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.
Best Answer
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 $B$.