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DEVELOPMENT OP A GEOMETRY PROM A SET OF AXIOMS

APPROVED:

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Major Professor

~rA.&~o Minor Professor

JL~~ Chn, Director of the Department of Mathematics

Dean Vof the Graduate School

Glasscock, Anita Louise, Development of a Geometry

from a Set of Axioms. Master of Science (Mathematics),

May 1973, ^8 pp.

The purpose of this paper is to develop a geometry

based on fourteen axioms and four undefined terras. The

first eight axioms in Chapter I are order axioms. Following

a sequence of theorems which can be proved from these eight

axioms, a ninth order axiom is added. Additional theorems,

which could not have been proved without the ninth axiom,

are then stated. The remainder of Chapter I deals with the

statement of four axioms of congruence and the theorems

that result.

Chapter II is a study of the consequences of adding a

"Dedekind-cut" type axiom, while Chapter III deals with

topological properties of linear point sets.

B "-}• 9 -

DEVELOPMENT OP A GEOMETRY FROM A SET OF AXIOMS

THESIS

Presented to the Graduate Council of the

North Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF SCIENCE

By

Anita Basshara Glasscock, B. S. E,

Denton, Texas

Kay, 1973

TABLE OF CONTENTS

Chapter Page

I. ORDER AND CONGRUENCE AXIOMS 1

II. COMPLETENESS AXIOM 17

III. TOPOLOGICAL PROPERTIES 39

CHAPTER: I

ORDER AND CONGRUENCE AXIOMS

The purpose of this paper is to develop a geometry

based on fourteen axioms and four undefined terms. The

first eight axioms in Chapter I are order axioms. Following

a sequence of theorems which can "be proved from these eight

axioms, a ninth order axiom is added. Additional theorems,

which could not have been proved without the ninth axiom,

are then stated. The remainder of Chapter I deals with the

statement of four axioms of congruence and the theorems

that result.

Chapter II is a study of the consequences of adding a

"Dedekind-cut" type axiom, while Chapter III deals with

topological properties of linear point sets.

Since this thesis is an outgrowth of a "Fundamentals

of Geometry" course, many of the theorems will be stated

without proof. It is necessary, however, to state them in

order to establish the background on which the remainder

of the thesis is developed. Proofs are included for some of

the more interesting background theorems.

The following terms are left undefined: point, line,

and between.

Order Axiom I. Every line is a set of points.

Order Axiom II. There exist three points such that

no line contains all of them. Such points are said to be

noncollinear.

Order Axiom III. If A and B are points, there is one

and only one line to which they both belong.

Order Axiom IT. If A and B are points, there is a point

G such that ABC. The notation WABCH means that B is between

A and C.

Order Axiom V. If ABC, then A, B, and C are different

and there is a line containing A, B, and C.

Order Axiom ¥1. If AEC is true, then CAB is not true.

Order Axiom VII* If A, B, and C are non-collinear

points, and D and E are points such that ABD and CEB, then

there exists a point P such that FED and AFC.

Order Axiom VIII. If A, B, and C are three collinear

points, then either ABC, ACB, or CAB.

Theorem 1.1. If A, B, and C are three collinear points,

then one and only one of the following holds: ABC, ACB, or

CAB.

Proof. At least one of the above holds by Order

Axiom VIII. Suppose two of the above hold. Since ABC and

CAB cannot both hold by Order Axiom VI, consider the

remaining cases::

Case 1. Suppose ABC and ACB in line AC, Order Axiom

VIII implies that since B, A, and C are in line AC, one of

the following holds: BAG, BCA, or CBA. Since ACB- is true,

then BAG is not true. Also, BCA implies that ABC is not true,

Dr* A 4 «-• ^ A 1 A r'UA 4 n 4-v>n A A f Q 4 n

is not true, thus CEA is not true. The assumption that ABC.

and. ACB leads to a contradiction.

Case 2. Suppose ACB and CAE in line AC.. Order Axiom

VIII implies that since B, C, and A are in line AC, one of

the following holds: BCA, BAC, or ABC. But ACB implies

that BAC' is not true and CAB implies that BCA is not true.

