Are $ \tau _{1}$ and $\tau _{2}$ same? 
Let $ \tau _{1}$be the product (standard) topology on $\mathbb{R}^{2}$ and $ \tau _{2}$ be the topology generated by the base $B_{2}=\{ C(a,r,R): a \in \mathbb{R}^{2}, r,R \in \mathbb{R}, 0<r<R \} $ where 
  $C(a,r,R)=\{(x,y) \in \mathbb{R}^{2}:r^{2}<(x-a_{1})^{2}+(y-a_{2})^{2} < R^{2}\}$ where $a=(a_{1},a_{2})$. 
  Then is $ \tau _{1} =  \tau _{2}?$

I feel that both are equal.
 Open sets in $ \tau _{1}$ are open rectangles and in $ \tau _{2} $ they are annuli. Given a rectangle we can find an annulus containing that rectangle and given an annulus we can always find a rectangle which contains that annulus. Hence the two are equal. Am I correct?
 A: Proof sketch: The standard way to show that two topologies are equal is to show that they are equal as sets. In other words, that any set which is open in one topology is open in the other.
This can be simplified to checking that any set in a base for one topology is open in the other. So, take your favourite base for the standard topology, show that every element there is a union of open annuli (equivalently that each point in it has an open annulus neighborhood). Then show that any open annulus is open in the standard topology.
A: All basic open subsets of $\tau_2$ are open in $\tau_1$. This implies that $\tau_2 \subseteq \tau_1$ right away. (unions of $\tau_2$-basic sets are also $\tau_1$-open.)
For every basic open ball $B(a,R)$ for $\tau_1$ we have that $C(a,r,R) \subseteq B(a,R)$ and this implies $\tau_1 \subseteq \tau_2$. ($O$ open in $\tau_1$ iff for every $a \in O$ we have some $R>0$ with $B(a,R) \subseteq O$; we can now also take $C(a,r,R)$ and get $O$ as a union of $\tau_2$-basic sets as well).
Hence we have equality, as you suggested.
