[PDF] The Betweenness Axioms For each pair of distinct points a and b

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The Betweenness Axioms.

For each pair of distinct pointsaandbthere is a subset s(a;b); called thesegment joiningatob, satisfying (B1)-(B3) below. Wheneveraandcare distinct points we say the pointbisbetweenaandcifb2s(a;c). (B1) Ifaandbare distinct points then (a)s(a;b)½l(a;b); (b)s(b;a) =s(a;b); and (c)s(b;a)\ fa;bgis empty. (B2) (a) Ifaandbare distinct points thens(a;b) is nonempty. (b) Ifa,bandcare distinct andfa;b;cgis collinear exactly one of the following holds: a2s(b;c); b2s(c;a); c2s(a;b): (c) Ifaandbare distinct points there is a pointcnot equal toaorbsuch thatb2s(a;c). SupposeLis a line andaandbare distinct points not lying onL. We sayaandbare on the same side ofLifs(a;b) does not meetLand we sayaandbare on opposite sides ofLifs(a;b) does meet L. From (B1)(b) we infer that these relations are symmetric inaandb. The previous axioms imply that a line is at most one dimensional in the sense of topology, whatever

that means. The following axiom says that the set of points is at most two dimensional in the sense of

topology, whatever that means. The continuity axiom, which will follow, will allow us to replace "at most"

with "exactly". (B3) SupposeLis a line anda,bandcare distinct points not lying onL. Then (a) ifaandbare on the same side ofLandbandcare on the same side ofLthenaandcare on the same side ofL. (b) ifaandbare on opposite sides ofLandbandcare on opposite sides ofLthenaandcare on the same side ofL. IfSis a segment we say a pointais anendpoint ofSif there is a pointbnot equal toasuch that

S=s(a;b).

Supposeaandbare distinct points;canddare distinct points; ands(a;b) =s(c;d). It is natural to

ask iffa;bg=fc;dg, or, equivalently, if a segment has exactly two endpoints. As we will show below, this

is indeed the case. Theorem.SupposeLis a line ando,a,bandcare distinct points lying onL. Then (B4)(c) Ifo62s(a;b) ando62s(b;c) theno62s(a;c). (B4)(d) Ifo2s(a;b) ando2s(b;c) theno62s(a;c). 1 Proof.Letdbe a point not lying onL; we have show that such a point exists as a consequence of the Incidence Axioms. LetMbe the line determined byoandd.

Lemma.SupposeXis a subset ofL. ThenX\M=X\ fog.

Proof.Sinceo2Mwe haveX\ fog ½X\M. Supposex2X\M. Werex6=owe the two member set fx;ogwould be a subset of each of the distinct lines linesLandMwhich contradicts (I1).

Now apply (B3) withLthere replaced byM.

The following Theorem is a fundamental tool in dealing with betweenness. The Splitting Theorem.Supposeaandcare distinct points andbis betweenaandc. Thenb62 fa;cg and fs(a;b);fbg;s(b;c)gis a partition ofs(a;c):

Remark.Although it is a bit tedious, the reader should check this proof very carefully. The writer claims

he has put inallthe steps! Proof.Thatb62 fa;cgfollows from (B1)(c). Supposedis a point on the line determined byaandc. Supposed2s(a;b). Thend62 fa;bgby (B1)(b). Sincec62s(a;b) by (B2)(b) we haved6=c. Thus the pointsa;b;c;dare distinct by (B1)(c). From (B2)(b) we infer thatb62s(a;d). Were it the case that b62s(c;d) (B4)(c) would implyb62s(a;c) sob2s(c;d). From (B2)(b) it follows thatd62s(b;c). Were d62s(c;a) we would haved62s(b;a) by (B4)(c) which in turn would imply by (B1)(b) thatd62s(a;b); thus d2s(c;a). By (B1)(b),d2s(a;c). Thuss(a;b)½s(a;c). Interchangingaandcand using (B1)(b) twice we infer thats(b;c)½s(a;c). Thus (1)s(a;b)[s(b;c)½s(a;c): Ifdis a point onl(a;c),d62 fa;cg,d62s(a;b) andd62 fa;b;cgthen, by (B4)(c),d62s(a;c). Thus s(a;c)½s(a;b)[ fbg [s(b;c): Ifd2s(a;b)\s(b;c) thend62s(a;c) by (B4)(c); this is incompatible with (1). Thatfs(a;b);fbg;s(b;c)g is disjointed now follows from (B1)(c) Remark.Having gotten through that we will now start being informal. Watch out! Many good mathe- maticians have gotten egg on their face in this subject. The following notion will shortly prove to be convenient.