Since ABC implies that CAB is not true, ABC is not true.

Thus, ACB and CAB do not both hold.

Therefore, at most one of the above holds, and the

theorem is proved.

Theorem 1.2. If ABC, then CBA.

Proof. Suppose ABC. This implies that A, B, and C

are collinear. By Theorem 1.2, exactly one of the following

holds: ABC, ACB , or CAB. Since ABC, then ACJEP and CAB. are

not true. Also, by the same theorem, since C, B, and A. are

collinear, exactly one of the following must hold: CBA,

CAB, or ACB. So, since ACB and CA& are not true, CBA is

true.

Theorem 1.3. If A and C are points, then there is a

point*B such that ABC.

Theorem 1.4. If ABC. and DCB, then ABD and ACD.

Proof. Suppose ABC and DCB:. There exists a point P,

noncollinear with A and C. Then, by Order Axiom IV, there is

a point E, such that BFE, and thus EFB by Theorem 1.2. The

points A, B, and E are noncollinear, and F and C are points

such that ABC and EB'B, then Order Axiom VII implies that there

exists a point G such that GFC and AGE. (See Figure 1.1)

Pig. 1.1

The points D, C, and G are noncollinear, and D and F •

are such that DCB and GFC, so there exists a point H such that

HFB and DHG. (See Figure 1.2)

Pig. 1.2

The points A, G, and D are noncollinear, and H and E

are points such that AGE and DHG, so there exists a point I

such that IHE and AID. (See Figure 1.3)

D

r--B.

Fig. 1.3

Since BFH, BFE, and IHE, then B, F, H, I, and E are

collinear. Line BF intersects line AC at only one point, or

else A, C, and F are collinear. But Aid and IHE imply

that I is in line BF and in line AC, Also, ABC and HFB:

imply that B is in line BP and in line AC, thus I = B, and

AID implies ABD.

Since from the above proof XYZ and WZY imply XYW, then

DCB and CBA imply DCA, which is ACD by Theorem 1.1. The

theorem is proved.

Theorem 1.5. If A, B, and C are noncollinear points,

then there do not exist three collinear points D, E, and F

such that ADB, BEC, and CFA.

Theorem 1.6. If ABC and AGD, then ABD and BCD.

Theorem 1.7. If ABD, ACD, and B 4 C, then either

ABCD or ACBD.

Theorem 1.8. If ABC, ABD, and D ^ C, then either

ABCD or ABDC.

Theorem 1.9. If A, B, and C are noncollinear points,

and D and F are points such that ABD and AFC, then there

exists a point E such that BEC and DEF.

Definition 1.1. If A and B are points, then

Segment AB = [X : AXBj,


Interval AB = fx : AXBj U £a, bJ. Interval AB will be

denoted by AB.


Ray AB = £x : ABx] U Interval AB. Hay AB will be denoted

by A ! .

Theorem 1.10. If A and B.are points, and C 4 A is a

point in AB, then AC is AB..

Theorem 1.11. If in is a line and. A is a point in m,

then there exist only two rays that start at A, and. lie in

m. The rays have only A in common.

Definition 1.4. Suppose m is a line and A is a point

not in m. The A-side of m is the set of all points X such

that either X = A, or X 4 A and m contains no point of AX.

Definition 1.5. Suppose m is a line and A is a point

not in m. The non-A-side of m is the set of all points X

such that X is not in m and there is a point of m between

A and X.

Theorem 1.12. If m is a line, A is a point not in m,

and X is a point, then exactly one of the following is true::

(1) X is in ra, (2) X is in the A-side of m, or (3) X is in

the non-A-side of m.


X is a point in m, and B is a point in the non-A-side of m,

then every point between X and B is in the non-A-side of m.

Proof. Suppose m is a line, A is a point not in m,

and X is a point in m. Suppose B is a point in the non-A-side

of m. Let C be a point such that XCB. Since B is in the

non-A-side of m, there is a point Q in m, such that AQB.

By Theorem 1.9, XCB and AQB imply that there is a point P

such that XPQ and CPA. Since X and Q are in m, P is in m.

