Classify Open Sets in $\mathbb R^2$ 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. 
 A: 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.
A: 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 $x=\mathfrak c$.
