the set $D\setminus V$ is self segment-dense in $D.$

In what follows we provide some examples of self segment-dense sets.

(see also Example 3.1, [@LYAK])\[EEE\] Let $V$ be the two dimensional Euclidean space ${{\mathbb R}}^2$ and define $U$ to be the set $$U :=\{(p,q) \in{{\mathbb R}}^2 : p\in \mathbb{Q},\, q\in\mathbb{Q}\},$$ where $\mathbb{Q}$ denotes the set of all rational numbers. Then, it is clear that $U$ is dense in ${{\mathbb R}}^2.$ On the other hand $U$ is not segment-dense in ${{\mathbb R}}^2,$ since for $x=(0, \sqrt{2})\in {{\mathbb R}}^2 $ and for every $y=(p,q)\in U$, one has $[x, y] \cap U = \{y\}.$

It can easily be observed that $U$ is also self segment-dense in ${{\mathbb R}}^2$, since for every $x,y\in U$ $x=(p,q),\,y=(r,s)$ we have $[x,y]\cap U=\{(p+t(r-p),q+t(s-q)): t\in[0,1]\cap\mathbb{Q}\},$ which is obviously dense in $[x,y].$

Let $V^{0}\left( \left[ 0,1\right] ,\mathbb{R}\right) $ be the space of piecewise constant $\mathbb{R}$-valued functions.

Obviously $V^{0}\left( \left[ 0,1\right] ,\mathbb{Q}\right) $ is dense in $V^{0}\left( \left[ 0,1\right] ,\mathbb{R}\right) $. So, it follows from the density of $V^{0}\left( \left[ 0,1\right] ,\mathbb{R}\right) $ in $L^{2}\left( \left[ 0,1\right] ,\mathbb{R}\right) $ that $$V^{0}\left( \left[ 0,1\right] ,\mathbb{Q}\right) \text{ is dense in }
L^{2}\left( \left[ 0,1\right] ,\mathbb{R}\right) .$$ Moreover, $V^{0}\left( \left[ 0,1\right] ,\mathbb{Q}\right) $ is 