Kuratowski-Ulam Theorem, nowhere dense set in product space

Question

Let $E$ be a nowhere dense subset of $\mathbb{R}\times \mathbb{R}$. For $x\in \mathbb{R}$, define $$E_x=\{ y\in\mathbb{R}\mid (x,y)\in E\}.$$ Let $D$ denote the set of $x$ for which $E_x$ is NOT nowhere dense in $\mathbb{R}$. By the Kuratowski-Ulam Theorem, we know that $D$ is of first cateogory in $\mathbb{R}$. My question is: can we strengthen this conclusion by saying that $D$ is nowhere dense in $\mathbb{R}$? If not, is there a counterexample?

Thanks in advance!

Answer 1

List the rational numbers as $\{q_n : n = 1, 2, \cdots \}$ and for each $n$, let $X_n = \{q_n\} \times [n, n+1]$. Finally, let $X = \cup_{n = 1}^\infty X_n$. Then $X$ is nowhere dense in $\mathbb R\times\mathbb R$ and for $x$ a rational number, $E_x$ is a non-degenerate closed interval, so the set $D$ is the set of rational numbers which, of course, fails to be nowhere dense in $\mathbb R$.