Let $(X,\tau)$ be a topological space, let $O\in\tau$, let $O^c$ be the complement of $O$ in $X$, let $\text{Int}[O]$ be the interior of $O$.

Which open sets $O$ satisfy $\text{Int}[O^c] \cup O = X$?

My guess is that it's precisely the clopen sets. I think the following is a proof. Is it correct?

Suppose $O$ is clopen. Then, in particular, $O$ is closed. So, its complement $O^c$ is open. Since the interior of an open set is again that open set, we have that $\text{Int}[O^c]$ is again $O^c$. Therefore, $\text{Int}[O^c] \cup O$ equals $O^c \cup O$, which equals $X$ by the definion of complement. This proves one direction.

Now suppose $\text{Int}[O^c] \cup O = X$. Since $O^c\cup O = X$ by definition, we must have that $O^c \subseteq \text{Int}[O^c]$, because otherwise the union $\text{Int}[O^c] \cup O$ wouldn't be enough to cover all of $X$. But the interior of a set is always a subset of that set, meaning $\text{Int}[O^c]\subseteq O^c$. Now we have $\text{Int}[O^c] = O^c$. This means $O^c$ is open, and so $O$ is closed. Also, $O$ is open by hypothesis.

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    $\begingroup$ That's correct. $\endgroup$ Jul 19 '17 at 13:26
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    $\begingroup$ In short: since $\text{int}[O^c]\subseteq O^c$ we have $\text{int}[O^c]\cup O=X=O^c\cup O\iff \text{int}[O^c]=O^c\iff O^c\text{ is open}$. $\endgroup$
    – drhab
    Jul 19 '17 at 13:42

Your proof is correct, but can be drastically shortened. Points of $X$ w.r.t. $E$, $E\subseteq X$, can only be of one of this type, exclusively: interior, boundary, or exterior. $\operatorname{Int}(E^c)$ are the exterior points of $E$. $\operatorname{cl}(E) =\operatorname{Int}(E^c)^c$

$$\operatorname{Int}(E^c)\cup E = X \iff E=\operatorname{Int}(E^c)^c \iff E \textrm{ is closed}$$

If you want to add on a closed set the condition that it must be open, then it is clopen.

I will repeat at a slower pace.

This $$\operatorname{Int}(E^c)\textrm{ are the exterior points of }E$$ comes from the definition of exterior points of a set. Follow this:

Points that are neither interior nor boundary points (that is points that are not adherence points) of a set have some neighborhoods disjont from the set, that is have some neighborhoods included in the complement of the set, that is are interior points of the complement of the set.

This $$\operatorname{cl}(E) =\operatorname{Int}(E^c)^c$$ come from what I have already written: a point can be one and only one among interior, boundary, and exterior of a set. Now the closure of a set is the union of the set of the interior points and the set of the boundary points that in turn for what just said is the complement of the set of the exterior points. But the exterior points of a set are the interior points of the complement of the set.

This $$\operatorname{Int}(E^c)\cup E = X \iff E=\operatorname{Int}(E^c)^c$$ comes from the fact that, given the sets $X$, $A\subseteq X$, $B\subseteq X$, with $A\cap B = \emptyset$, $A\cup B = X \iff B=A^c$ and the fact that $\operatorname{Int}(E^c)\cap E = \emptyset$ (that is exterior points of $E$ does not belong to $E$)

This $$E=\operatorname{Int}(E^c)^c \iff E \textrm{ is closed}$$ comes from the fact that $\operatorname{cl}(E) =\operatorname{Int}(E^c)^c$ and $E=\operatorname{cl}(E) \iff E \textrm{ is closed}$

  • $\begingroup$ This has too many steps that I cannot follow =( $\endgroup$ Jul 20 '17 at 23:47
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    $\begingroup$ @étale-cohomology too long answering with a comment. See my answer edited accordingly. $\endgroup$
    – trying
    Jul 21 '17 at 8:57
  • $\begingroup$ But then the proof isn't that much shorter, maybe? Anyone can make a proof artificially shorter by removing almost all the prose, using mostly symbols, and skipping "obvious" steps =). Of course, I'm accepting this answer because the proof was a pleasure to follow! If more proofs were written like this, one could almost read math like one reads a novel $\endgroup$ Jul 21 '17 at 11:14
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    $\begingroup$ @étale-cohomology I'm very flattered, thank you. What I've written in prose is something it is sedimented in my mind and was stirred up while reading the condition in title of your post, seeing clearly what the operands of the union and its result are. That's why the proof was shorter in my mind. I hope now you can see what I saw, leaving apart computations. $\endgroup$
    – trying
    Jul 21 '17 at 11:43
  • $\begingroup$ I can now, but I couldn't before! That's the magic of proofs $\endgroup$ Jul 21 '17 at 12:27

If $\operatorname{Int}(O^c) \cup O = X$ then

$$O^c = X \cap O^c = (\operatorname{Int}(O^c) \cup O) \cap O^c = (\operatorname{Int}(O^c) \cap O^c) \cup ( O \cap O^c) = (\operatorname{Int}(O^c) \cap O^c)$$

so $O^c \subseteq \operatorname{Int}(O^c)$ and $O^c$ is open and $O$ is closed.


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