Let $X$ be a topological space and let $\Gamma \mathcal O(X)$ be it's boolean algebra of clopen subsets. For compact totally disconnected space, show that $\Gamma \mathcal O(X)$ is complete (as a lattice) iff the closure of every open set is open.

Here is my proof, but I am not sure if it is valid, so I post it.

First I want to determine sup and inf in the lattice. For the open sets, the supremum is just set union, but the infimum is the interior of the intersection. By duality for closed sets, the supremum is the closure of set union, and the infimum is just intersection. So, in the lattice of all clopen sets, if $U_i$ are clopen ($I$ arbitrary index set), then $$ \inf_{i \in I} \{ U_i \} = \overline{ \bigcup_{i \in I} U_i } \quad \sup_{i \in I} \{ U_i \} = \left( \bigcap_{i \in I} U_i \right)^{\circ} $$ Note: Here is the first step where I am unsure, I derived the inf/sup by considering the inf/sup's in the lattices from which $\Gamma \mathcal O(X)$ is just the intersection, could that be done, maybe the inf/sup is totally different in the sublattice of clopen sets. The above mentioned inf/sup operations seem evident to me, but I have no idea how to make it mathematically sound.

Now let $\Gamma \mathcal O(X)$ be complete, that means inf/sup belongs to the set (or class). If $O$ is open, then it could be written as $$ O = \bigcup B_i $$ with $B_i$ clopen, because the space is totally disconnected it has a basis consisting of clopen sets. Now $\overline{O} = \overline{ \bigcup B_i }$ is just the Infimum of the sets $B_i$ in $\Gamma \mathcal O(X)$, so it belong to the class of clopen sets, and such must be open.

Note: I never used compactness of $X$, isn't it needed...

For the other direction, let $X$ have the property that the closure of every open set is open, I want to show that $\Gamma \mathcal O(X)$ as a lattice is complete. Let $O_i$ be clopen, then $\bigcup O_i$ is open, and so the closure $\overline{\bigcup O_i}$ must be open, and so it is clopen and belongs to $\Gamma \mathcal O(X)$, and it equals the Infimum. The completeness under one of inf or sup is enough to establish completeness under both, so the statement is established. Q.E.D

Note: Didn't needed compactness again, makes me wonder...


First of all, you interchanged $\inf$ and $\sup$ in your definitions, which should say that if $\mathscr{U}\subseteq\Gamma\mathcal{O}(X)$, then


It’s a reasonable idea, but it doesn’t work for arbitrary $X$, because the sets $\operatorname{cl}\bigcup\mathscr{U}$ $\operatorname{int}\bigcap\mathscr{U}$ and need not be clopen.

Example. Let $X$ be the space of irrationals with the usual topology. Fix $x\in X$, and let $\langle q_n:n\in\omega\rangle$ be a strictly decreasing sequence of rationals converging in $\Bbb R$ to $x$. For $n\in\omega$ let $$U_n=\{x\in X:q_{n+1}<x<q_n\}\;;$$ each $U_n$ is clopen in $X$. Let $G=\bigcup_{n\in\omega}U_n$; then $\operatorname{cl}_XG=G\cup\{x\}$, which is not open (and therefore not clopen) in $X$.

Of course the irrationals aren’t compact, though they are zero-dimensional and hence totally disconnected.

You also seem not quite to realize just what you have to prove. Let’s start with the easy direction: suppose that $X$ is a compact, totally disconnected space in which every open set has open closure. Let $\mathscr{U}\subseteq\Gamma\mathcal{O}(X)$; we need to show that $\mathscr{U}$ has a supremum in $\Gamma\mathcal{O}(X)$.

Let $G=\operatorname{cl}\bigcup\mathscr{U}$; then $G$ is clopen, and $U\subseteq G$ for each $U\in\mathscr{U}$, so $G$ is an upper bound for $\mathscr{U}$ in $\Gamma\mathcal{O}(X)$; to complete the proof that $G=\sup\mathscr{U}$, we must show that if $H\in\Gamma\mathcal{O}(X)$, and $U\subseteq H$ for all $U\in\mathscr{U}$, then $G\subseteq H$. If not, there is a point $x\in G\setminus H$. $H\supseteq\bigcup\mathscr{U}$, so $G\setminus H$ is an open nbhd of $x$ disjoint from $\bigcup\mathscr{U}$, and therefore $x\notin\operatorname{cl}\bigcup\mathscr{U}=G$, which is absurd.

Note that this direction did not require $X$ to be compact or totally disconnected.

Now suppose that $\Gamma\mathcal{O}(X)$ is complete, and let $V$ be a non-empty open set in $X$. Because $X$ is compact and totally disconnected, the clopen sets form a base for the topology, and there is a $\mathscr{U}\subseteq\Gamma\mathcal{O}(X)$ such that $V=\bigcup\mathscr{U}$. $\Gamma\mathcal{O}(X)$ is complete, so let $G=\sup\mathscr{U}$; clearly $U\subseteq G$ for each $U\in\mathscr{U}$, so $V\subseteq G$. We also know that $G$ is clopen in $X$, and that if $H$ is any clopen set in $X$ such that $U\subseteq H$, then $G\subseteq H$. To complete the proof, you must show that this somehow implies that $\operatorname{cl}V$ is open; the natural way to proceed is to try to show that $\operatorname{cl}V=G$. Since $G$ is closed, you do know that $\operatorname{cl}V\subseteq G$, so you want to try to get a contradiction by supposing that $\operatorname{cl}V\subsetneqq G$. I’ll leave that to you, but feel free to leave a comment if you get stuck.

  • $\begingroup$ Think I got it. Suppose that there exists $x \in G$ with $x \notin cl(V)$ (guess you mean $V$, not $U$). Then there exists an open set $W$ sucht that $x \in W$, $W \subseteq G$ and $W \cap cl(V) = \emptyset$. Using again that we have a basis consisting of clopen sets, $W$ contains an clopen neighborhood of $x$, so we can suppose $W$ is clopen from the start (if not just choose such an clopen nbhd inside $W$). Then $G \setminus W$ is clopen too, and because $W$ and $cl(V)$ are disjoint, still $V \subseteq cl(V) \subseteq G\setminus W$, but $G\setminus W$ strictly smaller as $G$, constradiction! $\endgroup$
    – StefanH
    Sep 17 '13 at 22:37
  • $\begingroup$ Are you sure I proofed the wrong direction... to proof $A\subseteq B$ you can suppose there exists $x\in A\setminus B$ and derive a contradiction... I supposed there exists $x \in G\setminus cl(V)$ and got a contradiction, proofing $G\subseteq cl(V)$... $\endgroup$
    – StefanH
    Sep 18 '13 at 11:26
  • $\begingroup$ @Stefan: Oops! You’re right; I must have been asleep when I read your comment. I’m going to delete my earlier comment, since it’s nonsense; the proof in your first comment is fine. $\endgroup$ Sep 18 '13 at 19:24

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