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 $p_0=0$.