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In $\mathbb R$, we know that connected open set is $(0,1)$ under homeomorphism. I am wondering what is the situation in $\mathbb R^2$.

From $\mathbb R^2-\text{pt}\simeq S^1$, we will have two open sets $\mathbb R^2$ and $\mathbb R^2-\text{pt}$. Similarly, $\mathbb R^2-\text{pt1},\dots,\text{ptn}$,$n\geq0$ are not homeomorphic.

So how many open sets are in $\mathbb R^2$? Any advice is helpful. Thank you.

gaoxinge
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  • Special case: if $A$ and $B$ are two Cantor sets, then are $\mathbb R^2 \setminus A$ and $\mathbb R^2 \setminus B$ homeomorphic? – GEdgar Jan 05 '16 at 13:17
  • @GEdgar Yes? It seems like the answer should be yes - I sort of more or less see how to prove it I think. But when you pose the question it seems like the answer must be no, since yes would be boring. ??? – David C. Ullrich Jan 05 '16 at 14:05
  • @David: I too have the feeling that indeed they are even biholomorphic if you think of $\mathbb R^2$ as $\mathbb C$ . What is your idea for a proof? – Georges Elencwajg Jan 05 '16 at 14:25
  • @GeorgesElencwajg Biholomorphic? Isn't it true that just the complement of four points need not be biholomorphic to the complement of four other points? Anyway. for homeomorphic: Write $A$ as the disjoint union of two Cantor sets $A_1$ and $A_2$; similarly for $B$. Take two "nice" disjoint (topological) disks $D_1$ and $D_2$ containing $A_1$ and $A_2$, respectively; similarly enclose $B_j$ in disks $E_j$. Now the complement of $D_1\cup D_2$ is homeomorphic to the complement of $E_1\cup E_2$. [continued] – David C. Ullrich Jan 05 '16 at 14:50
  • @GeorgesElencwajg Repeat: $D_1\setminus(D_{1,1}\cup D_{1,2})$ is homeomorphic to $E_1\setminus (E_{1,1}\cup E_{1,2})$; adjust that homeomorphism to agree with the first one on the boundary of $D_1$, etc. – David C. Ullrich Jan 05 '16 at 14:51
  • @David: why are the complements of four points not biholomorphic? One cannot extend the biholomorphism to the riemann sphere since it might have essential singularities at the missing points. Could you please explain? – Georges Elencwajg Jan 05 '16 at 16:40
  • @GeorgesElencwajg Can't have essential singularities at the missing points. A function is far from one to one near an essential singularity. Singularities must be removable (allowing infinity as a value of course)... – David C. Ullrich Jan 05 '16 at 17:00
  • @David: Of course! Thanks. For my punishment I shall not delete my stupid comments and will be shamed publicly here for eternity :-) – Georges Elencwajg Jan 05 '16 at 18:14

2 Answers2

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I dont think you can ever classified this, for an example $\mathbb R^2$ - cantor set is open, and infact $\mathbb{R^2}$- closed set is open, and closed set in $\mathbb R^2$ could be anything, very bad.

Atmost you can say that it is union of open ball of various radius.

Anubhav Mukherjee
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    Yes, of course the complement of any closed set is open, and this shows that there's not going to be any simple classification. But if we're being picky we note that the OP asked about connected open sets; the complement of a close set need not be connected. – David C. Ullrich Jan 05 '16 at 14:03
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The cardinality $x$ of the collection of open subsets in $\mathbb R^2$ is the continuum: $x=\mathfrak c=2^{\aleph_0}$.

a) $x\geq \mathfrak c$
Indeed just the open discs are already in number $\mathfrak c$ .

b) $x\leq \mathfrak c$
Let $\mathcal R$ be the denumerable collection of open discs with rational radius and center in $\mathbb Q^2$, which we will call rational discs.
Let $\mathcal T$ denote the set of open subsets of $\mathbb R^2$, whose cardinality $x=\operatorname {card} \mathcal T$ we are investigating, and let $\mathcal P(\mathcal R)$ denote the set of subsets of $\mathcal R$.
We have a map $U:\mathcal P(\mathcal R)\to \mathcal T$ associating to a subset $\mathcal S\subset \mathcal R$ of rational discs its union $U(\mathcal S)=\bigcup_{D\in S}D\subset \mathbb R^2$, an open subset of $\mathbb R^2$.
The map $U$ is surjective and since $\mathcal P(\mathcal R)$ has cardinality $\mathfrak c$ we get the required inequality $ \mathfrak c \geq x=\operatorname {card} \mathcal T$ .

c) From a) and b) we conclude (using the Cantor-Schroeder-Bernstein theorem ) that indeed $x=\mathfrak c$.

Georges Elencwajg
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  • I'm not sure what exactly a classification of the open subsets of $\mathbb R^2$ would be, but I am quite optimistic that a lot about them is known. In particular because they can be endowed with the structure of a Riemann surface and these have been extensively studied by analysts for more than 150 years. – Georges Elencwajg Jan 05 '16 at 14:20
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    Surely the question was really about the number of homeomorphism classes of open sets? Not that I suspect that changes the answer... – David C. Ullrich Jan 05 '16 at 14:52
  • @David: well, I'm not sure about "Surely" but I agree with your suspicion... – Georges Elencwajg Jan 05 '16 at 16:43