Exercise 26, § 2

In a Hausdorff topological vector space $E$, let $A$ be a connected closed set with the following property : for every $x \in A$, there exists a closed neighbourhood $V$ of $x$ in $E$ such that $V \cap A$ is convex. Show that $A$ is convex. For this establish the following statements. a) Show that any two points of $A$ can be joined by a broken line in $A$ (same method as GT, VI, § 1, exerc. 6). b) Show that, if two points in $A$ can be joined by a broken line in $A$ with $n > 1$ segments, then they can also be joined in $A$ by a broken line with $n - 1$ segments. (Induction on $n$ reduces to the case $n = 2$ which is equivalent to taking $\mathbf{R}^2$ as $E$; then let $T$ be a triangle with vertices $a, b, c$ such that the closed segments $ac, bc$ are contained in $A$, but the closed segment $ab$ is not; consider a point of the closure of the intersection of $\mathscr{C} A$ and the interior of $T$ that is farthest from the line $ab$, and show that the existence of such a point contradicts the hypothesis.)