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.


3 Answers 3


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 $B$.

  • $\begingroup$ What about my arguements for the interior. Were those ok? $\endgroup$
    – emka
    Feb 18, 2013 at 3:40
  • $\begingroup$ @abet: Yes: for any $S\supseteq A$, $S$ is open, so it’s its own interior. $\endgroup$ Feb 18, 2013 at 3:44
  • $\begingroup$ @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. $\endgroup$ Feb 18, 2013 at 3:51
  • $\begingroup$ Wow. I don't know what I was thinking. Glad my error helped you see yours. :P $\endgroup$ 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!

  • $\begingroup$ 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? $\endgroup$
    – Averroes2
    Sep 15, 2020 at 10:59
  • $\begingroup$ @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 $X$. $\endgroup$ 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 $\operatorname{cl}(A)=X$.


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