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For a topological space $X$, given any two open sets $A,B$, there is a set $S\subseteq X$ such that $\DeclareMathOperator{\ntr}{int}\ntr S=A$ and $\DeclareMathOperator{\ext}{ext}\ext S=B$.

Is this true for $X=\Bbb R^2$? If so, what topological properties on $X$ are needed for a general topological space to satisfy this? If not, are there any restrictions that I forgot to add?

Mario Carneiro
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  • What do you mean exactly by "interior" and "exterior"? – Daniel Robert-Nicoud Apr 13 '14 at 22:32
  • @Daniel The interior of a set $A$ is the union of all open sets contained in $A$; the exterior of $A$ is the interior of $X\setminus A$ (where $X$ is the topological space in question). – Mario Carneiro Apr 13 '14 at 22:33
  • Do you mean that the union of two disjoint open sets is the interior of one set and the exterior of another? Or do you mean that one open set is the interior of a set and the other is the exterior of the same set? – manthanomen Apr 13 '14 at 22:42
  • @manthanomen The second interpretation – Mario Carneiro Apr 13 '14 at 22:44
  • Pick two disjoint open disks. How can you make one disk the interior of a set and the other the exterior of the same set? It is not possible. Do you mean to say that there is an open set $A$ such that the first of your sets is in the interior of $A$ and the second in the exterior of $A$? And try not to ask the question in the title. – Beni Bogosel Apr 13 '14 at 22:46
  • @Beni I've clarified the question in the body. And I don't think you've quite grasped the subtlety here; I'm working on a proof for $\Bbb R^2$, and I'm pretty sure it's true. The idea is that the boundary of $S$ is everything except those two disks, by using a "pockmarked" set like $\Bbb Q^2$. – Mario Carneiro Apr 13 '14 at 22:50
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    In the simple case where $X=\mathbb{R}^2$, you can take $S=A\cup((\mathbb{R}\setminus B)\cap \mathbb{Q}^2)$ – Brandon Apr 13 '14 at 23:01
  • @Brandon I do expect this to be the case, but I'm not quite there with the proof yet (which I will post as an answer). – Mario Carneiro Apr 13 '14 at 23:26
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    I don't think Brandon's example works. Consider the case where $A$ is the unit disk minus the origin; then $int(S) \neq A$. – Adam Bjorndahl Apr 13 '14 at 23:33
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    Suppose the claim is true for $X$. Then taking $A=B=\emptyset$, we see that there must exist an $S$ with no interior and no exterior. $S$ having no interior means that it is co-dense (its complement is dense), and $S$ having no exterior means that it is dense. It follows that a necessary condition for $X$ to have the property you are interested in is the existence of a dense and co-dense subset. (This rules out, e.g., discrete spaces.) – Adam Bjorndahl Apr 13 '14 at 23:34
  • @Adam I'm currently at the point of trying to prove it under exactly that assumption (which I call a "scattered" top. sp. for want of a better name), but I think some separation axiom is also needed. – Mario Carneiro Apr 13 '14 at 23:41
  • @Adam To repair Brandon's example, does it suffice to assume $A,B$ are regular open? – Mario Carneiro Apr 14 '14 at 00:47
  • @Adam I've updated my answer to account for this case; I would take $S=A\cup(\Bbb Q^2\setminus B)\setminus{0}$ in this case. – Mario Carneiro Apr 14 '14 at 07:44

3 Answers3

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Here is the topological property that should suffice to make this statement true:

A topological space $X$ is resolvable iff there is a set $T\subseteq X$ such that $T$ and $T^c$ are both dense.

Then $\Bbb R^2$ is resolved by $\Bbb Q^2$.

