# Proof regarding cluster point in Brownian motion

I have a standard Brownian motion, $$B(t)$$, $$t\ge0$$.

I understand that every $$t > 0$$ is an accumulation point (i. e., cluster point) of $$(s: B(s) = B(t))$$ from the right, with probability $$1$$.

That is, for any fixed $$t > 0,$$

$$P[$$inf$$(s > t : B(s) = B(t)) = t] = 1$$

Now, I am trying to find out whether $$P[$$inf$$(s > t : B(s) = B(t)) = t$$ for all $$t \in [0, 1]] =$$ $$1$$ $$?$$

My heuristic is no:

$$P[$$inf$$(s > t : B(s) = B(t)) = t$$ for all $$t \in [0, 1]] \neq$$ $$1$$

But I would like to prove this mathematically using the last time in $$[0, 1]$$ when the Brownian motion hits $$0$$, instead of only trying to imagine or convince myself why. Any help would be most appreciated. Thanks a lot.

• Before asking if $P[$inf$(s > t : B(s) = B(t)) = t$ for all $t \in [0, 1]] \neq$ $1$ you should also worry about measurability of this event. Dec 6, 2023 at 9:07

(I assume $$B(0)=0$$.) Your intuition is on point. Let $$G:=\sup\{s\le 1: B(s) =0\}$$ denote the last-exit time from $$0$$ before time $$1$$. Then $$\Bbb P[G=0]=0$$ by your initial "accumulation point" observation (which is true even for $$t=0$$). Also, $$\left\{\inf(s>t: B(s) = B(t)) =t,\forall t\in[0,1]\right\}\subset\{G=1\}.$$ But $$\Bbb P[G=1] = 0$$ by continuity of $$t\mapsto B(t)$$ and the fact that $$\Bbb P[B(1)=0]=0$$.