# 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.

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 $B$.

• What about my arguements for the interior. Were those ok?
– emka
Feb 18, 2013 at 3:40
• @abet: Yes: for any $S\supseteq A$, $S$ is open, so it’s its own interior. Feb 18, 2013 at 3:44
• @Cameron: No, I meant to take complements. Unfortunately, I typoed it by leaving off the $X\setminus$ on the righthand side and then wrote the words for the symbols that I’d actually written instead of the ones that I’d meant to write. Thanks for catching it. Feb 18, 2013 at 3:51
• Wow. I don't know what I was thinking. Glad my error helped you see yours. :P Feb 18, 2013 at 15:49

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!

• I don't see what happens when if we investigate a subset $B\subset A$. Based on your implication regarding closed sets it would be an open set (since not disjoint with $A$. So it would be $int(B)=B$. But what would be its closure? Maybe $cl(B)=X$? How can we show this? Sep 15, 2020 at 10:59
• @Averroes2 if $B$ is a proper non-empty subset of $A$, then $B$ is not open and also not closed. The interior is empty and the closure again $X$. Sep 15, 2020 at 11:14

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:

1. $D$ is dense, i.e. if $U\neq\varnothing$ is open then $D\cap U\neq\varnothing$.
2. $\operatorname{cl}(D)=X$.

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 $\operatorname{cl}(A)=X$.