Closures and Interiors in a topology Let $X$  be a set.  Let $A$ be a proper non-empty subset of $X$.  Let $\tau = \{\emptyset\} \cup \{U \in P(X): A \subseteq U\}$

Question: Find the interior and closure of $A$ and prove that they are indeed the interior and closure.

(i)  Interior of $A$:  The interior of $A$ is $A$ since $A$ is an element of $\tau$ and hence it is an open set.
(ii) I'm stuck trying to figure out the closure of $A$.  In this topology, since any $U$ containing $A$ is an open set...does this mean that any closed set would be a set that does not contain $A$. So the closure of $A$ is the empty set.

Question: Suppose $B$ is a nonempty proper subset of $X$ such that $A$ is a nonempty proper subset of $B$.  Find the interior and closure of $B$ and prove your answers.

Again I think the interior of $B$ is $B$.  Similarly, the closure is empty.
 A: The closed sets are the complements of the open sets. The complement of $\varnothing$ is $X$. Now $U\supseteq A$ iff $X\setminus U\subseteq X\setminus A$, so the closed sets other than $X$ itself are precisely the sets disjoint from $A$. Thus, the only closed set containing $A$ is $X$, and $\operatorname{cl}A=X$.
You can also see this by looking at limit points. Suppose that $x\in X$, and $U$ is an open nbhd of $x$. Then $U\ne\varnothing$, so $U\supseteq A$. Thus, every open nbhd of $x$ contains a point of $A$ $-$ contains every point of $A$, in fact! $-$ so $x\in\operatorname{cl}A$. Since $x$ was an arbitrary point of $X$, $X=\operatorname{cl}A$.
The closure of $B$ is also $X$, by the same argument. The closure of a non-empty set $B$ can never be empty, because the whole space is always a closed set containing $B$.
A: You are right about $A$ being open (as $A \subset A$) so $\mbox{int}(A) = A$.
A set $F$ is closed iff $X \setminus F$ is open iff ($X \setminus F = \emptyset$ or $A \subset X \setminus F$) iff ($F = X$ or $A \cap F = \emptyset$).
So, in words, the closed sets are the whole space $X$ (as always) plus all sets disjoint from $A$ (and this includes the expected empty set as well).
$\mbox{cl}(A)$ is the smallest closed set that contains $A$, and the only closed set that actually contains $A$ can be $X$ (as $A$ is non-empty, a set cannot contain $A$ and be disjoint from $A$ at the same time, so there are no sets of the second type). So the closure equals $X$.
Now suppose $A \subset B$. Then $B$ is still open, by definition, so still $\mbox{int}(B) = B$, and by the same argument, the only closed set that contains $B$ (and thus $A$) is $X$, so $\mbox{cl}(B) = X$ still. The closure of any set $S$ contains at least $S$, so the only set with empty closure, ever, in any topology, is the empty set!
A: Indeed since $A$ is open it is equal to its own interior. For the second part we recall the following:

Theorem. Let $X$ be a topological space, $D\subseteq X$. The following are equivalent:
  
  
*
  
*$D$ is dense, i.e. if $U\neq\varnothing$ is open then $D\cap U\neq\varnothing$.
  
*$\operatorname{cl}(D)=X$.
  

From this it is obvious that $A$ is dense in this topology since not only $A\cap U\neq\varnothing$, in fact $A\subseteq U$ for every non-empty open set. Therefore $\operatorname{cl}(A)=X$.