Definition.SupposeXis a set of points. We let

b(X); theboundary ofX, be the set of pointsbsuch that there is a pointa2Xsuch thats(a;b)½X. Note that ifXis convex in the standard Euclidean model thenb(X) is the boundary ofXwith respect to the standard topology onR2.

Definition.We say a setCof points isconvexif

s(a;b)½Cwhenevera;b2C. Theorem.SupposeCis a nonempty family of convex sets of points. Then\Cis convex.

Proof.This is obvious.

Rays. Definition.Supposeoandaare distinct points. We let r(o;a) =fag [ fb2l(o;a)» fag:o62s(a;b)g 2 and we call this set theray emanating fromopassing througha. For any pointowe let

R(o) =fr(o;a) :ais a point anda6=og:

Theorem.Supposeois a point. ThenR(o) is a partition of the set of points not equal too. Proof.It follows directly from (B1)(b) and (B4)(c) that f(a;a) :ais a point anda6=og [ f(a;b) :aandbare points,a6=o,b6=o,a6=bando62s(a;b)g

is an equivalence relation on the set of points not equal too. It is immediate that the equivalence classes

are the raysr(o;a) corresponding to pointsanot equalo. Theorem.Supposeoandaare distinct points andLis a line. Then r(o;a)½L,L=l(o;a): Proof.It is an immediate consequence of the definition thatr(o;a)½l(o;a). By (B2)(a) there is a pointbbetweenoanda. By (B1)(c),a6=b. By (B2)(b),o62s(a;b). Thusr(o;a) contains at least two points and therefore can be contained in at most one line. Theorem.Supposeoandaare distinct points. Thenb(r(o;a)) =fog. Proof.We haveo62r(o;a). Supposeb2s(o;a). Theno62s(a;b) by (B2)(b) sob2r(o;a). That is, s(o;a)½b(r(o;a)). Thuso2b(r(o;a)). Supposep2b(r(o;a)). Thenp62r(o;a) and there isbsuch thatb2r(o;a) and (1)s(p;b)½r(o;a): Supposep6=o. Theno2s(p;a). Sinceo62s(a;b) we infer from (B4)(c) and (B1)(b) thato2s(p;b) which by (1) implieso2r(o;a) which is impossible. Definition.SupposeRis a ray. In view of the two preceding Theorems, we may define l(R) as the unique line containingRand we may define o(R) as the unique member ofb(R). We callo(R) theorigin ofR.

Definition.SupposeRis a ray. We set

R o=l(R)»(R[ fo(R)g): Theorem.SupposeRis a ray. ThenRois ray,o(Ro) =o(R) andl(Ro) =l(R). Proof.Supposeo;aare such thatR=r(o;a). LetL=l(o;a) and letS=L»(fog[R). By (B2)(c) there isb2l(o;a) such thato2s(a;b). I claim thatS=r(o;b). Supposec2S. Thenc6=oando2s(a;c). By (B1)(b) and (B4)(d),o62s(b;c) soc2r(o;b). Supposed2r(o;b). Thend6=oando62s(b;d). Since o2s(a;b) we infer from (B4)(c) thato2s(a;d) sob62R. Definition.SupposeRis a ray. We callRotheray oppositeR.We have shown thatfR;o(R);Rogis a partition ofl(R). 3 Definition. Geometric linear orderings.SupposeLis a line andS½L. We say a linear ordering< ofSisgeometricif

S\s(a;b) =fx2S:a < x < bgwhenevera;b2Sanda < b.

Note that the inverse of a geometric linear ordering ofSis a geometric linear ordering ofS. Theorem.SupposeLis a line,S½LandSupposex;y2S. Using (1) repeatedly we find that x < y < a,y2s(x;a),xÁyÁa; x < a < y,a2s(x;y),xÁaÁy; a < x < y,x2s(a;y),aÁxÁy:

It follows thatÁequals<.

Lemma.SupposeRis ray with originoandThen Proof.Supposea;b2Randa6=b. Thena < bif and only ifa2s(o;b) andb < aif and only ifb2s(o;a).

