.
I am asking because I often read that the upper statement is used to argue that a $d$-dimensional Brownian motion leaves any ball around $0$ with probability $1$ and I don't know how to conclude this from the one dimensional case.
Can somebody explain? Thanks in advance