First, we prove the theorem for regular open sets $A,B$ (note that a regular open set $U$ is one such that $\text{int }\bar U=U$). Now let $X$ be resolved by $T$, and let $A,B$ be disjoint regular open sets in $X$, and let $S=A\cup (T\setminus B)$. I claim that $\DeclareMathOperator{\ntr}{int}\ntr S=A$ and $\DeclareMathOperator{\ext}{ext}\ext S=B$. Since the interior function respects set inclusion, we have:

$$A=\ntr A\subseteq\ntr A\cup (T\setminus B)=\ntr S$$ $$B=\ntr B=\ntr A^c\cap B\subseteq\ntr A^c\cap(T^c\cup B)=\ext S$$

For the converse, note that $(\ntr\bar A\cup T)\setminus\bar A\subseteq (\bar A\cup T)\setminus\bar A\subseteq T$, and $(\ntr\bar A\cup T)\setminus\bar A=(\ntr\bar A\cup T)\cap\ext A$ is open, so $(\ntr\bar A\cup T)\setminus\bar A\subseteq\ntr T=\emptyset$. Therefore $\ntr\bar A\cup T\subseteq\bar A$, so $$\ntr S\subseteq\ntr A\cup T\subseteq\ntr\bar A\cup T=\ntr\ntr\bar A\cup T\subseteq\ntr\bar A=A$$ (since $A$ is regular open).

Similarly for $B$, $\ext S\subseteq\ntr T^c\cup\bar B\subseteq\ntr\bar B=B$ because $(\ntr T^c\cup\bar B)\setminus\bar B$ is an open set contained in $T^c$, which has empty interior because $T$ is dense.

Using the above as a lemma, we can generalize the theorem to any pair of open sets $A,B$. Let $\DeclareMathOperator{\irr}{irr}\irr U=\ntr\bar U\setminus U$. Then $U$ is regular iff $\irr U=\emptyset$ (which is to say, $\irr U$ is the set of "irregular points" of $U$). Now for arbitrary open disjoint $A,B$, let $Y=X\setminus(\irr A\cup \irr B)$ as a subspace. Then $\irr_YA=\irr_YB=\emptyset$, so there is a set $Q\subseteq Y$ such that $\ntr_Y Q=A\cap Y=A$ and $\ext_Y Q=B\cap Y=B$. Note that $\irr A$ is disjoint from $B$ and vice-versa, since $\irr A\subseteq\bar A$, and $B=\ntr B\subseteq \ntr A^c=\bar A^c$.

I claim that for $S=Q\cup\irr B$, $\ntr S=A$ and $\ext S=B$. Now $\ntr S$ is disjoint from $\irr B$ because $\ntr S$ and $B$ are disjoint (for the same reason that $A$ and $\irr B$ are disjoint), so $\ntr S\subseteq Y$ is open, and thus $\ntr S\subseteq\ntr_Y Q=A$. Similarly, $\ext S\subseteq B$. Conversely, $A\subseteq Q\subseteq S$, and since $A$ is open, we have $A\subseteq\ntr S$, and similarly $B\subseteq\ext S$. (Details of this paragraph borrowed from Karl's answer)