Thus Supposea;b;c2R,a < bandb < c. Thena2s(o;b) andb2s(o;c) so the Splitting Theorem implies thata2s(o;c) soSupposea;b2Landa < b. We need to show that

(1)s(a;b) =fx2L:a < x < bg: In case (a;b)2Ro£Ro(1) holds because the restriction ofProof.Simple exercise for the reader. Definition.SupposeLis a line,a consequence of the previous Theorem we find that there is exactly one geometric ordering of a given line

induced by a ray contained in that line. Theorem.SupposeLis a line;Ris a ray with origino;R½L;Proof.Simple exercise for the reader. Theorem.SupposeLis a line;Ris a ray with originLandR½L;Sis a ray with originpandS½L; andR6=S. Then exactly one of the following holds: (i) R½S; (ii)S½R; (iii)R\S=;; (i)R\Sis a segment:

Furthermore, (i) or (ii) hold if and only if the geometric ordering ofLinduced byRequals the geometric

linear ordering ofLinduced bySand (iii) or (iv) hold if and only if the geometric ordering ofLinduced by

Requals the inverse of the geometric linear ordering ofLinduced byS.

Moreover (i) holds if and only ifp2R; (ii) holds if and only ifo2R; (iii) holds if and only if either

o=pandS=Rooro6=pandL»(R[S[ fo;pg) =s(o;p); and (iv) holds if and only ifo6=pand

R\S=s(o;p).

Proof.Simple exercise for the reader.

The Continuity Axiom.There are reasons not to like this. (Cont) Any geometric linear ordering of a line is complete.

Halfspaces.

A lot of what we will do here will bear a resemblance to what we did with rays. Definition.SupposeLis a line andais a point not lying onL. For each pointanot lying onLwe let h(L;a) be the set of those pointsbsuch thateitherb=aorb6=a,b62Lands(a;b) does not meetL. Thush(L;a) is the set of points consisting ofaitself and the points which are on the same side ofLasa. We set

H(L) =fh(L;a) :ais a point anda62Lg:

We sayHis a halfspaceifH2H(L) for some lineL.

Theorem.SupposeLis a line. ThenH(L) has exactly two members and is a partition of the set of points not lying onL. 5

Proof.It follows directly from (B) that

f(a;a) :ais a point anda62Lg [ f(a;b) :aandbare points,a;b62L,a6=bands(a;b) does not meetLg

is an equivalence relation on the set of points not lying onLwhich has at most two equivalence classes.

Letabe a point not lying onL. such a point exists by (I). Letobe a point inL; such a point exists by (I). Letbbe a point not equal tooorasuch thatois in the segment joiningatob; such a point exists. By (B),aandbare not equivalent. ThusH(L) has at least two equivalence classes. Theorem.SupposeLis a line,aandbare distinct points,a62Lands(a;b) does not meetL. Then s(a;b)½h(L;a): Proof.Supposec2s(a;b). Thenc62Land, by the Splitting Theorem,s(a;c)½s(a;b) sos(a;c) does not meetL. Thusc2h(L;a). Theorem.SupposeHis a halfspace. Thenb(H) is a line and

H2H(b(H)):

Proof.LetLbe a line be such thatH2H(L). LetH0be such thatH(L) =fH;H0g. Supposeb2b(H). Leta2Hbe such thats(a;b)½H. Were it the case thatb62Lwe would have b2H0sos(a;b) would meetLwhich is impossible. Thusb2L. Supposeb2L. Thenb62H. Leta2H. Sincea62Lwe have thats(a;b) does not meetL. By the previous Theorem,s(a;b)½Hsob2b(H). Definition.SupposeHis a halfspace. In view of the preceding two Theorems we may define the halfspace H o by requiring thatH(b(H)) =fH;Hog. We callHothehalfspace oppositeH. Evidently, H oo=H: Theorem.SupposeLis a line,H2H(L),o2LandRis a ray with origino. Then eitherR½L,R½H orR½Ho. Proof.Letabe such thatR=r(o;a) and supposeb2R» fag. Ifa2Lthenl(a;b) =l(o;a) =Lsob2L. Ifa2Hthens(a;b) does not meetLsincel(a;b) can meetLin at most one point sob2H. Similarly, if a2Hoone shows thatR½Ho. Definition.SupposeLis a line,ois a point lying onL,Ris a ray with vertexoand

L\R=;:

Keeping in mind the previous Theorem, we let

h(L;R) be that memberHofH(L) such thatR½H. Theorem.SupposeHis a halfspace,Cis a convex set of points andC\b(H) =;. Then eitherC½H orC½Ho.

Proof.Straightforward exercise for the reader.

Theorem.Lines, segments and rays and halfspaces are convex. 6

Proof.Exercise for the reader.

