Prove/Disprove that :

$(i)$ Every open Set in $\mathbb R^p$ can be written as the union of countable number of disjoint open Sets.

$(ii)$ Every open subset of $\mathbb R^p$ is the union of a countable collection of closed sets.

I was able to look at some similar posts asking this problem; but one seemed to be using the other and vice versa and seem convoluted.

Unfortunately, I have no idea on how to move forward. Can anyone please help me in preparing a proof for both of these problems?

Thank you for your help.

  • $\begingroup$ An open set can be written as the union of itself. Maybe you require these disjoint open sets to satisfy a certain property. $\endgroup$ – Stefan Hamcke Sep 1 '14 at 20:47
  • $\begingroup$ But, it's not disjoint from itself? $\endgroup$ – MathMan Sep 1 '14 at 20:49
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    $\begingroup$ Maybe: (Pairwise) disjoint connected open sets? $\endgroup$ – Henno Brandsma Sep 1 '14 at 20:51
  • $\begingroup$ Only several sets $(U_i)_J$ can be disjoint, meaning that $U_i\cap U_j=\emptyset$ whenever $i\ne j$. $\endgroup$ – Stefan Hamcke Sep 1 '14 at 20:51
  • $\begingroup$ Consider for open $O \neq X$ and natural $n$: $F_n(O) = \{ x \in O: d(x, X \setminus O) \ge \frac{1}{n} \}$ $\endgroup$ – Henno Brandsma Sep 1 '14 at 20:53


  1. Every open set is union of balls with rational radius and rational center.
  2. Every open ball is a countable union of closed balls.

This gives (ii). For (i), given two points in your open set, say that they are equivalent iff there is a continuous path between them, completely contained in the open set. Argue that this is indeed an equivalence relation, and that its components (equivalence classes) are open. Now use that $\mathbb Q^n$ is dense in $\mathbb R^n$, so there can be no more than countably many equivalence classes.

  • $\begingroup$ Thank you but I am afraid that the text I am reading has not even introduced dense sets .. $\endgroup$ – MathMan Sep 1 '14 at 20:59
  • $\begingroup$ Surely you can show that every open set contains an element of $\mathbb Q^n$. $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 21:00
  • $\begingroup$ Uhm, so, to prove that every open set in $\mathbb R^p$ contains an element of $\mathbb Q^n$, we need to prove that every open set in $\mathbb R$ contains an element of $\mathbb Q$ ? This should be true since, any open interval contains infinite rational numbers. Am I correct? $\endgroup$ – MathMan Sep 1 '14 at 21:15
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    $\begingroup$ Sure, that's one way. And the stronger property you indicate is indeed true. $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 21:16
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    $\begingroup$ If you have a path from $ x $ to $ y $ in your open set, and $ z $ is in a ball centered at $ y $ and contained in the set, then there is a path from $ x $ to $ z $ in the set: First go to $ y $, and then go from $ y $ to $ z $ in a straight line. This shows the component that contains $ x $ is open. $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 22:59

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