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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.

The clock $A_t={\int }_0^t{e}^{2{\beta }_s}ds$ is continuous, strictly increasing and unbounded. Define ${\tau }_r=\inf \{t:A_t\ge r\}$. Part (a), with the initial vector retained, proves thatEvery finite $r$ has finite ${\tau }_r$, and $|{\mathcal{B}}_r|=e^{{\beta }_{{\tau }_r}}>0$. On the common event where the clock is a continuous bijection of $[0,\infty )$, this holds simultaneously for all times. Thus the origin is a polar point for planar Brownian motion started at $(1,0)$:To prove unbounded-time visits near zero, let $t_n=\inf \{t\ge n:{\beta }_t=-n\}$. Brownian recurrence makes each $t_n$ finite almost surely, and $t_n\ge n\to \infty$. Set $r_n=A_{t_n}$. Since $A_t\to \infty$, $r_n\to \infty$, whileConsequentlyFor every fixed neighbourhood of zero there are visits at arbitrarily large times. Since $\mathcal{B}$ has the law of planar Brownian motion with this initial value, both conclusions hold for that Brownian law. This complex exponential construction of planar Brownian motion proves the two properties together without confusing recurrence near a point with hitting the point.

First fix a deterministic point $y$ different from the initial point. By translation, work with $y=0$ and $B_0=x\ne 0$. For $R>|x|$, hitting $0$ before $T_R$ requires hitting every circle of radius $r\in (0,|x|)$ before $T_R$. Part (a) givesAny finite-time hit of $0$ lies before exit from some sufficiently large integer-radius circle, since a continuous path on a finite interval is bounded. A countable union over such radii therefore shows that this fixed point is never hit. This is the polar point for planar Brownian motion property; it is a statement about each fixed point, not about avoiding all points simultaneously.

For the initial point $x$, fix a deterministic $\epsilon >0$. The Gaussian distribution of $B_{\epsilon }$ has a density, so $B_{\epsilon }\ne x$ almost surely. Conditional on ${\mathcal{F}}_{\epsilon }$, the future is Brownian motion starting from $B_{\epsilon }$, and the preceding fixed-point result shows that it never hits $x$. Taking $\epsilon =1/m$ and a countable union proves there is no return to the starting point at any positive time.

In contrast, neighborhoods are revisited arbitrarily late. Choose $r>0$ so that the closed disk centered at $x$ of radius $r$ lies in the given open set $U$. From any point $z$ outside this disk, let $R\to \infty$ in part (a), centered at $x$. It givesAt each deterministic integer time $n$, condition on $B_n$. If it is in the disk there is already a visit to $U$ at time $n$; if it is outside, the Markov property and the last calculation ensure a later visit to that disk, hence to $U$. With probability one this holds for all $n$ simultaneously. Thus the set of visits to $U$ is unbounded. The recurrence of planar Brownian motion concerns neighborhoods even though every fixed point is polar.

Let $L$ be the fixed line and choose $y\in L$. Take orthonormal vectors $e_1,e_2$ spanning the plane perpendicular to $L$. The processis standard planar Brownian motion starting from $Y_0\ne 0$. Indeed the projected increments are centered Gaussian random vectors with covariance $(t-s)I_2$, are independent on disjoint intervals, and have continuous paths.

The event $B_t\in L$ is exactly $Y_t=0$. The polar point for planar Brownian motion result in (b) therefore givesThis is a projection criterion for Brownian avoidance of affine subspaces: avoiding a fixed point in a two-dimensional projection excludes hitting the original line.

For $k\ge 2$, substitute $r_k=k^{-k}$ and $R=k$ into the exit formula:Let ${\tau }_0$ be the first hit of the origin. On any path with ${\tau }_0<\infty$, continuity makes the radius bounded on $[0,{\tau }_0]$. For all sufficiently large $k$, that path reaches radius $r_k$ before ${\tau }_0$ and has not yet reached radius $k$, so it belongs to $\{S_{r_k}\le T_k\}$. Thus $\{{\tau }_0<\infty \}$ is contained in the liminf of these events. Fatou's inequality for their indicators gives probability at most ${\mathrm{liminf}}_{k}1/(k+1)=0$ for that liminf. Hence the polar point for planar Brownian motion conclusion isThe argument controls one event for all times, rather than merely proving zero hitting probability at each fixed time.

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.

No such conformal bijection exists. A prescribed point is a polar point for planar Brownian motion: a planar Brownian motion started away from zero hits zero with probability zero. For example, inside an annulus the probability of reaching radius $\epsilon$ before radius $R$ isby the harmonic function $\log |w|$ and the optional stopping theorem. For fixed $R>|z^{\prime }|$ this tends to zero as $\epsilon \downarrow 0$. Any finite-time visit to zero would occur before leaving some sufficiently large disc; the countable union over integer $R$ still has probability zero.

The exit time from $\mathbb{C}\setminus \{0\}$ is therefore infinite almost surely, contradicting part (b) if this were the image of a bounded domain. Equivalently, a hypothetical inverse $\psi :\mathbb{C}\setminus \{0\}\to D$ would be bounded and holomorphic. The Riemann removable singularity theorem extends it over zero, and the Liouville theorem makes it constant, contradicting bijectivity.

The map $z\mapsto z^{\pi /\alpha }$ takes a wedge of opening $\alpha$ to a half-plane and the circular boundary of radius $r$ to radius $r^{\pi /\alpha }$. Conformal invariance of planar Brownian motion preserves which boundary part is reached first, although it changes the clock. Use the branch defined by the wedge argument; the origin is a polar point for planar Brownian motion, so localization handles the vertex.

A Brownian motion starting outside a fixed affine subspace of codimension at least two almost surely never hits it. Choose a two-dimensional orthogonal projection annihilating the subspace's direction and with nonzero projected initial displacement. The projected process is planar Brownian motion starting away from zero, and a hit of the subspace would hit zero in that projection. The polar point for planar Brownian motion result excludes this. In three dimensions this shows avoidance of every fixed line when started outside it.