Can a countable set of parabolas cover the unit square in the plane? Can a countable set of parabolas cover the unit square in the plane?  My intuition tells me the answer is no, since it can't be covered by countably many horizontal lines (by the uncountability of $[0, 1]$).  Help would be appreciated.
 A: Like other people have said, you can use measure theory. Each parabola has measure zero, and a countable union of measure zero sets has measure zero.
However, you can also show this directly. If the parabolas cover the plane, then they also cover the intersection with the y-axis. But each parabola cuts the y-axis only once (or at most twice if you allow rotated parabolas), so the parabolas can only cover a countable subset of the y-axis.
A: An approach: Let the parabolas be $P_1,P_2,\dots$. Then by thickening each parabola slightly, we can make the area covered by the thickened parabola $P_i^\ast$ less than $\frac{\epsilon}{2^i}$, where $\epsilon$ is any preassigned positive number, say $\epsilon=1/2$.  Then the total area covered by the thickened parabolas is $\lt 1$.  
Remark: There are various ways to say that a set is "small." Cardinality is one such way. Measure is another.  
A: There is not a countable number of horizontal or vertical lines that can cover the unit square hence the following wouldnt work for parabolas either.
