Open subsets of $\mathbf{R}^n$

Question

In a math course, I read that the open subsets of $\mathbf{R}^d$ were the countable unions of hypercubes (i.e. the sets $A$ such that there exist elements $a_i$ and $b_i$ in $\bar{\mathbf{R}}$ such that $A = ]a_1 , b_1[ \times ... \times ]a_d , b_d[ $).

In order to prove this, I showed we can assume that the open subset is path connected.

Does anyone know the end of the proof ?

Answer 1

I will prove a more general result:

Proposition: If $E$ is a metric space endowed with the norm $||\cdot ||$, and $U$ is an open set containing a countable and dense subset $\{a_n\ |\ n \in \mathbb{N}\}$, then $U$ can be written as a countable union of balls centered at the $a_n$.

Proof: for all $n \in \mathbb{N}$, denote $r_n = \sup \{ r \in \mathbb{R}_+\ |\ B(a_n,r) \subset U\}$ where $B(a,r)$ is an open ball.

$r_n$ is well defined and $r_n \in ]0,+\infty]$ because $U$ is open. Then we obviously have $\bigcup \limits_{n \ge 1} B(a_n,r_n) \subset U$. Conversely, take $x \in U$. As $U$ is open, there exists some $\varepsilon > 0$ such that $B(x,\varepsilon) \subset U$. Using the density, there exists $m \in \mathbb{N}$ such that $a_m \in B(x,\frac{\varepsilon}{2})$. For all $y \in E$ such that $|y-a_m| < \frac{\varepsilon}{2}$, the triangular inequality yields $|y-x| \le |y-a_m|+|a_m-x| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}$ so $y \in B(x,\varepsilon) \subset U$. Thus $r_m \ge \frac{\varepsilon}{2}$, from which we conclude that $x \in B(a_m,r_m)$. Hence $U \subset \bigcup \limits_{n \ge 1} B(a_n,r_n)$.

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Coming back to your question, you simply have to consider $E = \mathbb{R}^n$ with the norm $||(x_1,...,x_n)||_{\infty} = \max(|x_1|,...,|x_n|)$. Every (open) set in E does have a countable dense subset, so the result of the proposition holds.

Answer 2

You do not need path-connectedness. The definition open sets says open sets are union of open balls centered at each point of an open set. More concretely, for each $x\in U$ there is $r_x>0$ such that $$U = \bigcup_{x\in U} B(x,r_x).$$ (Here $B(x,r)$ is the open ball centered at with radius $r$.) Let $I_x$ be an open hypercube such that $x\in I_x \subseteq B(x,r_x)$, then $U$ is the union of all $I_x$, so $U$ is a union of open hypercubes.

However, it does not guarantee you can express $U$ a countable union of hypercubes. However we can observe that we can take each endpoint of $I_x$ rational numbers. As there are only countably many hypercubes of rational endpoints, $U$ is a countable union of hypercubes.