Exercise 6, § 1

A subset $L$ of $\mathbf{R}^n$ is a broken line if there exists a finite sequence $(x_i)_{0 \leq i \leq p}$ of points of $\mathbf{R}^n$ such that, if $S_i$ denotes the segment with end-points $x_{i-1}$ and $x_i$ ($1 \leq i \leq p$), $L$ is the union of the $S_i$, which are called the sides of $L$. (In general there is an infinite number of finite sequences of points of $\mathbf{R}^n$ which define the same broken line.) A broken line is also the image of a mapping $u$ of $[0, 1]$ into $\mathbf{R}^n$ such that there exists a strictly increasing sequence $(t_j)_{0 \leq j \leq q}$ in $[0, 1]$ with $t_0 = 0$ and $t_q = 1$, with the property that $u$ is an affine linear mapping

(such a mapping $u$ is said to be "*piecewise linear*"). Given a non-empty subset $A$ of $\mathbf{R}^n$ we say that two points $a, b$ of $A$ can be joined by a broken line in $A$ if there exists a broken line $L \subset A$ defined by a sequence $(x_i)_{0 \leq i \leq p}$ such that $x_0 = a$ and $x_p = b$. If any two points of $A$ can be joined by a broken line in $A$, then $A$ is connected. Conversely, if $A$ is a connected open subset of $\mathbf{R}^n$, show that any two points of $A$ can be joined by a broken line in $A$ (consider the relation "*$x$ and $y$ can be joined by a broken line in $A$"* between points $x, y$ of $A$; show that this is an equivalence relation whose classes are open sets); we can even assume always that this broken line is a union of segments each of which is parallel to a coordinate axis (same method). Deduce that, if $A$ is a non-empty open set in $\mathbf{R}^n$, the component of a point $a \in A$ is the set of all points of $A$ which can be joined to $a$ by a broken line in $A$.