Can someone provide a proof or counterexample to the following conjecture?

Conjecture: if $X$ is a topological space such that every open set $U \subseteq X$ satisfies $\mathrm{int}( \mathrm{c\ell}(U)) = U$, then every open set is closed (i.e. clopen).

Context: for any topological space $X$, its lattice of open sets forms a "model" of intuitionistic propositional logic. We interpret propositions as open sets $U$, $\land$ as intersection, $\lor$ as union, $\varnothing$ as $\bot$ and $X$ as $\top$. Implication is defined $U \to V = \mathrm{int}( U^\complement \cup V)$ - see this nLab article - and negation is defined $\lnot U = U \to \bot = \mathrm{ext}(U)$.

It is known that, over intuitionistic logic, the law of excluded middle $P \lor \lnot P$ and double negation elimination $\lnot \lnot P \to P$ are equivalent. Saying $U \cup \lnot U = X$ for an open set $U \subseteq X$ is equivalent to saying $U$ is closed (clopen). Meanwhile, saying that $\lnot \lnot U = U$ is equivalent to saying $\mathrm{int}( \mathrm{c\ell}(U)) = U$.

Thus, I was trying to prove the equivalence of $\mathsf{LEM}$ and $\mathsf{DNE}$ using this topological interpretation. The forward direction is easy enough, but I couldn't prove the reverse easily, which is the conjecture above.

  • 2
    $\begingroup$ If $X$ is $T_1$, it’s easy to prove that it must have the discrete topology. There are non-trivial non-$T_1$ examples, and it turns out (non-trivially) that your conjecture is true; see this question, especially the second answer. $\endgroup$ Mar 10 at 4:32
  • $\begingroup$ @BrianM.Scott: I'm surprised at the amount of work there - I wouldn't think it would be harder than proving double negation elimination implies LEM in intuitionistic logic (see edit for context). $\endgroup$ Mar 10 at 4:50
  • $\begingroup$ And how would one prove DNE implies LEM in intuitionistic logic? $\endgroup$ Mar 10 at 15:44
  • 1
    $\begingroup$ @JordanMitchellBarrett While I in no way mean to disconsider the effort behind and the merits of the article referenced in the link suggested by Brian M. Scott and mathematrucker, the paragraph relevant to establishing the implication all open subsets are regular therefore all open subsets are clopen is quite short and I dare say it could be presented even more concisely than it is formulated in the article. $\endgroup$
    – ΑΘΩ
    Mar 11 at 6:27

As mentioned in my comment above, it is perhaps worth posting here a very succinct argument which establishes the OP's conjecture.

Proposition. If all the open subsets of the space $(X, \mathscr{T})$ are regular then they are all also closed (the topology is clopen).

Proof. For every $x \in X$ let us write $Z_x=\mathring{\overline{\{x\}}} \in \mathscr{T}$. By dualising via complementation the given hypothesis of regularity -- according to which $\mathring{\overline{U}}=U$ for any open subset $U \in \mathscr{T}$ -- we gather that $\overline{\mathring{F}}=F$ for any closed subset $F \subseteq X$. Applying this to the particular case of the closed subset $\overline{\{x\}}$ we infer that $\overline{Z_x}=\overline{\{x\}}$.

In order to proceed with this argument it is also essential that we establish the nontrivial relation $x \in Z_x$. This we achieve by considering an arbitrary point $x \in X$ and assuming the contrary, $x \in X \setminus Z_x$. For ease of notation let us write $\overset{\smile}{M}\colon=X \setminus \overline{M}=\left(X \setminus M\right)^{\circ}$ for the exterior of arbitrary subset $M \subseteq X$. Our assumption then means that $x \in X \setminus Z_x=\overline{\left(X \setminus \{x\}\right)^{\circ}}=\overline{\overset{\smile}{\{x\}}}$, whence we infer that $\overline{\overset{\smile}{\{x\}}} \supseteq \overset{\smile}{\{x\}} \cup \{x\}$. Since it is in general (in any topological space) true that $\overline{M \cup \overset{\smile}{M}}=X$ for any subset $M \subseteq X$, we gather in particular that $\overline{\overset{\smile}{\{x\}}}\supseteq \overline{\overset{\smile}{\{x\}} \cup \{x\}}=X$, which simply means that $\overline{\overset{\smile}{\{x\}}}=X$. We now apply the hypothesis of regularity by considering interiors in this last relation in order to obtain $X=\mathring{\overline{\overset{\smile}{\{x\}}}}=\overset{\smile}{\{x\}}$. This is an obvious contradiction as $\overset{\smile}{\{x\}} \subseteq X \setminus \{x\} \subset X$ (where $\subset$ denotes strict inclusion). In other words, it is in general true that exteriors of nonempty subsets are proper subsets and that under our particular assumption of regularity the adherence of a proper open subset must remain a proper subset.

Considering now an arbitrary closed subset $F \subseteq X$ and an arbitrary point $x \in F$ we clearly have $\overline{\{x\}} \subseteq F$. This however leads to $Z_x \subseteq \overline{Z_x}=\overline{\{x\}} \subseteq F$ and as $x \in Z_x \in \mathscr{T}$ the point $x$ is established to be an interior point of $F$. By virtue of the arbitrariness of $x$ this entails $F \subseteq \mathring{F}$, which means that $F=\mathring{F}$ and hence any closed subset is equally open. The topology is thus seen to consist of clopen subsets. $\Box$

  • $\begingroup$ To clarify, you're using $\mathring{A}$ to denote the interior of $A$, and $\overline{A}$ to denote the closure? $\endgroup$ Mar 11 at 9:39
  • $\begingroup$ @JordanMitchellBarrett Yes, I am using the standard infix (if you will) notation for the fundamental topological operators. $\endgroup$
    – ΑΘΩ
    Mar 11 at 13:36
  • $\begingroup$ @HennoBrandsma I will add an explanatory detail to the answer. $\endgroup$
    – ΑΘΩ
    Mar 11 at 13:37
  • $\begingroup$ This argument seems simpler than the one given on page 13 of this paper. Nice. $\endgroup$ Mar 11 at 15:12
  • $\begingroup$ @HennoBrandsma Thank you for taking my answer into consideration! I must tell all of you however that I have fooled you and even myself -- for a brief while -- not realising that I had omitted to give any justification to an essential claim, namely that $x \in Z_x$ for any point $x \in X$. I remedied this by supplying an additional paragraph in the presence of which my answer delivers essentially the same argument as the one in the article. I would dare say that one cannot simplify it any further. $\endgroup$
    – ΑΘΩ
    Mar 12 at 3:06

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