YuiTo Cheng
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Mario Carneiro
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  • I see that you edited in what you could of my idea. I noticed that my idea is wrong as I stated in chat. I think the slight modification $S\cup I(B)$ is more likely to work. – Karl Kroningfeld Apr 14 '14 at 07:49
  • @Karl Yep, I was working that in when I saw your comment. Figures it has to be symmetric w.r.t. complementation. – Mario Carneiro Apr 14 '14 at 07:56
  • Note that scattered is already notion in topology, meaning that every nonempty set has an isolated point. IMHO, there is really no need to introduce a new definition to say that $T$ is a dense co-dense subset of $X$. – user642796 Apr 14 '14 at 08:13
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    @Arthur I just knew that was going to happen. I swear, open a dictionary and point at an adjective, and some topologist has already given a definition to the word. Anyway, I wanted a way to eliminate $T$ from the definition (this is a topological property of the space, not a property of a subset of the space), and for continuity with the comments, I'd rather not change the name of the property on this page, unless there is already a name for this exact property. – Mario Carneiro Apr 14 '14 at 08:25
  • @MarioCarneiro: Yeah, topology is a tricky subject to come up with undefined notions. Even adding the random "pseudo-", "quasi-", or "meta-" prefix probably won't help you. (If one ever needs the property you define, I would assume that saying "$X$ has a dense co-dense subset" would suffice.) Incidentally, your notion of scattered contradicts the standard definition: if a space has an isolated point, then every dense set must contain that isolated point, meaning that no subset can be both dense and co-dense! – user642796 Apr 14 '14 at 08:32
  • @Arthur As you point out, every scattered space is perfect, because it has no isolated points. I wonder if the converse is true. That is, are there any spaces that are perfect but not scattered? – Mario Carneiro Apr 14 '14 at 08:50
  • @Mario: No, it's the opposite: scattered spaces cannot be perfect because they have isolated points. – user642796 Apr 14 '14 at 08:52
  • @Arthur Haha, as I said, I'll continue to use the term scattered under my definition on this page. Are there any perfect spaces with no dense co-dense subsets? – Mario Carneiro Apr 14 '14 at 08:53
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    The right term is resolvable! – Adam Bjorndahl Apr 14 '14 at 13:19
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Originally, this answer and edits to Mario Carneiro's answer were essentially the same, independently discovered, arguments. I subsequently noticed an equivalent approach that makes better use of my Lemma that I have elucidated in this revision. I refer to the portions of Mario's answer which treat the specific case in which the open sets are regular.

$\DeclareMathOperator{\interior}{int}\DeclareMathOperator{\exterior}{ext}\DeclareMathOperator{\irr}{irr}$For any subset $A\subseteq X$, define $\irr A=\{x\in\interior\bar A\colon x\notin A\}$. Now, suppose $A$ is open in $X$. Then $A\subseteq\interior\bar A$ means that $A\cup\irr A=A\cup(\interior\bar A\setminus A)=\interior\bar A$ is an open subset of $X$. The fact that it is contained in $\bar A$ means that it is regular.

Lemma. For any disjoint open sets $A,B$ in a topological space $X$, the sets $A\cup\irr A$ and $B\cup\irr B$ are also disjoint.
Proof. If $A\cap\irr B$ is nonempty, then $A$ has a nonempty (open) intersection with the interior of $\bar B$. But $A\cap B=\emptyset$, so $\bar B\setminus B$ contains an open set, contradiction. It follows that $A\cap\irr B=\emptyset=\irr A\cap B$. It only remains to show that $\irr A$ is disjoint from $\irr B$. If $x$ is in both, then let $U$ be a neighborhood of $x$ contained in both $\bar A$ and $\bar B$. We have, $$\emptyset\ne U\cap B\subseteq U\cap\interior A^c\subseteq\bar A\cap\interior A^c=\emptyset,$$ a contradiction. q.e.d.

Therefore, if $X$ is a topological space with a dense co-dense subset, then as Mario Carneiro showed there exists $S_0\subset X$ such that $\interior S_0=A\cup\irr A$ and $\exterior S_0=B\cup\irr B$ for any disjoint open sets $A,B\subseteq X$.

Observe that $A\subseteq\interior(S_0\setminus\irr A)\subseteq(\interior S_0)\setminus\irr A=A$. Define $S=S_0\cup \irr B\setminus\irr A$. If $U$ is an open subset of $S$, then $U\setminus\irr B$ is open in $S_0\setminus\irr A$, so $U\subseteq A$ by our initial calculation. It follows that $\interior S=A$. To see that $\exterior S=B$, just note the following symmetry: $S^c=S_0^c\cup\irr A\setminus\irr B$.

Karl Kroningfeld
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Let $X$ be a resolvable space with a dense/co-dense subset $D$, and $A, B$ disjoint open subsets of $X$. Then the set $S = A \cup (D\setminus \overline{A}\setminus\overline{B}) \cup (\overline{B}\setminus B)$ has $\operatorname{int} S=A$ and $\operatorname{ext} S=B$.

