I have a group $G$ with two topologies $\tau, \tau'$ on $G$ that makes it a Polish group (a completely metrizable and separable topological group). I need to show that if $\mathcal{B}(\tau)=\mathcal{B}(\tau') $ , that is they define the same borel sets, then we have $\tau=\tau'$.
I just tried the obvious, I suppose a $U\in \tau$ then $U\in \mathcal{B}(\tau)=\mathcal{B}(\tau')$ but I can't go anywhere from here...
Could you help me?