Exercise 15, § 6

Let $X = \mathbf{R}^n$, $M$ a closed subset of $X$. A point $x \in M$ is said to be linearly accessible if there exists a point $y \in X - M$ such that the open segment with endpoints $x, y$ is contained in $X - M$. Prove that the set $L \subset M$ of linearly accessible points of $M$ is Souslinian. (Let $d(x, y)$ be the euclidean distance in $X$, and for every $z \in X$, let $f_z(x, y) = d(x, z) + d(z, y) - d(x, y)$. Remark that, in $X \times X \times X$, the set of $(x, y, z)$ such that $x \in M, y \in X - M, z \in M, z \neq x$ and $f_z(x, y) = 0$ is a countable union of compact sets, and consequently that the same is true of its projection on the product of the two first factors.)