Sec. 2.2 Open and closed half-planes 27
(a)G 1 ⊂G 1 ′ or (b)G 1 ⊂G 2 ′ or (c)G 1 ∩G 1 ′= 0 /,G 1 ∩G 2 ′=0./In (c) we haveP∈G 1 ,P∈G 1 ′andQ∈G 1 ,Q∈G 2 ′for somePandQ.Butthenwe
have[P,Q]⊂G 1 ,byA 3 (ii) applied toG 1 ,and[P,Q]∩l=0, by A/ 3 (iii) applied to
{G 1 ′,G 2 ′}. This gives a contradiction asl∩G 1 =0. Thus (c) cannot happen./
By similar reasoning, we must have either
(d)G 1 ′⊂G 1 or (e)G 1 ′⊂G 2.Now (a) and (d) giveG 1 =G 1 ′and it follows thatG 2 =G 2 ′as
(G 1 ∪G 2 )\G 1 =G 2 ,(G 1 ′∪G 2 ′)\G 1 ′=G 2 ′.Similarly (b) and (e) giveG 1 =G 2 ′and it follows thatG 2 =G 1 ′.
Finally, we cannot have (a) and (e) as that would implyG 1 ⊂G 2. Neither can we
have (b) and (d).
TERMINOLOGY. If two points
are both inG 1 or both inG 2 they
are said to beon the one side
of the linel, while if one of the
points is inG 1 and the other is in
G 2 they are said to beon differ-
ent sides ofl.
A
B
Figure 2.5. A closed half-plane shaded.2.2.3 Closed half-planes
Definition.IfG 1 ,G 2 are open half-planes with common edgel, we call
H 1 =G 1 ∪l,H 2 =G 2 ∪l,closed half-planes with common edgel.
Closed half-planesH 1 ,H 2 with common edge l have the properties:-(i)H 1 ∪H 2 =Π.(ii)H 1 ∩H 2 =l.(iii)Each ofH 1 ,H 2 is a convex set.(iv)If A∈l and B=AisinH 1 ,then[A,B⊂H 1.Proof.
(i) By A 3 (i),Π=G 1 ∪G 2 ∪l=(G 1 ∪l)∪(G 2 ∪l)=H 1 ∪H 2.
(ii) For(G 1 ∪l)∩(G 2 ∪l)=(G 1 ∩G 2 )∪(G 1 ∩l)∪(G 2 ∩l)∪(l∩l)=l∩l=l.
(iii) We prove thatH 1 is convex; proof forH 2 is similar.
LetA,B∈H 1 ; we wish to show that[A,B]⊂H 1.
CASE (a). LetA,B∈G 1. Then the conclusion follows from A 3 (ii).