I would like to know whether a random walk defined by a Brownian motion is topologically irreducible. I mean the following. Let $(X_t)_{t\geq 0}$ be a stochastic process that is just an standard Brownian motion in $\mathbb{R}^2$. This stochastic process is Markovian, i.e. it is also a Markov process defined via the Markov kernel $$ P_t(x,A) = \mathbb{P}[X_t\in A\mid X_0=x] $$ for all $t\geq 0$, $x\in\mathbb{R}$ and $A\subset\mathbb{R}^2$ Borel. We say that a Markov process is topologically irreducible if for every $x\in\mathbb{R}^2$ and every $A\subset\mathbb{R}^2$ open there exists some $t\geq 0$ such that $P_t(x,A)>0$, i.e. every region with positive measure is accesible from every initial point.
I have the intuition that a if $(X_t)_{t\geq 0}$ is an standard Brownian motion then its corresponding Markov semigroup will be irreducible in this sense, however I didn't manage to prove it and couldn't find the result anywhere. Does anybody know whether this is true and how to prove it? Any helpful reference would be very much appreciated as well :)
For each time $t>0$, the probability density function is strictly positive everywhere in $R^2$, so it gives strictly positive to every open ball of $R^2$.
https://en.wikipedia.org/wiki/Brownian_motion