Theorem.SupposeLis a line andHis a halfspace. Then exactly one of the following holds: (i)L½H; (ii)L½Ho; (iii)L=b(H); (iv) there is a pointosuch thatL\b(H) =fogandL\His a ray with origino. Proof.SinceLis convex, ifLdoes not meetb(H) then either (i) or (ii) holds. IfLmeetsb(H) in more than one point then (iii) holds. Supposeois a point such thatL\b(H) =fog. Use (B2)(b) to choose a pointainL\H. We leave it as a simple exercise for the reader to verify thatL\H=r(o;a).

Fundamental sets.

We now study intersections of finite families of halfspaces. Definition.We say a set of pointsFisfundamentalifFis nonempty and F=\F for some finite nonempty familyFof halfspaces. Theorem.SupposeFis a fundamental set of points,a2b(F) andb2F. Thens(a;b)½F. Proof.LetFbe family of halfspaces such thatF=\F. Sincea2b(F) there isc2Fsuch thats(a;c)½F. LetH2 F. Thens(a;c)½Hso eithera2b(H)[H. This impliess(a;b)½H. Definition.We say a familyEof halfspaces isefficientifEis finite and nonempty,\E 6=;and\F 6=\E wheneverFis a nonempty proper subfamily ofE. Theorem.SupposeFis a nonempty family of halfspaces and\F 6=;. Then there is an efficient subfamily

EofFsuch that\E=\F.

Proof.LetGbe the class of nonempty subfamiliesGofFsuch that\F=\G. Note thatF 2GsoGis nonempty and therefore m= minfcardG:G 2Gg is a well defined positive integer. Any memberEofGsuch thatcardE=mis efficient. Remark.We will show below that such anEi; i= 1;2;are efficient families of halfspaces and\E1=\E2 thenE1=E2.

Corollary.SupposeFis a fundamental set of points. Then there is an efficient familyEof halfspaces such

thatE ½ FandF=\E. Theorem.SupposeHandIare distinct halfspaces. Then exactly one of the following holds: (i)b(H) meetsb(I) in a single point andfH;Igis efficient. (ii)b(H) =b(I),I=Ho,H\Iis empty, andfH;Igis not efficient. (iii)b(I)½H,b(H)½IandfH;Igis efficient. (iv)b(I)½H,b(H)½Io,H½I, andfH;Igis not efficient. 7 (v)b(I)½Ho,b(H)½I,I½H, andfH;Igis not efficient. (vi)b(I)½Ho,b(H)½Io,H\Iis empty, andfH;Igis not efficient.

Remark.Draw a picture.

Remark.(i) implies thatH\Iis not empty wheneverHandIare halfspaces such thatb(H) andb(I) meet in a single point. Proof.Supposeois a point such thatb(H)\b(I) =fog. Choose a pointainb(H)\Iand a pointbin b(I)\H; such points exist by a previous Theorem. Thens(a;b)½H\Iby a previous Theorem soH\I is nonempty. By (B) there are pointsc;dnot equal toaorbsuch thatb2s(a;c) anda2s(b;d). We have c2H»Iandd2I»HsoH\Iis not equal to eitherHorI. ThusfH;Igis efficient and (i) holds. In caseb(H) meetsb(I) in more that one point we haveb(H) =b(I) soI=Hoand (ii) holds. Supposeb(H) does not meetb(I). Keeping in mind that lines are convex we find thateither(vii) b(I)½Handb(H)½I;or(viii)b(I)½Handb(I)½Ho;or(ix)b(I)½Hoandb(I)½H;or(x) b(I)½Hoandb(I)½Ho. If (vii) holds we proceed as in the first paragraph to show thatfH;Igis

efficient. We leave it as an exercise for the reader to show that (iv), (v) and (vi) hold if (viii), (ix) or (x)

hold, respectivley. The Crossbar Theorem.Supposeois a point andHandIare halfspace such thatb(H)\b(I) =fog.

