1.6: Half-lines and segments

Assume there is a line $l$ passing thru two distinct points $P$ and $Q$. In this case we might denote $l$ as $(PQ)$. There might be more than one line thru $P$ and $Q$, but if we write $(PQ)$ we assume that we made a choice of such line.

We will denote by $[PQ)$ the half-line that starts at $P$ and contains $Q$. Formally speaking, $[PQ)$ is a subset of $(PQ)$ which corresponds to $[0,\infty)$ under an isometry $f: (PQ) \to \mathbb{R}$ such that $f(P) = 0$ and $f(Q) > 0$.

The subset of line $(PQ)$ between $P$ and $Q$ is called the segment between $P$ and $Q$ and denoted by $[PQ]$. Formally, the segment can be defined as the intersection of two half-lines: $[PQ] = [PQ) \cap [QP)$.

Exercise $\PageIndex{1}$

Show that

(a) if $X \in [PQ)$, then $QX = |PX - PQ|$;

(b) if $X \in [PQ]$, then $QX + XQ = PQ$.

Hint

Fix an isometry $f: (PQ) \to \mathbb{R}$ such that $f(P) = 0$ and $f(Q) = q > 0$.

Assume that $f(X) = x$. By the definition of the half-line $X \in [PQ)$ if and only if $x \ge 0$. Show that the latter holds if and only if $|x - q| = ||x| - |q||$. Hence (a) follows.

To prove (b), observe that $X \in [PQ]$ if and only if $0 \le x \le q$. Show that the latter holds if and only if $|x - q| + |x| = |q|$.