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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
Paul Frost
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joseparreiras
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    You left out the word "disjoint," I think. If they don't have have to be disjoint, this is pretty easy, I think. – saulspatz Jul 13 '18 at 17:27
  • I'm neither a topologist nor a mathematician but from a layman's point of view I think there's no uncountable set of isolated points in $\Bbb R^n$ so I think having a think about the vertices of your hypercubes in this context might give you some insight into this problem. – it's a hire car baby Jul 13 '18 at 17:47
  • In fact I think you'd need a continuum to form the boundary of any disc or circle and that's going to require an uncountable number of vertices belonging to some set of hypercubes. Since each hypercube is finite-dimensional, uncountably many hypercubes will be required. – it's a hire car baby Jul 13 '18 at 17:56
  • Yes, I think you are right. Consider an open disk. For each point $x$ on the bounding circle, there must be a hypercube whose closure has $x$ as a vertex. Only $2^n$ disjoint hypercubes can share a vertex, so we must have uncountably many hypercubes. I made a comment earlier saying I didn't believe you could cover a disk, but I couldn't come up with a proof, so I deleted the comment. Good work! – saulspatz Jul 13 '18 at 18:03

2 Answers2

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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.

Lukas
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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 $U$.

Paul Frost
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