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Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Put $\tau =T_r\wedge T_R$. First establish almost sure finiteness. The Itô formula gives the martingale $|B_t{|}^2-2t$ for planar Brownian motion. Stop it at $t\wedge T_R$. The stopped path has norm at most $R$, soBy monotone convergence, $\mathbb{E}{T}_R<\infty$, and hence $\tau <\infty$ almost surely.

The function $h(y)=\log |y|$ is a harmonic function away from $0$: in polar coordinates its Laplacian is $h^{\prime \prime }(\rho )+h^{\prime }(\rho )/\rho =-{\rho }^{-2}+{\rho }^{-2}=0$. The Itô formula shows $h(B_{t\wedge \tau })$ is a local martingale. It is bounded between $\log r$ and $\log R$, so it is a true martingale and passage to $\tau$ is justified by dominated convergence. Writing $p={\mathbb{P}}_x(T_r<T_R)$ and using continuity at exit givesSolving yields the planar Brownian annulus hitting probabilityThe stopping argument stays away from the singularity of the logarithm until the calculation is complete.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

Translate the centre of the ball to zero. Choose an inner radius $\rho$ strictly smaller than the ball's radius. If the starting point already lies in the ball there is nothing to prove; otherwise write $d=|z|>\rho$ and choose $R>d$. Let $H_{\rho },H_R$ be the first hits of the two circles.

The Brownian exit time from the annulus $\rho <|z|<R$ is finite almost surely: it is no later than exit of the first coordinate from $(-R,R)$, which is finite by part (a). The function $\log |z|$ is harmonic on this annulus. The Itô formula and optional stopping theorem apply to its bounded stopped values, yieldingSolving gives the planar Brownian annulus hitting probabilityAs $R\to \infty$ this tends to one, while the event in the formula is contained in $\{H_{\rho }<\infty \}$. Thus the inner circle is hit almost surely. That circle lies strictly inside the original open ball, proving the recurrence of planar Brownian motion in the form requested:A probability-one intersection over balls with rational centres and positive rational radii makes this conclusion simultaneous for all nonempty open balls, since every such ball contains one from that countable family.

Planar Brownian motion is recurrent. More explicitly, the planar Brownian annulus hitting probability giveswhen $|x|>1$. Thus the unit disc is hit almost surely from every starting point. Applying the Strong Markov property after each departure and return shows that such returns occur after arbitrarily large times. Therefore

Let $\tau =T_r\wedge T_R$. The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r<|z|<R$, so Itô formula shows thatis a bounded martingale. By the optional sampling theorem for a supermartingale applied to this martingale,where $p={\mathbb{P}}_x(T_r<T_R)$. Solving gives the planar Brownian annulus hitting probability

The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r_1<|z|<r_2$. The Itô formula therefore makes $\log |B_{t\wedge \tau }|$ a bounded martingale. The optional sampling theorem for a supermartingale giveswhere $p=\mathbb{P}(|B_{\tau }|=r_1)$. Solving this linear equation yields the planar Brownian annulus hitting probability