Sec. 3.1 Distance 37
(iv) We haveA=Bby definition of[A,B,andA=Cas 0<|A,B|<|A,C|so
that 0<|A,C|.WealsohaveB=Cas|A,B|<|A,C|combined withB=Cwould
give|A,B|<|A,B|, whereas once(A,B)is known|A,B|is uniquely determined. We
cannot haveA∈[B,C]asC∈[A,B. Then by 2.1.4 eitherB∈[C,A]orC∈[A,B].
But by (ii) of the present result, ifC∈[A,B]we would have|A,C|≤|A,B|.Asthisis
ruled out by assumption, we must haveB∈[A,C].
Segments and half-lines have the following further properties:-(i)Let l∈Λbe a line, A∈l and≤la natural order on l. Then there are points B
and C on l such that A≤lB and B=A, and such that C≤lA and C=A.(ii)If A=B, there are points X∈[A,B]such that X=A and X=B.(iii)If[A,B=[C,DthenA=C.C
A
B
Figure 3.2.A
X
B
Proof.
(i) By A 4 (iv) with anyk>0,thereissomeB∈lsuch thatA≤lBand|A,B|=k.
As|A,B|>0, we haveA=B. Proof for the existence ofCis similar.
(ii) Let≤lbe the natural order onl=ABfor whichA≤lB.AsA=Bwe have
|A,B|>0 and then with anyksuch that 0<k<|A,B|, there is a pointX∈lsuch that
A≤lXand|A,X|=k.As|A,X|=0, we haveA=X;as|A,X|<|A,B|thenX=B.
AsX∈[A,Band|A,X|<|A,B|,wehaveX∈[A,B].
(iii) With the notation of (ii),P∈[A,Bif and only ifA≤lP.NowC∈[C,D=[A,B
soA≤lC; similarlyA≤lD.
CASE 1. LetC≤lD,sothat[C,D={Q∈l:C≤lQ}.AsA∈[A,B=[C,Dwe
haveC≤lAand this combined withA≤lCimpliesA=C.
CASE 2. LetD≤lC.Then[C,D={Q∈l:Q≤lC}. By (i) of the present result
there is anX∈lsuch thatX≤lAandX=A.ThenX≤lA,A≤lCsoX≤lCand thus
X∈[C,D.HoweverX∈[A,Bas otherwise we would haveA≤lXwhich combined
withX≤lAimpliesX=Aand involves a contradiction. ThenX∈[C,D,X∈[A,B
which contradicts the fact that[A,B=[C,Dand so this case cannot occur.