Classification of Open Subsets of $\mathbb{R}^2$

Question

Is there some nice-ish characterization of open subsets of $\mathbb{R}^2$? For example, open subsets of $\mathbb{R}$ can be represented as the countable union of disjoint open intervals, so they can be mapped to sequences $(a_1,b_1,a_2,b_2,...)$ in $\mathbb{R}\cup\{\pm\infty\}$. It seems like it would be far more complicated for the 2 dimensional case, but is there some other way to describe each open set in $\mathbb{R}^2$ using a sequence of numbers as in the 1 dimensional case?

Answer 1

Every open set in $\Bbb R^2$ can be written as a union of open balls with rational centers and rational radii $U =\bigcup_{n=0}^\infty B_{r_n}(x_n,y_n)$, so you can map it to $(x_0,y_0,r_0,x_1,y_1,r_1,x_2,y_2,r_2,\ldots)$.

Answer 2

As jjagmath said, since $\mathbb{R}^2$ is separable (by $\mathbb{Q}^2)$, you can express $U$ as a countable union of balls of center and radius that are rational. The only issue is that you will lose the unicity of this union.

In $\mathbb{R}$, if you have an open set $U$, you can find $J \subset \mathbb{Q}$ such that $$U = \bigsqcup_{q \in J} ]a_q,b_q[$$ With $p<q \implies b_p < a_q$.

This decomposition is unique up to increasing bijection, and the sets $ \{a_q, q \in J\}$ and $ \{b_q, q \in J\}$ are both unique and describe $U$.

You unfortunately can't get this with higher dimensional spaces since the topology doesn't match any complete order. You could get an equivalent by looking to the connected parts of $U$ (which are also path-connected) but this will never be as strong as the property in dimension $1$.