# Uncountable subcollection of open sets with nonempty intersection

Given an uncountable collection of open intervals in $\mathbb{R}$, can we choose an uncountable subcollection which will have unempty intersection? I read that indeed, we could do that.

But on the other hand, I don't quite see why would this be impossible to have uncountably many open sets which all but for inifnitely many have an empty intersection. I know only countably many can be disjoint, but that doesn't mean that uncountably many are all NOT disjoint. It could happen there are these two in every subfamily of cardinality continuum that are disjoint, at least that's how I see it (bad intuition though).

Is there any constructive proof for this problem?

• Any open set contains a rational. A countable union of countable sets is countable... Jun 18, 2016 at 17:35
• Yeah, as I wrote I know that. Jun 18, 2016 at 17:36
• I mean -- empty intersection of some amount of sets is not equivalent to all of them being disjoint, that's why I don't see where to use the fact you stated. Jun 18, 2016 at 17:40
• As you wrote? Somewhere else maybe - neither of those two assertions appears in your post. If you know both those things then you should see the simple proof that follows - what's the problem? Jun 18, 2016 at 17:40
• I'd explain if you had a better attitude - "Yeah, as I wrote I know that"? Good luck Jun 18, 2016 at 17:41

Let $\mathcal{U} = \{A_\lambda \mid \lambda \in \Lambda\}$ be a collection of open subsets of $\mathbb{R}$ where $\Lambda$ is uncountable.
For a rational $x\in \mathbb{Q}$, let $\mathcal{U}_x\subset \mathcal{U}$ be the set of sets appearing in $\mathcal{U}$ which contain $x$. Note that for every $A_\lambda\in\mathcal{U}$, there exists an $x$ such that $A_{\lambda}\in \mathcal{U}_x$ because every open set contains a rational number. Hence $\bigcup_{x\in\mathbb{Q}}\mathcal{U}_x = \mathcal{U}$.
Now, suppose every $\mathcal{U}_x$ was only countable, then $\mathcal{U}$ is a countable union of countable sets and so is itself countable. This is a contradiction and so there must exist some $x$ for which $\mathcal{U}_x$ is uncountable, and further by construction $x \in \bigcap\mathcal{U}_x$ so the set is non-empty.