# The union of family of open disjoint non empty subsets is dense.

The statement I'm gonna write down here is a proposition I conjectured in order to prove a detail in Hartogs theorem; thus I won't write all the context, otherwise it would become the longest post ever in MSE! On the other side this statement is quite general, thus it shouldn't be a problem.

Consider the unitary open disk $\Delta$ in $\Bbb C$. $\Delta$ is a topological space, with topology induced by $\Bbb C$. We'll work in this topological space $\Delta$, thus, "open", "the closure of" etc are all related to this topological space.

Let $\Lambda$ be a family of index more than countable. Let $\{E_{\lambda}\}_{\lambda\in\Lambda}$ be a family of disjoint open non empty subsets of $\Delta$. Can we say that $$\operatorname{cl}_{\Delta}\left(\bigcup_{\lambda\in\Lambda}E_{\lambda}\right)=\Delta\;\;\;\;?$$ I'd say yes, and it seems natural, but honestly I don't know neither where to start to prove it.

• There is no such family of disjoint open sets. (Hence the answer to your question is actually yes, but only because if $P$ is false then anything whatever follows from $P$...) Jul 13, 2015 at 18:56
– Joe
Jul 13, 2015 at 18:58
• ??? More explicit about what exactly? The fact that any collection of disjoint nonempty open sets is countable, or that silliness about one falsehood implying anything? The second isn't that important, and the first is clear beacuse $\Delta$ is separable... Jul 13, 2015 at 19:00
• What is so special about it being the unit disk? Could not all these open sets belong to the disk of radius 1/2? Jul 13, 2015 at 19:04
• @DavidP: It's clear that exists such a family but my question is "does my argument work for every such a family?".
– Joe
Jul 13, 2015 at 19:32

To see this, let $\Lambda$ be an uncountable index set and suppose that $(E_{\lambda})_{\lambda\in\Lambda}$ is a family of disjoint, non-empty open subsets of $\Delta$. Also, let $$\mathbb Q_{\mathbb C}\equiv\{a+bi\,|\,\text{a and b are rational numbers}\}$$ be the set of numbers on the complex plane with “rational grid points” (that is, both the real and the imaginary parts are rational). It is not difficult to show that $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$ and also that $\mathbb Q_{\mathbb C}$ is countable.
For each $\lambda\in\Lambda$, pick $x_{\lambda}\in E_{\lambda}$ (you need the axiom of choice here). Since $E_{\lambda}$ is open and $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$, one can pick $q_{\lambda}\in\mathbb Q_{\mathbb C}$ such that $q_{\lambda}\in E_{\lambda}$. Because of disjointness, all the $(q_{\lambda})_{\lambda\in\Lambda}$ are distinct, so that the function $\lambda\mapsto q_{\lambda}$, mapping from $\Lambda$ to $\mathbb Q_{\mathbb C}$ is injective. This is implies that the set $\mathbb Q_{\mathbb C}$ is uncountable, which is impossible.