Exercise 11, § 6

Let $X = \mathbf{R}^n$ and let $M$ be 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 extremities $x, y$ is contained in $X - M$. Show that the set $L \subset M$ of linearly accessible points of $M$ is a Souslin set. [Let $d(x, y)$ be the Euclidean metric in $X$, and for each $z \in X$ let $f_z(x, y) = d(x, z) + d(z, y) - d(x, y)$. Note that the set of all $(x, y, z) \in X \times X \times X$ 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 so is its projection on the product of the first two factors.]