Supposea2b(H)\Iandb2b(I)\H. Then

(1)H\I=[fr(o;e) :e2s(a;b)g: Remark.aandbexist by virtue of a previous Theorem. Remark.As we shall see, in hyperbolic geometry it isnotthe case that ifd2H\Ithen there is a segment containingdjoining a point ofb(H)\Ito a point ofb(I)\H. Proof.LetL=b(H) and letM=b(I). Supposee2s(a;b). Note thats(a;b) does not meet eitherL orM. By the Splitting Theorem,s(a;e) ands(b;e) are subsets ofs(a;b). Sinces(b;e) does not meetL it follows thate2h(L;b) =Hand sinces(a;e) does not meetMit follows thate2h(M;a) =I. Thus e2H\I. Sincee62L[Mwe have thatr(o;e) does meetLorM. Ifx2r(o;e) thens(x;e) does not meet eitherLorMbecauses(x;e)½r(o;e). Thusx2H\I. Thus the right hand side of (1) is a subset ofH\I. Supposed2H\I. LetN=l(o;d) and note thatNdoes not meetH\Iobecause werex2N\H\Io we would haveo2s(x;d) becaused2Iando62s(x;d) becaused2H. Choosec2L\Io. Sinces(b;c) meets neitherLnorMit is a subset ofH\Io. Thuss(b;c) does not meetN. Buts(a;c) meetsNinoso (B) impliess(a;b) meetsNin a pointe. Sinces(d;e)½H\Iwe find thatd2r(o;e). ThusH\Iis a subset of the right hand side of (1). Corollary.SupposeHandIare distinct halfspaces;Nis a line;ois a point such thatb(H)\b(I)\N=fog; and

J=h(N;b(H)\I):

TheneitherH\I½JorfH\J;I\Jogis a partition of (H\I)»N. Proof.SupposeH\I\N=;. Then, asH\Iis convex and nonempty, there is a unique halfspaceJsuch thatb(J) =NandH\I½J. SupposeH\I\N6=;. Leta;bbe such thata2I\b(J),b2H\b(I). By the Crossbar Theorem,

H\I=[fr(o;c) :c2s(a;b)g:

H\NandI\Nare rays with originowhich are subsets ofNand have a point in common so they both equalH\I\N. Thus there is a unique membercofs(a;b) such thatr(o;c) =H\I\N. LetJ=h(N;a) and note thatJo=h(N;b). Using the Crossbar Theorem twice we find that H\J=[fr(o;d) :d2s(a;c)gandI\Jo=[fr(o;e) :d2s(c;b)g:

Thus the Theorem is proved.

8 Theorem.SupposeH,IandJare distinct halfspace andb(H),b(I) andb(J) meet in the pointo. Then fH;I;Jgis not efficient. Proof.Applying the previous Theorem withN=b(J) we find thateither(i)H\I½J,or(ii)H\I½Jo, or(iii)fH\J;I\Jogis a partition ofH\I»b(J) or (iv)fH\Jo;I\Jgis a partition ofH\I»b(Jo). In case (i) holds we haveH\I=H\I\J. In case (ii) holds we haveH\I\J=;. In case (iii) holds we haveH\I\J=H\J. In case (iv) holds we haveH\I\J=I\J. Corollary.SupposeEis efficient family of halfspaces andois a point. Then fH2 E:o2b(H)g contains at most two members. Theorem.SupposeLis a line,Fis a finite family of halfspaces,

S=\fL\b(H) :H2 Fg

andS6=;. ThenSis a segment, a ray or the line itself. Furthermore, b(S)½ [fL\b(H) :H2 Fg: Proof.LetGbe the set of thoseHinFsuchL\b(H) is a ray with origin inLand letObe the set of origins of these rays. By a previous Theorem,L\b(H) =LifH2 F » G. IfGis empty thenS=Land the Theorem holds, so let us assumeGis not empty. LetFurthermore, fs(F;H) :H2 Fg [ f[fb(s(F;H)) :H2 Egg is a partition ofb(\F). Proof.The Theorem holds trivially ifFhas exactly one member, so let us supposeFhas at least two members.

SupposeH2 Fand letS=s(F;H).

SupposeSis empty. LetG=F » fHg. Then\Gis a convex set which does not meetb(H) so is either a subset ofHorHo. Were it a subset ofHowe would haveF=H\(\G)½H\Ho=;which is impossible. ThusF=H\(\G) =\G. Now supposeSis nonempty. ThatSis a line, a ray or a segment follows from the previous Theorem.

Choose a pointcinF.

9 Supposeb2S. Thenb2b(H) sob62P. By a previous Theorem,s(b;c)½H. SupposeG2 F » fHg. SinceGis a halfspace andb2Gwe haves(b;c)½G. Thusb2b(P). Supposea2b(S). Sincea2b(H) we havea62P. By a previous Theorem,s(b;c)½H. Suppose G2 F » fHg. Choose a pointbinS. Thens(a;b)½S½G. Sinceb;c2GandGis a halfspaces(b;c)½G.

Thuss(a;b)½Gsoa2b(P).