Clearly $A \subset \operatorname{int} S$. To see the reverse inclusion, note that $S \subset A \cup \operatorname{ext} A \cup \overline{B}$. Obviously a point in $\overline{B}$ cannot be interior to any subset of $X\setminus B$. Because the complement of $D\setminus \overline{A}\setminus\overline{B}$ is dense and the complement of $\overline{B} \setminus {B}$ is dense and open, their intersection is everywhere dense, so a point in $S \cap \operatorname{ext} A$ cannot be interior to $S$. Thus we have shown that $\operatorname{int} S = A$.

Because $X\setminus\overline{A}\setminus\overline{B}$ is open, we have the inclusion $$ X\setminus\overline{A}\setminus\overline{B} \subset \overline{X\setminus\overline{A}\setminus\overline{B}} = \overline{D \cap (X\setminus\overline{A}\setminus\overline{B})} = \overline{D \setminus\overline{A}\setminus\overline{B}} $$ which implies that $X\setminus B \subset \overline{S}$, and because $B$ is open we have $$ \overline{D \setminus\overline{A}\setminus\overline{B}}\subset \overline{D\setminus \overline{B}} \subset \overline{D}\setminus \operatorname{int}B = X\setminus B $$ hence $\overline{S} \subset X\setminus B$. Thus we conclude that $\operatorname{ext} S = B$.

Niels J. Diepeveen
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  • I question the statements "Because the complement of $D\setminus\bar A\setminus\bar B$ is dense and the complement of $\bar B\setminus B$ is dense and open, their intersection is everywhere dense" and "$\overline{X\setminus\overline{A}\setminus\overline{B}} = \overline{D \cap (X\setminus\overline{A}\setminus\overline{B})}$" – Mario Carneiro Apr 14 '14 at 21:52
  • As far as I can see both are true because the intersection of a dense set and an open set is dense in the open set, and therefore the intersection has the same closure as the open set. – Niels J. Diepeveen Apr 14 '14 at 21:59
  • +1 Great answer! I'm happy that now there are three essentially different proofs of the same statement here. – Mario Carneiro Apr 14 '14 at 22:07
  • It seems like from the last statement you can only conclude $\ext S\supseteq B$. Also, you could probably skip the second part entirely if you take advantage of the obvious complementation symmetry. – Mario Carneiro Apr 14 '14 at 22:13
  • I'm not sure what you mean by your latest comment. I proved first that $X\setminus B \subset \overline{S}$, and then the reverse inclusion. Together that gives $B = X \setminus \overline{S} = \operatorname{ext}S$. I must admit that the symmetry was not that obvious to me. – Niels J. Diepeveen Apr 14 '14 at 22:23
  • Oh, I missed the other subset statement and though you were saying "$\bar S\subseteq X\setminus B$ implies $\ext S=B$". – Mario Carneiro Apr 14 '14 at 23:40
  • To make the complementation symmetry more manifest: $$S^c=B\cup(D^c\setminus\bar B\setminus\bar A)\cup(\bar A\setminus A\setminus\bar B),$$ and since $(D^c\setminus\bar B\setminus\bar A)\cup(\bar A\setminus A)$ has empty interior, so does $(D^c\setminus\bar B\setminus\bar A)\cup(\bar A\setminus A\setminus\bar B)$. Thus $\ntr S^c=\ext S=B$ by the same argument. – Mario Carneiro Apr 14 '14 at 23:57
  • I see what you mean. It is a pity that the whole thing cannot be made perfectly symmetrical because of the common boundary of $A$ and $B$. – Niels J. Diepeveen Apr 15 '14 at 00:56
  • Your approach is different from mine and Karl's only in that you use $\partial A$ in place of $\irr A$. If you use $\irr A$, it is actually completely symmetric since $\irr A$ and $\irr B$ are disjoint (whereas the boundaries may not be). – Mario Carneiro Apr 15 '14 at 02:50