Show that if $A$ is a basis for a topology on $X$,then what will be the topology generated by $A$ 
Show that if $A$ is a basis for a topology on $X$,then the topology generated by $A$ equals the intersection of all the topologies on $X$ that contain $A$.Prove the same if $A$ is a sub basis.

My attempt:
If $A$ is a basis that generates the topology $\mathcal{T}$ then $\mathcal{T}=\cup A_{\alpha}$ where $A_{\alpha}$ denotes the elements of $A$.
Now, let $\cap \mathcal{T}_{\alpha}$ denote the topologies whose bases contain the basis $A$.Then $A \subset B_{\alpha}$ where $B_{\alpha}$ denotes the basis of $\mathcal{T}_{\alpha}$.
Then $\mathcal{T} \subset \cap \mathcal{T}_{\alpha}$.
To show the reverse inclusion, we note that $\mathcal{T}$ is also an element of the family of topologies $\{\mathcal{T}_{\alpha}\}$ .Using this we can conclude that $\cap \{\mathcal{T}_{\alpha}\}=\mathcal{T}$
How do I prove it when $A$ is a sub basis?I know that every basis element can be written as the finite intersection of the sub basis element .Then?Also is my above proof ok?
 A: The idea of the proof is basically correct, but I would use the criterion that to show two topologies are equal, then we pick an open set of one topology and show it belongs to the other one as well. You are working with bases, and this is fine as well. I would justify a bit more in the point that "every open set can be written as union of basis elements" and the collection of all open sets is indeed the topology.
For the subbasis part, as you recall, try to go back to the proposition you have just proved. I sketch here what I would do:
Let $\tau$ be the topology that consists of all unions of finite intersections of sets in $A$, and let $\tau^\prime$ be the topology which is the intersection of all topologies that contain $A$.

*

*$\tau'\subseteq \tau$.

Suppose $U \in \tau'$. Then $U$ is an element of all the topologies which contain $A$, and so also $\tau$, which is one of these topologies. Hence, $U \in \tau$, which implies $\tau'\subseteq \tau$.


*$\tau \subseteq \tau'$:

Let $W \in \tau$. Then $W = \bigcup W_i$ with each $W_i$ a finite intersection of sets in $A$, and define $W_i = W_{i1} \cap \dots \cap W_{in}$, with $W_{ij} \in A$, with $i \in I$ generic index set and $j \in \mathbb{N}$. Since $W_{ij} \in A$, and $A \subseteq \tau'$, and since $\tau'$ is a topology (and hence it is closed under finite intersections), $W_i \in \tau'$, and so (since $\tau'$ is also closed under arbitrary unions) $W \in \tau'$, concluding that $\tau \subseteq \tau'$.