Supposeo2b(P). Choose a pointdinpsuch thats(o;d)½P. Choose a halfspaceHinFsuch that o2b(H); were such a choice impossible we would haveo2P. LetS=s(F;H). I claim thato2S[b(S). If this were not the case, there would be pointsa;bsuch thatb(S) =fa;bgandb2s(a;o). But, by the preceding Theorem,bwould lie onb(G) for someG2 F » fHgand we would haveo2Gowhich is incompatible withs(o;d)½P½G. Let

A=fs(F;H) :H2 Fg [ f[fb(s(F;H)) :H2 Egg:

We have shown thatF=[A. ThatAis disjointed follows directly from definitions and the fact thatb(S)\S is empty wheneverSis a a segment. Corollary.SupposeEi; i= 1;2;are efficient families of halfspaces and\E1=\E2. ThenE1=E2. Proof.The familiesfs(Ei;H) :H2 Eig; i= 1;2;must be equal so the familiesfb(H) :H2 Eig; i= 1;2; must be equal. It follows thatE1=E2.

Corollary.SupposeFis a fundamental set. There is one and only one efficient familyEof halfspaces such

thatF=\E. Definition.SupposeFis a fundamental set. LetEbe that efficient family of halfspaces such thatF=\E. The members ofEare calledbounding halfspaces forF. The members offs(E;H) :H2 Egare called

sides ofF. A point is avertex ofFif it is either an endpoint of a segment which is side ofFor it is the

origin of a ray which is a side ofF. Note that f(S;H) :Sis a side ofF,H2 EandS½b(H)g

is a univalent function which carries the set of sides ofPonto the set of bounding halfspaces. Thus we

may speak of theside ofFcorresponding to a bounding halfspaceor thebounding halfspace corresponding to a side. Theorem.SupposeFis a fundamental set,ois a vertex ofFand

H=fH:His a bounding halfspace forFando2b(H)g:

ThenHhas exactly two members. Furthermore, ifSis the side ofFcorresponding to a member ofH then eitherSis a segment andois an endpoint ofSorSis a ray andothe origin ofS. Proof.Were there three distinct bounding halfspacesH;I;JofFsuch thatb(H)\b(I)\b(J) =fogthe family of bounding halfspaces ofF, by virtue of a preceding Theorem, would not be efficient. LetSbe a side ofFsuch thato2b(S). LetHbe that bounding halfspace forFsuch thatS½b(H). By virtue of a preceding Theorem, there is bounding halfspaceIofFsuch thatb(H)\b(I) =fog. LetT be the side ofFcontained inI. Ifo62b(T) then there are distinct pointsa;bsuch thatT=s(a;b) and a2s(o;b). LetJbe the a bounding halfspace forFsuch thata2b(J); such a bounding halfspace exists. Note thatJcannot equalIorJ. This implies thatS[T½Jwhich is impossible sinceoandbare on opposite sides ofb(J).

Definition.LetFbe a fundamental set.

Letobe a vertex ofF. A sideSofFis calledadjacent tooifo2b(S); otherwise it is calledopposite too. A vertexpofFnot equal toois calledadjacent tooifs(o;p) is a side ofF; otherwise it is called opposite too. The preceding Theorem says a vertex of a fundamental set has exactly two adjacent sides. Ifoandpare opposite vertices ofFthe segments(o;p) is called adiagonal ofF. Two distinct sides ofFareadjacentif they have an endpoint in common; otherwise they areopposite. 10 Theorem.SupposeFis a fundamental set anda;bare distinct points inb(F). Thens(a;b)½Fif and only ifs(a;b) is not contained in any side ofF. Proof.LetEbe the efficient family of halfspaces such thatF=\E Supposeaandbare in different sides ofF. Then there are distinct membersH;IofEsuch that a2b(H) andb2b(I). LetGbe the emptyset ifE=fH;Igand letGequal\¡E » fH;Ig¢otherwise. Thena2I\Gandb2H\G. It follows thats(a;b)½H\I\G=F: Supposeais a vertex ofFandbis in an opposite side ofF. Then there are distinct membersH;I;J ofEsuch thata2b(H)\b(I) andb2b(J). LetGbe the emptyset ifE=fH;I;Jgand letGequal \¡E » fH;I;Jg¢otherwise. Thena2J\Gandb2H\I\G. It follows thats(a;b)½H\I\J\G=F. Supposeaandbare opposite vertices ofF. Then there are distinct membersH;I;J;KofEsuch that a2b(H)\b(I) andb2b(J)b(K). LetGbe the emptyset ifE=fH;I;J;Kgand letGequal\¡E » fH;I;J;Kg¢otherwise. Thena2J\K\Gandb2H\I\G. It follows thats(a;b)½H\I\J\K\G=F. Definition.Anangleis a fundamental set with one vertex and two sides By virtue of the preceding

Theorem, these sides must be rays with origin equal to this vertex. By virtue of a preceding Theorem, ifH

andIare distinct halfspaces thenH\Iis an angle if and only ifb(H) meetsb(I) in a point. Theorem.SupposeRandSare distinct rays with origino. Then there is one and only angle with sides

RandS.

