# will there be pairwise disjoint open sets in $\mathbb{R}^2$

Suppose $S$ is a collection of pairwise disjoint open sets in $\mathbb{R}^2$

1. $S$ can not be finite

2. S can not be countably infinite.

3. S can not be uncountably infinite

4. S is empty.

1 is wrong I can take any finite no of disjoint open sets by housdorff property I can find right?

2 is also wrong I can take points from $\mathbb{N}\times \mathbb{N}$ and seperate them by those pairwise disjoint open sets so here S is countably infinite,

3 is also wrong I will do the same thing by putting $\mathbb{Q}^c\times \mathbb{Q}^c$

so 4 is right. Is my arguments are ok?

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$4$ can not be right. Consider $S$ consisting of two disjoint balls. – Norbert Jul 8 '12 at 23:20
No, $\mathbb{Q}^c \times \mathbb{Q}^c$ is dense. In fact, since 4 is wrong, I think the answer is 3. – user12014 Jul 8 '12 at 23:21
ah!! right thank you :) – Un Chien Andalou Jul 8 '12 at 23:23

Any disjoint collection of open sets in $\mathbb{R}^n$ is countable. You argue as follows. Let $\mathcal{K}$ be such a collection. For each $k\in K$, choose an element of $\mathbb{Q}^n$ lying in $k$. This is a 1-1 mapping from the elements of $K$ into a subset of the countable set $\mathbb{Q}^n$. Therefore $K$ is countable.
In your notation, $S$ can be countable or empty. It cannot be uncountable.