Classification of open subset of $\mathbb{R}^{3}$ [closed]

Question

There is a theorem which gives a classification of connected open sets of $\mathbb{R}^{2}$. Unfortunately, I don't remember the correct statement, but it looks like this

Theorem ? Let $U\subset\mathbb{R}^{2}=\mathbb{C}$ be a connected open subset, then $U$ is conformally equivalent to $\mathbb{R}^{2}-\cup_{n\in \mathbb{N}}I_{n}$, where $I_{n}$ is either empty or an interval of the form $[z,z+x_{n}]$ where $z$ is a complex number and $x_{n}$ is a positive real number.

conformally equivalent means that there exists a $C^{1}$-diffeomorphism $f:U\rightarrow \mathbb{R}^{2}-\cup_{n\in \mathbb{N}}I_{n}$ such that $df_{u}$ preserves angles for any $u\in U$.

1. I would like to know if the statement is correct ?
2. Does it admit a generalization in dimension 3. I mean a (EDIT Not necessarily conformal ?) classification of open connected subsets of $\mathbb{R}^{3}$

Answer 1

The theorem in $C$ is stated incorrectly. The exact result is the following.

Let $D$ be an open connected set in $C$. Then $D$ is conformally equivalent to $\overline{C}\backslash E$, where $E$ is a closed set whose connected components are horizontal segments of the form $[z_j,z_j+x_j],\; x_j\geq 0$. In particular, these components can be points. The sets of components can be uncountable.

For a proof, see Jenkins, Univalent functions and conformal mapping. Springer 1958. The case of finite connectivity is due to Grotsch, and in this case there is uniqueness.

As stated, in general the region with horizontal slits may not be unique. To make it unique, one has to impose an additional condition on the set $E$ stated in Jenkins' book.

In $R^3$ this makes no sense because there are too few conformal mappings. So your question about "classification of open sets in $R^3$ with not necessarily conformal maps" is meaningless. To ask about a classification you should define which regions are considered equivalent.

Answer 2

The result in two dimensions the OP is thinking of is Grotsch's uniformization by slit domains (I think Grotsch only did finitely many boundary components). Koebe has conjectured that any planar domain is conformally equivalent to a circle domain (the complement of a set of closed disks and points). As far as I know, this is still open in full generality, but He and Schramm showed that this is true when the number of boundary components is at most countable:

Fixed points, Koebe uniformization and circle packings
Zheng-Xu He, Oded Schramm*
Annals of Math, 1993.

As Adam Goucher states, it would be too much to expect a direct analogue in one dimension higher.

Answer 3

1. You're thinking of the Riemann mapping theorem, which is similar to what you said but with the additional constraint that the domain must be simply-connected.

2. There isn't an analogue in three dimensions, because conformal transformations are too constrained:

https://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)

In particular, the only conformal transformations are the obvious ones (Möbius maps).