Is any open subset of $R^n$ a union of countable hypercubes?

Question

I'm trying to extend the proof seen here that any open subset of $\mathbb{R}$ can be written as a countable union of disjoint open intervals. Apparently it seems like an analogous proof for $\mathbb{R}^n$. However I can't think of any equivalence relation that would allow me to proceed with my sketch.

The sketch is like:

1. Define an equivalence relation $\sim$ on $\mathbb{R}^n$
2. Show that the equivalence class $\sim(x)$ is non-empty and open
3. Prove that $\sim(x)\bigcap \sim(y) \neq \emptyset \Rightarrow \sim(x)=\sim(y)$
4. Demonstrate that $\sim(x)$ is an hypercube $\prod_{i=1}^N(a_i,b_i)$
5. Show that the set of equivalence classes D is countable
6. Deduce that $\bigcup D$ is equal to our open subset

Answer 1

It seems that you suppose the hypercubes to be open. In this case your statement is definetely not true. If n > 1 take the open ball of radius 1 $B_1(0)$ around the origin (given by the standard euclidean metric). This ball is definitely not a hypercube. So if there were disjoint open hypercubes $W_n$, such that $B_1(0) = \cup W_n$, there would be at least two. Since $B_1(0)$ is (path) connected, this is a contradiction.

If you don't suppose the cubes to be open, you can look here for an answer Every open subset $O$ of $\Bbb R^d,d \geq 1$, can be written as a countable union of almost disjoint closed cubes.

Answer 2

Lukas has explained why it is impossible for $n > 1$. But it is true for $n = 1$, see Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs].

So what is the difference between $n > 1$ and $n = 1$?

This is very simple. The components (= maximal connected subsets) of an open $U \subset \mathbb{R}$ are open intervals. But $U$ is the union of its components which are disjoint open intervals. For $n > 1$ the components of an open $U \subset \mathbb{R}^n$ are in general no open hypercubes. In fact, no connected open $U \subset \mathbb{R}^n$ which is not an open hypercube can be the disjoint union of open hypercubes contained in $U$.