Thus, CPA implies that C is in the non-A-side of m, so

every point between X and B is in the non-A-side of m.

(See Figure 1.4)

Pig. 1.4

Theorem 1,14. If m is a line, A is a point not in m,

and X and Y are points in the non-A-side of i, then

there is not a point of m between X and Y,


and B and C are in the non-A-side of m, then every point

between B and C is in the non-A-side of m.

Consider the Euclidean 3-space. It is found to satisfy

the first eight axioms and, therefore, the theorems proved

thus far apply to this set of points. Notice that if m is

a line and A is a point not in m, the non-A-side of m

consists of the points which are in the same plane as A

and m and lie in the opposite side of m from A. The A-side

of m is not just a plane, though. In fact, it includes all

points in the 3-space which are neither in m nor in the

non-A-side of m. It is thus impossible to prove a propo-

sition such as the following: If X is in the A-side of m,

and Y is in the non-A-side of m, then there is a point of m

between X and Y. However, after adding the ninth axiom,

the set of points which satisfy the axioms is restricted

to all the points in a plane and many previously unprovable

8

propositions, including the one just mentioned, become

theorems•

Order Axiom IX. If A, B, and C are noncollinear

points and m is a line which contains a point between A and

B, then either m contains A, or m contains B, or m contains

C, or m contains a point between B and C, or m contains a

point between A and C.


and B and C are points in the A-side of m, then there is no

point of m between B and C.

Theorem 1.1?. If m is a line, A is a point not in m,

X is in the A-side of m, and X is in the non-A-side of m,

then there is a point P in m such that XPY.

Definition 1.6. The point set M is a side of line m

provided there exists a point A not in m such that M is the

A-side of m.

Theorem 1.18. If m is a line, then m has only two sides

and they have no point in common.

Definition 1.7. If A, B, and C are noncollinear points,

then Angle ABC is the union of BA and BC. Angle ABC will be

denoted by /.ABC.

Definition 1.8. The interior of /.ABC is the set of all

points lying both in the A-side of line BC and in the C-side

of line AB.

Definition 1.9. The exterior of /.ABC is the set of all

points that lie neither in /ABC nor in the interior of

£ABC.

Theorem 1.19* If X and. Y are points in the interior

of ZABC, then there is no point of £ABC which lies between

X and Y.

Theorem 1.20. If X is in the interior of Z.ABC, and I

is in the exterior of Z.ABC, then there is a point of Z.ABC

lying between X and X.

Four axioms which relate to the undefined term,

congruence, will now be stated. The notation AB e CD will

denote AB is congruent to CD.

Congruence Axiom I. If A' and B' are points and B and

C are points, then there exists one and only one point C* such

that A'B'C' and BC a B'C'.

Congruence Axiom II. If AB a CD and CD = EF, then AB » EF.

Congruence Axiom III. If ABC, A'B'C', AB s A'B', and

BC s B'C', then AC = A'C'.

Congruence Axiom IV. If A, B, and C are noncollinear

points, A', B', and C' are noncollinear points, D is a point

such that ABD, D' is a point such that A'B'D' and AB s A'B',

BC s B'C1 , AC s A'C' , BD e B'D» , then CD £ C D 1 .

Theorem 1.21. Every interval is congruent to itself.

Proof. Suppose BC is an interval. Let A be a point

such that ABC. Congruence Axiom I implies that there exists

one and only one point P such that ABP and BC s BP, since

A and B are points and B and C are points. Suppose C / P.

Two cases are implied by ABP and ABC.

Case 1. Suppose ABCP. Since A and B are points, and

B and P are points, .there is one and only one point Q such

10

that ABQ and BP s pQ, Since B and P are points, and C and P

are points, there exists exactly one point H such that

BPR and CP = PR. Congruence Axiom III implies that since

BCP, BPS, BC s BP, and CP = PR, then BP = BR. By Congruence

Axiom I one and only one point S satisfies ABS and BC s BS.

Thus B = R, since ABP and BC = BP, and also, ABR and BC s BH.

This contradicts BPR, so this case does not hold.

