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!



Kuratowski-Ulam Theorem, nowhere dense set in product space
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 $\mathbb R$.


