As I was trying to exhibit new interesting(?) path transformations of Brownian motion, I became interested in
the (random) set of times $t$ such that $B(t)=B(t+1)=0$, where $B(t)$ denotes a standard Brownian motion.
Not surprisingly, I found that this set is almost surely empty. By scaling invariance of Brownian motion, the same result holds if $1$ is replaced by any positive real $s$.
Precisely, for all $s>0$, 
$$P[\exists t, B(t)=B(t+s)=0]=0.$$
Let me rephrase this a bit differently. Let $Z$ be the set of Brownian motion zeros and $D(Z)$ be the associated algebraic self-difference set
$$D(Z)=Z-Z=\{u-v : (u,v)\in Z^2\}.$$
In this setting, the preceding result says that for all $s\not=0$,
$$P[s\in D(Z)]=0.$$
Integrating this relation over $s$ with respect to Lebesgue measure $\lambda$ and exchanging the order of integration gives
$$E[\lambda(D(Z))]=0.$$
In other words, the set $D(Z)$ is almost-surely negligible.


I've never seen any property of this set in the literature, thus my question is:


*Is this a well-known result and, if so, what other properties are known, regarding its Hausdorff dimension for example ?*


**[EDIT]**
It seems that some of you doubt that the first probability is zero, so let me sketch a proof. (I apologize if there is a stupid mistake I've not been able to see). First, it is clear that,
$$P[\cup\_{n\geq 0}\{B\_n=B\_{n+1}=0\}]=0$$
since this is a countable union of zero probability events. Therefore,
$$P[\exists t\geq 0, B\_t=B\_{t+1}=0]\leq\sum\_{n=0}^{\infty}P[\exists t\in (n,n+1),B\_t=B\_{t+1}=0]=\sum\_{n=0}^\infty p\_n.$$
I asserts that each term $p\_n$ inside the sum is equal to zero. For simplicity, I will do it only for $n=0$. (For other values of $n$, it suffices to condition first on the value of $B\_n$ and then apply the same argument). Set $X\_t=B\_t$ and $Y\_t=B\_{t+1}-B\_{1}$. Then,
$$p\_0=P[\exists t\in(0,1),(X\_t,X\_1+Y\_t)=(0,0)],$$
and it suffices to show that
$$P[\exists t\in(0,1),(X\_t,x+Y\_t)=(0,0)\vert X\_1=x]=0$$
for (almost) all $x\in\mathbb{R}$. Now, $X=(X\_t,0\leq t\leq 1)$ and $Y=(Y\_t,0\leq t\leq 1)$ are two independent Brownian motions, hence conditionally to $X\_1=x$ the two-dimensional process $(X,x+Y)$ is a Brownian ``bridge'' (starting from $(0,x)$ and ending on the line $\{(x,r):r\in\mathbb{R}\}$). This process inherits from the standard two-dimensional Brownian motion the property that it does not hit $(0,0)$. Hence $p\_0=0$.



Properties of the algebraic self-difference set of Brownian motion zeros
Regarding Hausdorff dimension $\dim\_H$, the following holds:


$$
\dim\_H(Z-Z)=1 \quad\text{a.s.}
$$


This is essentially a consequence of Marstrand's projection theorem, the (stochastic) self-similarity of $Z$, and the well known fact that $\dim\_H Z=1/2$ (and the box counting dimension is also $1/2$).


Recall that Marstrand's projection theorem says that for a Borel set $E\subset\mathbb{R}^2$, the orthogonal projection onto *almost every* line has Hausdorff dimension $\min(\dim\_H E,1)$. In particular, taking $E=Z\_1\times Z\_2$, where $Z\_1, Z\_2$ are independent realizations of the zero set of Brownian bridge (see below for def.), then
$$
\dim\_H(Z\_1-\lambda Z\_2)=1 \quad\text{for almost all }\lambda.
$$
(For "nice" sets such as $Z\_1, Z\_2$, we have $\dim\_H(Z\_1\times Z\_2)=\dim\_H(Z\_1)+\dim\_H(Z\_2)$).


Now let us look at $Z-Z$. By scale invariance of Brownian motion, without loss of generality we can look at the zeros of the Brownian bridge instead, i.e. Brownian motion started at $0$ conditioned on $B(1)=0$. In this case, $Z$ can be constructed by replacing $[0,1]$ by the random union $[0,\tau\_1]\cup [\tau\_2,1]$, where $\tau\_1=\max\{ t\le 1/2: B\_t=0\}$ and $\tau\_2=\min\{ t\ge 1/2: B\_t=0\}$, and continuing inductively on these intervals. See the Chapter 14.3 of Lyons and Peres' [Probability on trees and networks](http://mypage.iu.edu/~rdlyons/prbtree/book.pdf) for the details.


Conditioned on $\tau\_1,\tau\_2$, $Z\_1$ has the law of $\tau\_1 Z$, $Z\_2$ the law of $(1-\tau\_2)Z$ up to a translation, and $Z\_1, Z\_2$ are independent. On the other hand, the distribution of $(1-\tau\_2)/\tau\_1$ is absolutely continuous (it has some explicitly computable density). This means that $Z-Z\supset Z\_1 - Z\_2$ which, conditioned on the abs. continuous random variable $\lambda=(1-\tau\_2)/\tau\_1$ is homothetic to $Z'-\lambda Z''$, where $Z',Z''$ are two independent realizations of the zeros of the Brownian bridge. But we have seen before that for a.e. realization of $(Z', Z'')$, we have $\dim\_H(Z'-\lambda Z'') = 1$ for almost all $\lambda$, so we are done (after Fubini).


In fact, using newer (and deeper) techniques such as those in M. Hochman's paper "[Dynamics on fractals and fractal distributions](http://arxiv.org/abs/1008.3731)" and [my own paper with Hochman](http://arxiv.org/abs/0910.1956), one can show that a.s.
$$
\dim\_H(Z-\lambda Z) = 1 \quad\text{for all }\lambda\in\mathbb{R}\setminus\{0\}.
$$