Case 2. Suppose ABPC. Since A and B are points, and

B and P are points, there exists one and only one point Q

such that ABQ and BP s BQ. Since B and Q are points, and

P and C are points, there is exactly one point S such that

BQS and PC = QS. Since BPC, BQS, BP = BQ, and PC = QS, then

BC = BS. Congruence Axiom II implies that since BC « BP

and BP = BQ, then BC = BQ. Congruence Axiom I Implies that

there is one and only one point R such that ABR and BC = BR.

Thus P = S = Q, since ABP, ABS, ABQ, BC £ BP, BC 5 BS, and

BC = BQ. But BQS implies that Q ^ S, and thus a contradiction

is found.

Since the supposition that C 4 P leads to a contra-

diction, C = P, which implies that BC = BC.

Theorem 1.22. If AB = CD, then CD = AB.

Theorem 1.23. If AB = CD and AB = EF, then CD s EF.

Theorem 1.24. If ABC, A'B'C', AB h A'B1, and AC £ A'C',

then BC = B'C'.

Theorem 1.25. If AC = A'C1 and ABC, then there exists

one and only one point B' such that A'B'C', AB = A'B', and

nn ~ t-i l r% I

11

Definition 1.10. The statement that AB is less than CD

means that there exists a point X such that CXD and either

AB = CX or AB = XD, The notation AB < CD will be used.

Theorem 1.26. If AB < CD, then there exists a point X

such that CXD and AB s CX.

Theorem 1.27. If AB < CD and CD e EF, then AB < EP.

Theorem 1.28. If AB < CD and CD < EP, then AB < EF.

Theorem 1.29. If A and B. are points and C and D are

points, then one and only one of the following holds:

AB b CD, AB < CD, or CD < AB.

Definition 1.11. The statement that /ABC £/A'B'C'

means that there exist points X and I, both distinct from

B', one lying in B'A' and the other in B'c"' such that

BA m B'X, BC s B'Y, and AC « XY.

Theorem 1.30. Every angle is congruent to itself.

Theorem 1.31. If /ABC s/A'B'C', then /A'B'C' s/ABC.

Theorem 1.32. If /ABC s /A'B'C' and /A'B'C' i/A,,B,,C"l

then ZABC s Lkx »B' 'C''.

Theorem 1.33. If /ABC a /.A'B'C' , BA e B'A' , and

BC m B'C', then AC b A'C'.

Definition 1.12. The statement that M is a midpoint

of AB means that AMB and AM h MB.

Theorem 1.3^. No interval has two midpoints.

Definition I.13. The statement that/.AXB andZA'Xa1 are

vertical angles mean that /AXB and /.A'XB' are angles such

that either AXB' and BXA1 or AXA' and BXB'.

Theorem 1.35. Vertical angles are congruent.

12

Definition 1.14. The statement that ZABC and Z.A'B'C1

are supplementary angles means that there exist points X

and Y, both distinct from B, one lying in Bk* and the other

in BC* such that either ABX and Y is in or CBX and Y is

in B^.

Theorem 1.36. Supplementary angles of congruent angles

are congruent.

Definition 1.15. The statement that ZABC is a right

angle means that ZABC is congruent to one of its supplements.

Theorem 1.3?. If ZABC s ZA'B'C' and ZA'B'C1 is a right

angle, then ZABC is a right angle.

Theorem I.38. If D is a point in the interior of

ZABC, then BD contains a point between A and C.

Proof. Suppose D is a point in the interior of ZABC.

Let G be such that ABG and F be such that DBF. Theorem 1.12

implies that since A is not in line BD that exactly one of

the following holds:

(1) C is in the A-side of line BD#

(2) C is in line BD, or

(3) C is in the non-A-side of line BD.