Proof.Choose a pointainRand a pointbinS. Thenh(l(R);b)\h(l(S);a) is the desired angle. Definition.SupposeFis a fundamental set andvis a vertex ofF. Theangle ofFcorresponding to the vertexvisH\IwhereHandIare the members of the efficient family of halfspaces whose intersection isFsuch thatfvg=b(H)\b(I). Definition.Apolygonis a fundamental set all of whose sides are segments. Ann-gonis a polygon with nsides. Atriangleis a 3-gon. Aquadrilateralis a 4-gon. And so forth. Definition.SupposePis a an polygon andVis the set of its vertices. Aboundary pathofPis a function v:f0;1;:::;lg !V such that (i)l¸2; (ii)viandvi+1are adjacent whenever 0·i < l; (iii)vi6=vjif 0·i < j < l.

We sayvstartsatv0andendsatvl.

Theorem.SupposePis an-gon,Vis the set of vertices ofPandaandbare are (not necessarily distinct) vertices ofP. Then (i) There are exactly two boundary paths v:f0;1;:::;lg !Vandw:f0;1;:::;mg !V which start ataand end atb. (ii)l+m=n+ 2. (iii) Ifa=bthenl=mandvi=wn+1¡i; i= 0;1;:::;n+ 1 and the ranges ofvandwequalV. (iv) Ifa6=bthen (rngv\rngw)» fa;bgis empty. 11

Proof.Simple combinatorics.

Theorem.The number of sides of a polygon equals the number of vertices.

Proof.Simple combinatorics.

Theorem.A polygon has at least three sides.

Proof.Exercise for the reader.

Theorem.Supposea;bandcare distinct andfa;b;cgis noncollinear. LetHa=h(l(b;c);a),Hb= h(l(c;a);b)gandHc=h(l(a;b);c).

ThenfHa;Hb;Hcgis efficient and the intersection of these halfspaces is a triangle with verticesa,band

cwith corresponding anglesHb\Hc,Ha\Hc,Ha\Hb. Theorem.SupposePis a polygon with at least four vertices;Vis the set of vertices ofP; andv2V. Letuandwbe the vertices ofPsuch thats(u;v) ands(u;w) are the sides ofPadjacent tov. letTbe the triangle with verticesu,vandw; letH=h(l(u;w);v); letQ=P\Ho; letEbe the efficient family of halfspaces whose intersection isP; and let

F= (E » fh(l(u;v);w);h(l(u;v);w)g)[ fHog:

ThenfT;s(u;w);Qgis a partition ofP;Fis an efficient family of halfspaces;Qis a polygon with verticesV» fvg; andQ=\F. RemarkThis theorem can be used to prove other theorems about polygons by inducting on the number of vertices.

Proof.Exercise for the reader.

Theorem.Two polygons with the same vertices are equal. In particular, there is one and only one triangle

whose set of vertices is a given three point noncollinear set.

Proof.Use the previous Theorem.

Theorem.SupposePis a polygon andHis a halfspace. ThenP½Hif and only if the set of vertices of

Pis a subset ofH[b(H).

Proof.SupposeP½H,a2Hoandais a vertex ofH. Letb2P. We have shown earlier thats(a;b)½F.

But, asb2F½H,s(a;b) would meetb(H).

Suppose the set of vertices ofPis a subset ofH[b(H). Leta;b;cbe vertices ofPsuch thataandb are the vertices adjacent toc. LetI=h(l(a;c);b) and letJ=h(l(b;c);a). ThenF½I\J, asJandHare

members of the efficient family of halfspaces whose intersection isF. It is a simple matter which we leave

