# In $\mathbb R^p$:Every open subset is the union of a countable collection of closed sets & every open set is the countable union of disjoint open sets

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?

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

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.
• Surely you can show that every open set contains an element of $\mathbb Q^n$. – Andrés E. Caicedo Sep 1 '14 at 21:00
• 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? – MathMan Sep 1 '14 at 21:15
• 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. – Andrés E. Caicedo Sep 1 '14 at 22:59