Consider each case:

Case 1. Suppose C is in the A-side of line BD. There

is a point X such that GXD and X, is in ZABC, by Theorem

1.20. Thus, X is in or Bit. Since G is in the non-A-side

of line BD, D is in line BD, then Theorem 1.13 implies X

is in the non-A-side of line BD. But since every element

of BA and every element of BC is in the A-side of line BD,

13

X is not in £ABC. This is a contradiction, so C is not in

the A-side of line BD. (See Figure 1.5)

Fig. 1.5

Case 2. Suppose C is in line BD. This case does not

hold because if (a) C is in BB then BC = BD and thus D is

not in the interior of LkBC, or (b) if C is in BF, then CBD,

but this is a contradiction because there does not exist a

point of Z.ABC between a point of the angle and a point of the

interior.

Case 3. Since neither of the first two cases is true,

then C is in the non-A-side of line BD. Then, by the defi-

nition of the non-A-side of line BD, there exists a point Y

in line BD between A and C. Suppose Y is in BF. Then FBD

and YFB (or FIB) imply YBD. But since Y is in the interior

of ZABC, then D is in the exterior of ZABC, which is a

contradiction. Thus, Y is in HS and Y is such that AYC.

Theorem 1.39. If D is a point in the interior ofLABC,

then every point of BB, except B, is in the interior of £ABC.

Theorem 1.^0. If ^ABC s A'B'C' and X is in the interior

of ABC, then there is a point Y in the interior of A'B'C'

such that ZABX s M'B'Y.

14

Proof. Suppose /ABC s /A'B'C1 and X is in the interior

of ZABC. Let Q be a point such that QB'A' and let H be a

point such that RB'C*. Thus, Q and B1 are points, and

B and A are points, so Congruence Axiom I implies that there

exists one and only one point A M such that QB'A'1 and

BA s B'A". Since B and B1 are points and B and C are points,

there exists one and only one point C M such that RB^ 1'

and BC m B'C". Theorem I.38 states that because X is in

the interior of ZABC, there exists a point X1 in BX such

that CX'A. Let S be a point such that SA'^". Because

S and A'1 are points, A and X' are points, there exists one

and only one point Y such that SA''Y and AX1 = A''Y.

Point Y is in the interior of M'B'C' since

A"Y s AX1 < AC s A " C M . Also, BX1 s BY since

£BAX' = IBAC s ZB'A' 'C' ' = £ B1 A1 'I, BA = B'A'1 , and AX' = A11Y,

Thus, Y is in the interior of /A'B'C and ABX = A"B'Y

since BA s B'A11, BX' b B'Y, and X'A w YA''. (See Figure 1.6)

Fig. 1.6

Definition 1.16. The statement that XZ is the sum of

AB and CD means that there exists a point Y in XZ such that

AB s XY and CD £ YZ, or AB s YZ and CD s XY.

Theorem 1.41, If XZ is the sum of AB and. CD, then

there exists a point P in XZ such that AB = XP and

CD = PZ.

Proof. Suppose XZ is the sum of AB and CD. One of

the following holds:

Case 1. There exists a point Y in XZ such that AB = XI

and CD s XZ. By letting P = I, the theorem is proved for

this case.

Case 2. There exists a point I in XZ such that AB s YZ

and CD s XY. Let E be such that EXZ. Thus, E and X are

points, A and B are points, so there exists a point P such

that EXP and AB = XP. Since XP = AB < XZ and EXP, then

EXPZ. Since XPZ, ZYX, XP H AB = ZY, and XZ s ZX, then

Theorem 1.24 implies PZ s YX s CD. Thus, P is in XZ such

that AB s XP and CD = PZ, and the theorem is proved.

Theorem 1.42. If AB is the sum of PQ and RS, and CD is

the sum of PQ and RS, then AB s CD.

Proof. Suppose AB is the sum of PQ and RS, and CD is

the sum of PQ and RS. It is implied by the previous theorem

that there exists a point X such that AXB and PQ e AX and

RS * XB. Also, there exists Y such that CYD and PQ a CY

and RS s YD.

Since AXB, CYD, AX s PQ = CY, and XB 5 RS a YD, then

AB s CD by Congruence Axiom II. The theorem is proved.

Theorem 1.43. If EF = GH and EF is the sum of AB and

CD, then GH is the sum of AB and CD.

