2 added 169 characters in body

(Too long to be a comment, but:) It is impossible to, say, make the union of $C$ either the complement of an open concave region (trivial) or the complement of an open circle (there's only one possible choice of line for each point on the boundary of the circle, and then any other line has to be parallel to one of these tangent lines).

On the other hand, I think it should be possible - certainly with $AC+CH$ - to make the union of elements of $C$ be, say, the complement of an open triangle $T$. Work in $\omega_1$-many stages, where at each stage $\eta$ you have countably many lines already drawn, and you're trying to get a new line which goes through the point $(x_\eta, y_\eta)$ (assuming this point isn't in $T$ and doesn't lie on an existing line already) and is "legal" (i.e., doesn't intersect $T$, doesn't intersect an existing point of intersection between already-chosen lines, and isn't parallel to any given line). Initialize the construction by starting with the three lines containing the three edges of the triangle; then it seems that, by our assumption that $CH$ holds, we can continue this construction to hit every point not in $T$. [EDIT: As Emil points out below, CH can be dropped by considering instead the cardinality of the set of points to be avoided at each given stage. So only $AC$ is used.]

A key step in this proof is that the set of lines through a given point not in the closure of $T$ can be identified with a set of reals of positive measure, so the countably many lines we've already determined can't get in the way of continuing the construction; and the boundary of $T$ is covered by finitely many lines, so that won't cause unavoidable trouble either. This last condition fails if we try to avoid an open disc.

On the other hand, it does seem possible to avoid a closed disc, by basically the same argument as above. I think, in general, it is possible to have the union of the lines in $C$ be the complement of any convex, bounded, open region; and it is possible to have the union of lines in $C$ be the complement of any closed, convex region whose boundary is a union of countably many line segments; so long as we assume $AC+CH$. ($CH$ can be replaced by the appropriate cardinal characteristic requirement that any set of reals of cardinality $<2^{\aleph_0}$ has measure 0; see http://mathoverflow.net/questions/8972/do-sets-with-positive-lebesgue-measure-have-same-cardinality-as-r for why this isn't a consequence of ZFC alone.) I have no idea how much choice is required for any of this.

1

(Too long to be a comment, but:) It is impossible to, say, make the union of $C$ either the complement of an open concave region (trivial) or the complement of an open circle (there's only one possible choice of line for each point on the boundary of the circle, and then any other line has to be parallel to one of these tangent lines).

On the other hand, I think it should be possible - certainly with $AC+CH$ - to make the union of elements of $C$ be, say, the complement of an open triangle $T$. Work in $\omega_1$-many stages, where at each stage $\eta$ you have countably many lines already drawn, and you're trying to get a new line which goes through the point $(x_\eta, y_\eta)$ (assuming this point isn't in $T$ and doesn't lie on an existing line already) and is "legal" (i.e., doesn't intersect $T$, doesn't intersect an existing point of intersection between already-chosen lines, and isn't parallel to any given line). Initialize the construction by starting with the three lines containing the three edges of the triangle; then it seems that, by our assumption that $CH$ holds, we can continue this construction to hit every point not in $T$.

A key step in this proof is that the set of lines through a given point not in the closure of $T$ can be identified with a set of reals of positive measure, so the countably many lines we've already determined can't get in the way of continuing the construction; and the boundary of $T$ is covered by finitely many lines, so that won't cause unavoidable trouble either. This last condition fails if we try to avoid an open disc.

On the other hand, it does seem possible to avoid a closed disc, by basically the same argument as above. I think, in general, it is possible to have the union of the lines in $C$ be the complement of any convex, bounded, open region; and it is possible to have the union of lines in $C$ be the complement of any closed, convex region whose boundary is a union of countably many line segments; so long as we assume $AC+CH$. ($CH$ can be replaced by the appropriate cardinal characteristic requirement that any set of reals of cardinality $<2^{\aleph_0}$ has measure 0; see http://mathoverflow.net/questions/8972/do-sets-with-positive-lebesgue-measure-have-same-cardinality-as-r for why this isn't a consequence of ZFC alone.) I have no idea how much choice is required for any of this.