I call a subset of $\mathbb{R}^2$ $0$-dimensional if it has a (countable) basis made of clopen sets wrt its relative topology. For example, the subset $\mathbb{Q}\times\{0\}$ is $0$-dimensional.
Now my questions are:
- Is there a subset $X$ of $\mathbb{R}^2$ such that $X$ is connected and $X\cap(\{r\}\times\mathbb{R})$ is $0$-dimensional for every $r\in\mathbb{R}$?
- In case there is, do we need choice to construct such a set?
Thanks!
As suggested, I expand a bit my comment and post it as an answer. In fact, to give an example satisfying $1$, it would suffice to take just any horizontal line.
But I guessed from $\mathbb Q$ given as an example of a $0$-dimensional set that density might be a desired property too. Then we can define $X$ to be the union of all lines with rational slope through the origin which forms a dense subset of $\mathbb R^2$. Moreover, the intersection of $X$ with any (but one) vertical line is (linearly) homeomorphic to $\mathbb Q$.
If you want to fix the "but one" above, ie. to obtain all the intersections dense and $0$-dimensional, consider the following $X$: the union of the antidiagonal $\{(x,-x) \,|\, x \in \mathbb R \}$ and all the parallel rational shifts of the diagonal $\{(x,x+q) \,|\, x \in \mathbb R \}$, $q \in \mathbb Q$.