prove that if X is a countable set of lines in the plane then the union of all lines in X can't cover the plane here's my try:
Let $X$ be a countable set of lines in the plane. the cardinality of the set of all lines in the plane with a slope between $0$ and $2\pi$ is $\aleph$ so there must be some line in the plane, $\Gamma$, with a slope $\alpha$ that is not in $X$. so now, every line in $X$ must intersect $\Gamma$ at only one point. because $X$ is a countable set then there is only a countable number of points on $\Gamma$ that the lines of $X$ cover so there is an uncountable set of points on $\Gamma$ that the lines of $X$ don't cover (because $|\Gamma |=\aleph$). Hence, the union of lines in $X$ don't cover the plane.
first of all, is this proof correct? and second, can anyone give me another proof, maybe an easier one?
 A: Yes it look's correct, basically what you construct is a piece of the space that has the same cardinality as $\mathbb(R)$ and that is only covered with countably number of points. On the same theme would be that each line touches the unit circle at at most two points, making those points where a line touches the unit circle countable, but the unit circle consist of uncountable points.
I would run into measure theory and realize that the integral of the characteristic function of a line (ie a function being one for every point the line crosses and zero every where else) is zero. Then so is the integral of the supremum of those function. That is the line misses almost every point in the plane.
A: Here is another proof: Let $(\ell_n)_{n\geq1}$ be the given family of lines. Enclose each $\ell_n$ between two Gaussians with area $2^{-n}$ between them. The total area of these bubbled double-spikes is then $1$, and their union $U$, which contains all the $\ell_n$, has area $\leq1$.  It follows that most of the plane is uncovered.
