Meager set and disjoint with line $y=ax$

Question

Lemma. Let $X$ and $Y$ be second countable. If $A\subset X$ and $B\subset Y$, then $A\times B$ is meager iff at least of $A,B$ is meager.

Assume $f\colon \mathbb R\to \mathbb R$ and $D$ be a meager set subset of $R.$ Consider the graph of $f$ on $D$, that is. $$W:=\{\langle x, f(x)\rangle\colon \ x\in D\setminus\{0\}\}$$ By lemma above $W$ is meager in $\mathbb R^2$. Now, $\mathbb R^2\setminus W$ is co-meager in $\mathbb R^2$. I want to find a line $\ell_a(x)=ax$ such that $\ell_a\cap W=\{\langle x,ax\rangle\ \colon \ x\in\mathbb R\}\cap W=\emptyset.$ Is it possible in $\text ZFC$? Do we need a set-theortical assumption, like the union of less than continuum many meager set will not cover $\mathbb R.$ Any help will be apprciacted greatly

Answer 1

It is not necessarily possible. Let $D$ be a meagre subset of $\Bbb R$ of cardinality $2^\omega$, e.g., the middle-thirds Cantor set. Then we can index $D=\{x_a:a\in\Bbb R\}$ by the real numbers. Now let

$$f:D\to\Bbb R:x_a\mapsto ax_a\;;$$

then the graph of $f$ meets every line $\ell_a$ through the origin.