TITLE: Does such a family of sets exist?
QUESTION [2 upvotes]: Here is a question on set theory:
Does there exist a family $S$ of closed discs in $\mathbb{R^2}$ having positive radii such that each of these discs has at most one point in common and $\mathbb{R^2}- {\cup D}$ is countable.
I have no idea how to approach this question.
Please tell me a hint how to start this one

REPLY [3 votes]: It's a well known theorem (of Sierpiński I believe) that the real line does not admit a nontrivial partition into countably many closed sets; see my answer to this question.
Now assume for a contradiction that the plane is the union of a (necessarily countable) collection of closed discs (of positive radius) with disjoint interiors, plus countably many single points.
Let $L$ be a line in the plane which does not pass through any of the countably many points where two of those closed discs touch. The intersection of each of the given discs with $L$ is either a closed interval or a single point. Thus $L$ is the union of a countable disjoint collection of closed intervals and singletons, contradicting the theorem mentioned above.
Remark. If $S$ is a family of closed discs of positive radius with disjoint interiors in $\mathbb R^2$, then $\mathbb R^2\setminus\bigcup S$ is an uncountable $G_\delta$ set, so it has cardinality $2^{\aleph_0}$.
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