Covering $\mathbb R^2$ with function graphs

Suppose we have a countable family of function graphs (each function is $\mathbb R\to\mathbb R$, not necessary continuous). Obviously, they cannot cover the whole plane $\mathbb R^2$, because they cannot even cover every of uncountably many points on a single vertical line.

But suppose now we are allowed to rotate each graph from the family by arbitrary angle around an arbitrary point of the plane (the total number of graphs is still countable). Is it possible to cover the whole plane $\mathbb R^2$ in this case?

Solution 1:

You can construct just one function $f: R \rightarrow R$ such that countably many rotations of the graph of $f$ cover the plane. This is due to R. O. Davies. See theorem 31 and the remark below it in Arnie's notes which has many other interesting results (without proofs) about set theory of plane.

Solution 2:

The idea below does not work: I wrote "empty interior" when I meant "nowhere dense". But graphs do not need to be nowhere dense.

We cannot cover the plane this way. The reason is that the graph of any $f:\mathbb R\to\mathbb R$ has empty interior (as a subset of the plane), since every vertical line meets $f$ in at most one point. Having empty interior is not affected by rotations. The Baire category theorem gives us that $\mathbb R^2$ is not a countable union of sets with empty interior, and this concludes the proof.