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Let us consider subsets in $\mathbb R^n$. A set $E$ is regular open if $E = \text{int}(\bar{E})$. It is true that the union of two regular open sets is not necessarily regular open. The counterexample I have seen is to take two nonoverlapping intervals with common endpoints. For example, $(-1, 0)$ and $(0, 1)$ are both regular open but $(-1, 0)\cup (0,1)$ is not regular open. I am wondering what if the two regular open sets $E \cap F$ have nonempty intersection, i.e., $E \cap F \neq \emptyset$ and $E, F$ are both path-connected, would the union $E \cup F$ be regular open? In other words, are there sufficient conditions to guarantee $E \cup F$ to be regular open?

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No. Consider these subsets of $\mathbb{R}^2$:$$E=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge0<\theta<\frac{3\pi}2\right\}$$and$$F=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge\frac\pi2<\theta<2\pi\right\}.$$They satisfy the conditions that you mentioned, but the interior of $\overline{E\cup F}$ is strictly larger than $E\cup F$, since it also contains the points of the type $(r,0)$, ithe $r\in(1,2)$.

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  • $\begingroup$ Thanks. Are there sufficient conditions to guarantee the conclusion? $\endgroup$ – user1101010 Jul 6 '18 at 18:07
  • $\begingroup$ @iris2017 I don't know. $\endgroup$ – José Carlos Santos Jul 6 '18 at 18:13
  • $\begingroup$ @iris2017 See my answer below. $\endgroup$ – giobrach Jul 6 '18 at 18:40
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To answer your question in the comments to José's answer: it is sufficient to require $E,F$ to be regular open and for $E \cup F$ to have an open complement, i.e. $\mathrm{et}(E \cup F) = X \setminus (E \cup F)$.

Proof. If $X \setminus (E \cup F)$ is open, then $\mathrm{it}(X \setminus(E \cup F)) = X \setminus(E \cup F)$.

Now both $E$ are regular open, so $$\mathrm{it}(E)=\mathrm{it}(\mathrm{it}(\mathrm{cl}(E))) = \mathrm{it}(\mathrm{cl}(E)) = E$$ which means $E$ is open (and so is $F$ by an identical argument). Therefore their union is also open and $\mathrm{it}(E \cup F) = E \cup F$. Therefore, by exploiting the fact that, for all $A \subseteq X$, $$\mathrm{cl}(A) = X \setminus \mathrm{it}(X \setminus A), $$ we get $$\mathrm{it}(\mathrm{cl}(E\cup F)) = \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E \cup F))) = \mathrm{it}(X \setminus (X \setminus (E \cup F))) = \mathrm{it}(E \cup F) = E \cup F.$$ Hence $E \cup F$ is regular open. $\qquad \square$


Addendum. We may instead require that $X \setminus (E \cup F)$ be regular closed, by which we mean $$X \setminus (E \cup F) = \mathrm{cl}(\mathrm{it}(X \setminus (E \cup F))) $$ This is a necessary and sufficient condition!

Proposition. Let $E,F \subseteq X$ be regular open. Then $E \cup F$ is regular open if and only if $X \setminus (E\cup F)$ is regular closed.

Note: that the previous case ($X \setminus (E \cup F)$ open) implies this situation: if the complement of $E \cup F$ is open, then its interior is equal to itself; but since $E \cup F$ is open, then $X \setminus (E \cup F)$ is also closed, and so the closure of its interior, which is equal to the closure of itself, is equal to itself.

Proof. As we've seen above, if $E,F$ are regular open, then they are open. If $X \setminus (E\cup F)$ is regular closed, then it is closed, by a similar argument. Hence, $$\begin{split} X \setminus (E\cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= \mathrm{cl}(\mathrm{it}((X \setminus E)\cap (X \setminus F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F)) \\ &= X \setminus \mathrm{it}(X \setminus (\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}((X \setminus \mathrm{it}(X \setminus E)) \cup (X \setminus \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E) \cup \mathrm{cl}(F)) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)), \end{split}$$ which implies $E \cup F = \mathrm{it}(\mathrm{cl}(E\cup F))$.

Another, shorter way to see this is $$\begin{split} X \setminus (E \cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)). \end{split}$$

This chain of equalities, followed in the reverse order, also takes care of the other implication: if $E \cup F$ is regular open, then $$\begin{split} X \setminus (E \cup F) &= X \setminus \mathrm{it}(\mathrm{cl}(E\cup F)) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X\setminus (E \cup F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))). \\ \end{split} $$ Hence $X \setminus (E \cup F)$ is regular closed. $\qquad \square $

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  • $\begingroup$ Thanks. Could you give an example? I could not see how $X \setminus (E \cup F)$ could be open. $\endgroup$ – user1101010 Jul 6 '18 at 18:50
  • $\begingroup$ Both $E\cup F$ and its complement need to be both open and closed. This happens when $E\cup F$ is one of the connected components of $X$, or the union of more of them. Can you think of an instance of this? $\endgroup$ – giobrach Jul 6 '18 at 19:01
  • $\begingroup$ Do we only need to guarantee that $X \setminus (E \cup F)$ is connected? We can take two open balls in $\mathbb R^n$ for $n \ge 2$. Is this right? $\endgroup$ – user1101010 Jul 6 '18 at 19:26
  • $\begingroup$ My theorem also works in the case where $X \setminus (E\cup F)$ decomposes into separate connected components. Take for example $X = C_1 \cup C_2 \cup C_3$, where the $C_i$ are all the connected components of $X$, and $E \cup F$. Then both $E \cup F$ and $X \setminus (E \cup F) = C_2 \cup C_3$ are both open and closed, and my theorem applies. $\endgroup$ – giobrach Jul 6 '18 at 19:35
  • $\begingroup$ Of course, we are working in a topological space $(X,\tau)$ where $X$ is the total space. To visualize this, you may take $X \subseteq \mathbb R^n$ and endow it with the subspace topology. $\endgroup$ – giobrach Jul 6 '18 at 19:36
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If $\overline A$, $\overline B$ disjoint,
then int (A $\cup$ B) = int A $\cup$ int B.

Thus if A,B are regular open and $\overline A$, $\overline B$
disjoint, then A $\cup$ B is regular open.

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