# Can one cover the plane with less than continuum of lines?

I will be working in ZFC, but I am not assuming the Continuum Hypothesis (or Martin's Axiom). I know that it is consistent with ZFC that one can cover the real line with less than continuum of meager sets. My questions are about coverings by sets which are "nicer" than just meager (for instance, their proper intersections are finite). By "plane" below I mean the Euclidean plane and by "line" I mean an affine line. Of course, by Baire's theorem, Questions 1 and 2 both have negative answer if the sets $A$ are countable.

Question 1. Let $A$ be a set so that ${\aleph }_0=|{\mathbb N}|<|A| < 2^{{\aleph }_0} =|{\mathbb R}|$. Can one cover the plane by a family of lines $L_\alpha$, $\alpha\in A$?

In case I am missing some simple geometric consideration,

Question 2. For $A$ as above, can one cover the plane by a family of (real, irreducible) algebraic curves $X_\alpha$, $\alpha\in A$?

It feels as if these questions are related to the list compiled in this post by Joel David Hamkins, but I cannot quite figure out how.

• The paper "Set Theory of the Plane" by Miller (see math.wisc.edu/~miller/old/m873-05/setplane.pdf) contains many results concerning the covering of the plane. Oct 22, 2013 at 6:53
• For your question 1, the answer is no. If $\{X_{\alpha}\}_{\alpha\in A}$ cover a circle, then $|A|=2^{\aleph_0}$ (just because each line can cover at most two points on the circle).
– 喻 良
Oct 22, 2013 at 6:53
• @LiangYu Wouldn't your nice argument also apply to a curve like the graph of $\exp$, so that the answer to 2. is also 'no'? Oct 22, 2013 at 6:58
• @ToddTrimble, I think you are right.
– 喻 良
Oct 22, 2013 at 7:00
• @LiangYu: yes, this is indeed a very nice and easy geometric argument. The same circle argument works for algebraic curves since once can take the continuum of concentric circles. Would you like to write your argument as an answer so that I can close the question? Oct 22, 2013 at 7:06

For your question 1, the answer is no. If $\{X_{\alpha}\}_{\alpha\in A}$ cover a circle, then $|A|=2^{\aleph_0}$ (just because each line can cover at most two points of the circle).
• You could even use a straight line instead of a circle. If $Y$ is any line not appearing among the $X_\alpha$, then each $X_\alpha$ meets $Y$ in (at most) one point, so the $X_\alpha$ do not cover $Y$. Oct 22, 2013 at 7:14