16

Proof. Suppose EF = GH and EP is the sum of AB and CD,

Theorem 1.41 implies that there exists a point X such that

EXP and EX = AB and XF = CD. Let J be such that JGH. Since

J and G are points, and E and X are points, there exists a

point Y such that JGY and GY £ EX. Since JGH, JGY, and

GY s EX < EF s GH, then GYH. Also, EXF, GYH, EX £ GY,

and EF = GH imply by Theorem 1,2k that XF. s YH.

Therefore, Y is such that GYH and GY = EX = AB and

YH = XF = CD, so GH is the sum of AB and CD, and the theorem

is proved.

CHAPTER II

COMPLETENESS AXIOM

Chapter II is a continuation of Chapter I with the

addition of a completeness axiom. This axiom greatly

increases the number of provable propositions to include

such important results as the following three:

(1) Every line segment has a midpoint.

(2) Right angles are congruent.

(3) An angle can be reconstructed.

Completeness Axiom. If m is a line and and are

nonempty subsets of m such that S2 = 0 and S1 U S2 - a,

and if A. and Ag are in S^, either A.jA2 C S^ or A^k^C. S^,

and if B1 and B2 are in S2> either BjB2 C S2 or C S2;

then there exists a point X such that if A is in and BL

is in S2, then either XA = or XB = S2#

Theorem 2.1. If C is a point not in line n and B. is

a point in n, then there exists a point Y such that CY

does not intersect n and if X is a point in the interior of

/.BCY, then CX does intersect n.

Proof. Suppose n is a line, A and B are points in n,

and C is a point not in n. Let D be a point such that ACD.

Consider line m = line DB.

Let = [w in i ; CW does not intersect nj and

S 2 » . (z in m : CZ does intersect nj.

17

18

Note that S 2 « 0 and U S 2 = m. Also, note that S 1

and Sg are each nonempty since D is in or else A, G, and

B are collinear, and A is in S2» (See Eigure 2#1)

Pig. 2.1

It must be shown that if Ax and Ag are in S 1 § either

AÂ2 C or A ^ x C Suppose A ^ and Ag are in S^.

Consider these cases: A^BA^, AÂgB., and AÂ^B.

Case 1. Suppose A^BA^. This implies that A^ is in

the non-A^-side of line n. If C is in the A^side of n,

then there exists a point of n between C and A2, which is a

contradiction. If C is in the non-A^side of n, then there

exists a point of n between C and Alf which is also a con-

tradiction, The assumption that A^BAg does not hold.

Case 2. Suppose AÂ^B:. Let D' be such that .

Every point of CD' is in the non-B-side of line CAg. If — ^

E in m is a point such that CE intersects n at F, then E

is in the non-A-side of m since the C-side of m = A-side

of m. Every point of n in the non-A-side of m is in the — y ^

B-side of line CA2 or else CA2 intersects n. Thus CD'

does not intersect n. Thus A2A^ C S^.

Case 3. Suppose AgA^B. A similar argument to Case 2

considering the B and non-B-sides of line CA^ leads to the

19

conclusion that AÂ^C. S^. Thus, if A^ and A^ are in S^,

either AÂ^ C S^ or AÂ^ C S^.

Similarly, if and B^ are in S^, then either

B1 B2 S2 o r B2B1 ^ S2*

The Completeness Axiom implies that there exists a

point Y such that if I is in and J is in S^, then either

a - . , - * . . , .

Suppose YJ * Sg. Then Y is in S^, so CY intersects n

at a point P. Let G be a point such that BFG, Note that

CYF implies that C is in the non-F-side of m, so there

exists H in m such that CHG. Thus, H is in S^. But CHG

implies that H is in the G-side of line CF and BFG implies

that B is in the non-G-side of line CF, so H is not in YJ

and must be in S^. A contradiction has been found. — ^

Therefore, YI = S^ and Y is a point such that CY does

not intersect n, and if X is in the interior of ZBCY, then

CX does intersect n.

Lemma 2.1. If AC is a segment, then there is a point X

between A and C such that AX < XC,

Proof. Let AC be a segment and Y be a point such that

AYC. Consider these cases: (1) YC < AY, (2) AY £ YC, or

(3) AY < YC. The following shows that in each of these

cases a point X such that AXC and AX < XC can be constructed.

