Prove that $\mathbb{R}^2$ cannot be a subset of the union of a countable collection of lines in $\mathbb{R}^2$

Question

I'm not really sure about this. I think the proof relies on the theorem: If $U$ is a subset of $A$, and $U$ is uncountable, then $A$ is uncountable.

The problem statement says a collection of lines is countable. $\mathbb{R}^2$ is not countable so $\mathbb{R}^2$ cant be a subset of a countable collection of lines in $\mathbb{R}^2$.

Answer 1

Let $\;L=\{\ell_i\}_{i\in I}\;,\;\;|I|\le \aleph_0\;$ , be a countable collection of lines in the plane, and let also

$$\;Y:=\left\{y\in\Bbb R\;:\; \exists i\in I\;\text{such that}\; (0,y)\in\ell_i\right\}\;$$

If $\;\;\bigcup\limits_{i\in I}\ell_i=\Bbb R^2\;$ then it must be that $\;Y=\mathcal Y:=\{(0,y)\;:\;\;y\in\Bbb R\}=\;$ the $\;y -$ axis , but this can't be since $\;|Y|\le\aleph_0\;$ whereas $\;|\mathcal Y|=2^{\aleph_0}=\mathfrak c :=\;$ the continuum cardinal.

Answer 2

Consider the unit circle. Each line has at most two points in common with it, but the circle has uncountably many points.

Answer 3

I suppose that one can also use the following argument:

Let if possible, $ \mathbb{R}^2 = \cup_{i \in I} L_i$ where $I$ is some countable set and $L_i$ denotes a line in the plane. Now compute the Lebesgue measure on both sides. $$ m_2(\mathbb{R}^2) \leq \sum_{i} m_2(L_i) =0. $$ A contradiction.