problems on topology involving basis and sub-basis 
Need some help guys, The question was: Can you define a basis generated by a sub-basis? 
I tried to prove this today in class. My prof said that everything is correct except B3 (the 3rd basis). I didn't declare or define it properly and without that I claimed that there exists a basis B3 which follows the definition or 2nd criteria of a basis. How should I define that in my proof?
Any help will be greatly appreciated. Thanks in advance.  
 A: You did every thing correctly. 
For the $B_3$ you looking for, note that elements of the basis $B$ are finite intersections  of subbasis $S$ and you reach to $B_2\cap B_2$ is in the basis $B$ because $B_1\cap B_2$ is a finite intersection of elements of $S$. 
Now you looking for a $B_3$ such that $B_3\subseteq B_1\cap B_2$? Take $B_3=B_1\cap B_2$.
A: The base could also be written as: 
$$\mathcal{B}=\{ \bigcap \mathcal{S}': \mathcal{S}' \subseteq \mathcal{S} \text{ finite }\}$$
with the convention (follows from void truth considerations) that $\bigcap \emptyset = X$, which then handles the first demand on candidate bases, that $\bigcup \mathcal{B}=X$. This way we don’t even have to ask for $\mathcal{S}$ to cover $X$, as some texts do, superfluously. 
$\mathcal{B}$ is then by construction closed under finite (or dual) intersections, as the union of two (or finitely many) finite subfamilies of $\mathcal{S}$ is also finite, and this implies the second demand on bases (as e.g. Munkres defines them).
We can then just take $B_3= B_1 \cap B_2$ for all $x$ in that intersection. Maybe you should have made that fact more explicit.
