I have a problem with showing the fact which states that none of nonempty subset of plane is not contained in countable sum of segments. It seems to be trivial, but maybe my intuition is wrong. Anyway I have no idea how to write it down formally.
Before I will describe the second problem I have to remind that a set is very dense when $G\cap U$ is uncountable for every nonempty open $U\subseteq\mathbb{R}^2.$
With this knowledge I was trying to prove that there exists such borel set $G$ which is very dense and its complemention is very dense, too.
My attempt:
We assume that there exists borel set $G$ which is very dense, that is $G\cap U$ is uncountable for every nonempty open $U\subseteq\mathbb{R}^2.$ We have to show that $G^C$ is very dense, that is $G^C\cap U$ is uncountable for every nonempty open $U\subseteq\mathbb{R}^2.$ Knowing that none of nonempty subset of plane is not contained in countable sum of segments we have that $U$ must be uncountable, so $G^C\cap U$ is uncountable, too.
Is my reasoning correct? It seems to be too easy, so I think it's not correct.
I appreciate any help and advices.
Why not just let $\Bbb P=\Bbb R\setminus\Bbb Q$ and take $G=\Bbb P\times\Bbb P$? Every non-empty open set in $\Bbb R^2$ contains a set of the form $\{x\}\times(a,b)$ for some $x\in\Bbb P$, which clearly intersects $G$ in an uncountable set, and a set of the form $\{q\}\times(a,b)$ for some $q\in\Bbb Q$, which is uncountable and contained in $\Bbb R^2\setminus G$, so $G$ and $\Bbb R^2\setminus G$ are very dense in $\Bbb R^2$. And
$$\begin{align*} \Bbb P\times\Bbb P&=(\Bbb P\times\Bbb R)\cap(\Bbb R\times\Bbb P)\\ &=\bigcap_{q\in\Bbb Q}\big((\Bbb R\setminus\{q\})\times\Bbb R\big)\cap\bigcap_{q\in\Bbb Q}\big(\Bbb R\times(\Bbb R\setminus\{q\})\big) \end{align*}$$
is clearly a $G_\delta$ in $\Bbb R^2$, hence Borel.