Orientation on a line and equivalence relation

Question

In order to define orientation on a line, my textbook defines half-line as the set of points that are on the same side of a given point on the line. Then, they define the relation of "pointing in the same direction" on the set of half-lines of a given line. They point in the same direction if one is subset of the other. Now, the orientation of a line is just one of two elements of the factor set. But while proving that relation is equivalence relation, they skipped the part that it is transitive. They say that if $a, b$ and $c$ are half-lines, then from $a\subset b$ and $c\subset b$ follows that $a\subset c$ or $c\subset a$, but they don't prove it, which is essentially the reason why it's transitive. I tried proving it using definition but can't work it out. It seems that I need to suppose the order of the points on the line and there is just too many cases. So the question is, how to prove that given relation is transitive ?

Answer 1

One way to do this is to first prove the equivalence:

Given halflines $a,b$ contained in the same line with origins $A,B$ respectively such that $A\neq B$ : $$a\subset b \iff ( A\in b \wedge B\notin a)$$

Having proven this we can continue your case like this:

Assume $a\subset b, c\subset b$. If $A=B$ or $C=B$, we are done. Assume $A\neq B,C\neq B$. By the equivalence above $$ A\in b \wedge B\notin a \wedge C\in b \wedge B\notin c$$ Since $A,C$ belong to the same halfline with origin $B$: $$B-A-C \vee A=C \vee B-C-A$$ ($P-Q-R$ means that $Q$ lies between $P$ and $R$)

In the first case $B-A-C$ from $B\notin a$ we get $C\in a$ and from $B\notin c$ we get $A\notin c$. Hence, again by the equivalence $c\subset a$.

In the second case, $c=a$.

In the third case, we get $a\subset c$ (very similarly as in case 1)