Covering $R^2$ by open half planes Let $\{s_\iota, \iota\in I\}$ be any collection of vectors in $\mathbb{R}^2$ and let $U_\iota=\{\pi\in \mathbb{R}^2;\pi\cdot s_\iota >0\}$. Then, $U_\iota$ is either empty or an open half plane in $\mathbb{R}^2$ for any $\iota\in I$. Is it true, that we can find countably many $\iota_k\in I$, $k\in\mathbb{N}$ such that $\bigcup_{\iota\in I} U_\iota=\bigcup_{k=1}^\infty U_{\iota_k}$?
 A: It is not always true, but if they cover $\mathbb R^2 \setminus \{(0,0)\}$, then it is true there exists a finite subcover.
There is a quotient map $\mathbb R^2 \setminus \{(0,0)\} \to S^1$ by identifying vectors on a common half-line ($s \sim t $ if there is $\alpha > 0$ such that $s = \alpha.t$). It turns out that your open half-planes are compatible with $\sim$ (if $\pi$ and $\pi'$ are congruent, the sign of their product with $s_\iota$ is the same, so one of them is in the half plane if and only if the other is too), and they correspond to open intervals of $S^1$ (open semi-circles, even). But $S^1$ is compact, so if you have an open cover of $S^1$, you can extract from it a finite cover.
However, if they don't cover the whole circle, there may not be a finite covering.
For an example, pick $s_\iota = (1,1/\iota)$ for $\iota \in I = (0;1]$. Any finite subcover will be included in $\cup_{\iota> \epsilon} U_\iota$ for some $\epsilon > 0$, which is strictly smaller than $\cup_{\iota \in I} U_\iota$ : the vector $(\epsilon,-1)$ is in the latter but not in the former. 
A: I do not think the result as stated is true.
Counterexample(?):
Consider the set of outward facing unit tangent vectors on the unit circle $x^2+y^2=1$. Then the area that your open half planes($U_i$) cover is $\mathbb{R}\backslash D$ where $D$ denotes the closed unit disk. This clearly cannot be expressed as a finite union of open half planes. 