to the reader to show thatI\J½H. Theorem.SupposePis a polygon,Lis a line ando2L\P. ThenL\Pis a segment with endpoints in b(P). Proof.Suppose no side ofPmeetsL. Then there would beH2H(L) such that the vertices ofPare all inH. The previous Theorem would then imply thatP½Hwhich is impossible becauseo2P»H. Leta;bbe adjacent vertices ofPsuch thats(a;b) meetsLin a pointc. Letvbe that boundary path starting ataand ending atbwhose range does not equalfa;bgand which therefore hasnmembers. Were it the case thats(vi;vi+1)\Lwere empty for eachi= 0;:::;n¡1 we would conclude by induction thats(v0;vi)\Lis empty for eachi= 0;:::;n¡1 and this would imply thats(a;b) =s(v0;vn¡1) did not meetL. So there is sideS=s(d;e) ofPnot equal tos(a;b) which meetsLin some pointf. Since o2h(l(a;b);f)\h(l(d;e);c) we find thato2s(c;f). Theorem.SupposePis a polygon ando2P. ThenPis the disjoint union offog; the segmentss(o;v) wherevis a vertex ofP; and the triangles with verticeso;a;bwheres(a;b) is a side ofP. Proof.Use the previous Theorem and the the Crossbar Theorem. 12 Theorem.SupposePis a polygon andvis a boundary path whose range is the set of vertices ofP.

Suppose

0·i < j < k < l < n

wherei;j;k;lare integers andn¸4 is the number of vertices ofP. Then s(vi;vj)\s(vk;vl) =; and there is a pointo2Psuch that s(vi;vk)\s(vj;vl) =fpg: Proof.LetL=l(vk;vl). Since none of the segmentss(vp;vp+1); p=i;:::;j¡1 meetLwe infer thatvi andvjare on the same side ofLand sos(vi;vj) does not meetL. LetM=l(vi;vk). Arguing as we did in the preceding paragraph we find that the pointsvp; i < p < k; are on the same sideHofMand that the pointsvq; k < q < lorl·q < nor 0< q < iare on the same sideIofM. WereH=Iwe would infer from a previous Theorem thatP½H. This is impossible because s(vi;vk) meetsPby a previous Theorem. ThusI=Hoand therefores(vj;vl) meetsMin a pointp. In a previous Theorem we found thats(vj;vl)½P.

More on angles.

Definition.SupposeHandIare halfspace that intersect in a pointo. We set a(H;I) =H\I and note that this set is an angle with vertexoand sidesb(H)\Iandb(I)\H. Suppose that ifRandSare distinct rays with the same origino. There is one and only one angle a(R;S) whose sides areRandS. Its vertex iso. Ifa,b,care distinct points andfa;b;cgis noncollinear we let a(a;b;c) =h(l(b;a);c)\h(l(b;c);a); note that the sides of this angle arer(b;a) andr(b;c) and that its vertex isb. Definition.LetAbe an angle and letHandIbe its bounding halfspaces. The anglesHo\IandH\Io aresupplementaryto the angleH\I. The angleHo\Ioisverticalto the angleH\I. Note that these relations are symmetric becauseJoo=Jfor any halfspaceJ. Definition. Flags.Aflagis an ordered pair (H;R) whereHis a half space andRis a ray contained in theb(H). Definition.Suppose (H;R) is a flag andois the origin ofR. We let

A(H;R) =fS2R(o) :S½Hg:

Theorem.Supposeois a point;Ris a ray with origino; (H;R) is a flag; and Á is the set of ordered pairs (S;T) such thatS;T2A(H;R) and

S½a(R;T):

13

ThenÁis a complete linear ordering ofS.

Proof.SupposeS;T2 SandS6=T. LetIbe the halfspace such thatS½b(I) andR½I. Suppose T6½a(R;S) =H\I. SinceS6=TandT½Hwe haveT½Io. Choose a pointainRand a pointbinT. Sincea2Iandb2Iothe segments(a;b) must meetb(I) in a pointc. Sinces(a;b)½Hwe havec2S.

From the Crossbar Theorem we infer thatS=r(o;c)½a(R;T) soSÁT. It follows thatÁis trichotomous.

SupposeS;T;U2 S,SÁTandTÁU. Choose a pointainRand a pointdinU. By the Crossbar

Theorem,

a(R;U) =fr(o;x) :x2s(a;d)g: BecauseTÁUthere is a pointc2T\s(a;b). By the Crossbar Theorem, a(R;T) =fr(o;x) :x2s(a;c)g:

BecauseSÁTthere is a pointb2S\s(a;c) from which it follows thatS½a(R;U). ThusÁis transitive.

SupposeTis a nonempty subset ofSanduis a member ofSwhich is an upper bound forT. IfU2 T thenUis a least upper bound forTso let us supposeU62 T. Choose a pointainRand a pointbinU. LetS2a(R;T),T2a(S;U): Proof.Straightforward exercise in the Crossbar Theorem. 14

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