Proving equivalence of topologies using subbases Suppose I have two topologies $\mathcal{T}$ and $\mathcal{T}'$ on a set $X$.  Furthermore suppose $\mathcal{T}$ is generated by a collection $\mathcal{E} \subseteq \mathscr{P}(X)$ and $\mathcal{T}'$ is generated by a collection $\mathcal{F} \subseteq \mathscr{P}(X)$.
I recall from my general topology course that in order to show that $\mathcal{T} = \mathcal{T}'$, it suffices to show that for each $E \in \mathcal{E}$, there is $F \in \mathcal{F}$ such that $E \subseteq F$, and that for each $F \in \mathcal{F}$, there is $E \in \mathcal{E}$ such that $F \subseteq E$.
What I don't recall is why this suffices.  Can anyone help refresh my memory?  Or perhaps correct my memory if this is incorrect?  Thanks!
 A: Since this is too long to post as a comment, I'll post it here. I do not think what you have said about bases suffices to show that the topologies are equivalent. For example, let $X = \Bbb{R}$, let $\mathcal{T}$ be the standard topology on $\Bbb{R}$ with the basis generated by all open intervals, $\mathcal{E} = \{(a,b): a,b \in \Bbb{R} , a<b \}$ and let $\mathcal{T}'$ be the lower limit topology on $\Bbb{R}$ generated by all half open intervals $\mathcal{F} = \{[a,b): a,b \in \Bbb{R} , a<b \}$. Then clearly for any basic open set $[a,b) \in \mathcal{F}$ we know $(a,b) \subset [a,b)$ with $(a,b)\in \mathcal{E}$. And for any basic open set $(c,d) \in \mathcal{E}$ we know there is a basic open set $\left[\frac{c+d}{2},d \right) \in \mathcal{F}$ such that $\left[\frac{c+d}{2},d \right) \subset (c,d)$. However, it is well known that the lower limit topology is a refinement of the standard topology on $\Bbb{R}$ so the two topologies are not equivalent just because the relation you mentioned holds for the respective bases.
What you can do to show that two topologies are equal is show that for any open set in one topology, it can be found in the other, and vica versa. The argument works by double containment.
