TITLE: When I prove two topologies are same, is it sufficient to prove they have same basis? If so,Why?
QUESTION [7 upvotes]: $(X, \mathcal{T}_X), (Y,\mathcal{T}_Y)$ be topological space, and let
  $A, B$ be subset of $X,Y$ respectively.
Now, $(X \times Y, \mathcal{T}_{X \times Y})$ be product topology. $A
 \times B \subset X \times Y$. Consider subspace topology
  $\mathcal{T}_1$ on $A \times B$
Also, $(A, \mathcal{T}_A), (B, \mathcal{T}_B)$ where $\mathcal{T}_A$
  and $\mathcal{T}_B$ are subspace topologies on $A, B$
  respectively,consider product topology $\mathcal{T}_2$ on $A \times B$
Prove $\mathcal{T}_1 = \mathcal{T}_2$

My idea is to show that the bases of $\mathcal{T}_1$ and $\mathcal{T}_2$ are same.
where $\mathcal{T}_1$ basis is {$(A \times B) \cap (U \times V)  | (U \times V) \in \mathcal{T}_{X \times Y}$}
and $\mathcal{T}_2$ basis is {$U_0 \times V_0 | U_0 \in \mathcal{T}_A,  V_0 \in \mathcal{T}_B$}. I know how to show these two bases are equal. However, I don't if this is sufficient to prove two topologies are equal.

REPLY [6 votes]: It is sufficient, and here is why. 
Let $(X,\mathcal{T}_1)$ and $(Y,\mathcal{T}_2)$ be topological spaces with the same basis $\mathcal{B}$. For the sake of contradiction, suppose $\mathcal{T}_1 \neq \mathcal{T}_2$. Then $\mathcal{T}_1$ contains an element $V$ such that $V\notin \mathcal{T}_2$. Since $V \in \mathcal{T}_1$ there exists a collection of basic elements $\{B_\alpha \}_{\alpha \in \lambda} \subseteq \mathcal{B}$ such that $$V = \bigcup_{\alpha \in \lambda} B_\alpha $$ And since $\mathcal{B}$ is a basis for $\mathcal{T}_2$ then $\bigcup_{\alpha \in \lambda} B_\alpha$ is an element of $\mathcal{T}_2$, a contradiction.