Basis For A Topology Let $X$ be a non empty set and $A$ is a subset of $X$.
Show that the family of all subsets of $X$ which contains $A$, together with the empty set, forms a topology on $X$. 
(Use definition of basis for a topology)
My take on this problem is to let the set $T = \{\emptyset, B \}$ where $B$ is the family of all subsets of $X$ which contains $A$. Then I will proceed on proving that T forms a topology on $X$ i.e.
a) Show that the elements of $T$ are open. The empty set is open trivially. Then I will claim that the set $B$ is a basis for a topology on $X$ and prove that elements of $B$ are open. Is that correct?
b) Then proceed with proving $T$ is a topology on $X$ i.e. the empty set and $X$ must be in $T$, arbitrary union of elements of $T$ must be in $T$ and finite intersection of elements of $T$ must be in $T$. Is that correct?
Your insights will be appreciated.
 A: It appears your terminology might be a bit off (which is respectable in my opinion if you're new to topology). The above comments should help straighten that out, but if you want me to try and explain it let me know.
Anyhow, given $X$ a set and $A \subset X$ and putting $B = \{ Y \subset X \mid A \subset Y \}$ if you want to prove that $B \cup \emptyset$ is a topology on $X$ by using the definition of a basis I would proceed as follows:


*

*Prove that $B$ is in fact a basis. That is

  
*
  
*$\forall x \in X \; \exists U \in B$ s.t. $x \in U$
  
*$\forall U,V \in B, \forall x \in U \cap V \; \exists W \in B$ s.t. $x \in W \subset U \cap V$
  


*Prove that any set that is open by the topology generated by this basis is in the basis.

*Conclude that $B$ is in fact a topology since any basis is a subset of the topology it generates (why?) and the topology formed by a basis is in fact a topology.


As for some hints:


*

*$A \subset X \implies X \in ?$

*If $U,V \in B$ then $A \subset U, A \subset V \implies ?$

*A set $U$ is considered in the topology if $\forall x \in U$ there's a basis element $V \in B$ s.t. $x \in V \subset U$. Note that $A \subset V$ so then $A \subset ?$.

*If $U \in B$ then $\forall x \in U x \in U \subset U, U \in B$.

A: To show that $T$ is a topology on $X$ you need to show:


*

*$\emptyset,X\in T$

*$\bigcup_{i\in I}A_i \in T$ if $A_i \in T$ for every $i\in I$

*$\bigcap_{i=1}^{n}A_i \in T$ if $A_i \in T$ for every $1\leq i\leq n$


obviously it all satisfies since $A$ is contained in all those sets (except for the empty set) and therefore since $B$'s sets are from the original topology you get what you want. 
in the cases that the empty set takes part it's trivial to show that it satisfies as well.
