0

I am trying to either prove or disprove the assertion that every open set in $\mathbb{R}^2$ is a countable union of open balls.

I am leaning toward "no" because of the use of the word "countable." If $A \subset \mathbb{R}^2$ is an open set, then for every $\vec{v} \in A$, I can choose an open ball $B_{r_a} (\vec{v}) \subset A$ by the definition of "opennesss" in the standard topology, then $A = \bigcup\limits_{r_a} B_{r_a} (v)$. But if $A$ is uncountable, there are uncountably many open balls. If the set is compact, I can limit this to finitely many (i.e., countably many) open balls, but there's no guarantee I can do this, and the only compact, open sets in $\mathbb{R}^2$ are the empty set and $A$ itself.

I'm not sure if my intuition is correct. With that said, I am struggling to find a counterexample or to establish that it works, especially since I'm not in the point of Munkres's book that compactness is introduced. Is there something I'm missing?

Cardinality
  • 1,097
  • 3
  • 7
  • I’m not sure. In $\Bbb R$ you prove the answer is: “yes”, by saying each interval contains a rational and thereby finding an injection into $\Bbb Q$, implying the interval set is rational – FShrike Sep 14 '22 at 07:12
  • 1
    Perhaps try to use the fact that $\mathbb{Q}^2$ is countable and dense in $\mathbb{R}^2$? – Lorago Sep 14 '22 at 07:14
  • 1
    @coffeemath The property for $\mathbb{R}$ is basically local connectedness of $\mathbb{R}$ and the fact that connected open sets are intervals. If you take any connected open set in $\mathbb{R}^2$ that is not a ball, then you can't write it as a disjoint union of open balls (also because of connectedness). – Jakobian Sep 16 '22 at 10:34

0 Answers0