Suppose Y is such that YC < AY. By definition, this

implies that there exists a point Z such that AZY and AZ = YC,

Since AC = CA and AZ = YC, then by Theorem 1.24, AY = CZ.

Thus, AZ h YC.< AY =.ZC. Let Z = X.

20

Suppose Y is such that AY = YC. Let X be such that

AXY. Because AYC, AXYC is implied. Thus AX < AY s YG; < CX.

Suppose AY < YC. Let Y = X.

In each case it has been shown that if AC is a segment,

then there is a point X between A and C. such that AX < XC.

Theorem 2.2. If AC is a segment, then there exists a

point B such that ABC and AB = BC.

Proof. Suppose AC is a segment. Let

S x = {X : AXC and AX s xc j U {w : WAc} U (a) and

S 2 = (Y : AYC and GY < Ya3 U [Z : ACZ} U [c).

Note that S 1 U S 2 = line AC and S 1 A S 2 = 0. Also, S 1

and Sg are nonempty subsets of line AC.

Suppose A^ and A^ are in S^. Then AÂÂ, AÂÂ, AÂA^,

A^ = A, or Ag = A. Consider each case:

Case 1. Suppose AjAÂ. If A^ is such that AA^C, then )

AAÂ^C implies that A-jA,, C S^. If A 2 is such that AÂC, then

AÂÂC implies that AÂ^ C S^.

Case 2. Suppose AÂÂ. A similar argument to Case 1

leads to the desired conclusion.

Case 3. Suppose AÂA,,. If A^ is such that AA^C, then

AjAA^C implies that AgA^ C s ^ I f A 2 i s s u c h t h a t A2 A G»

then AÂA^C implies that AÂ^ C. S- .

Case b. Suppose A^ = A, If is such that AA^C, then

AÂ^C implies that AÂ^ C S^. If A^ is such that AÂC, then A2 A1 C i mP l i e s that A ^ g C S^.

Case 5. Suppose A 2 = A. The argument in this case is

similar to the one in Case 4.

21

B7 such that if P is an element of and Q, is an element of

Thus in each case for A^ and A^ points in S^, either «" •> -

A1A2 C SI or A^A^ A similar argument shows that for

and Bg points in S^, either B^B2 C S 2 or B^B^ C S2.

The Completeness Axiom implies that there exists a point

at iJ

Sg, then B? = Sx or BQ = S2»

Suppose BQ = S^. Let P be such that FAC. Then B is

such that ACB, B = C, or ABC and BC < AB. Note that ACB

is not true since FAC and ACB imply FCB, but since C is in

Sg, then F is in BC. A contradiction is found since F is

in S^. A contradiction also arises from the assumption

that B = C. If it is supposed that B = C, then the lemma

implies that there exists a point X such that BXA and BX < AX,

Thus, X is in S^. But since A is in BX, then A must be in

Sg, which is a contradiction.

If BC < AB1 and ABC, then by definition there exists a

point H such that AHB and AH s CB. Since AH = CB < AB, then

AHB. The lemma establishes that there exists a point K such

that BKH and BK < KH. Because AHB, BKHf and ABC, then AHK

and CBK. Congruence Axiom III and Theorem 1.27 imply that

since AHK and CBK, CB 5 AH and BK < KH, then CK < AK. This

implies that K is in S^ but CBK implies that K is in S^.

Because of the contradiction, BQ 4 S„. *

The Completeness Axiom implies that BP = S^. By def-

inition of S1, B is such that BAC, B = A, ABC and AB < BC,

or AB ar BC. Similar arguments to those for BQzSg rule out

all cases except AB s BC. The theorem is proved.

22

Theorem 2.3* If C is not in line AB, then there does

not exist a point C' £ C in the C-side of line AB such that

AC s AC' and BC a BC".

Proof. Let A, B, and C be noncollinear points. Suppose

there exists G* 4 C in the C-side of line AB such that

AC = AC' and BC = BC1. Consider the following two cases:

Case 1. Suppose there exists a point P such that P is

in line C C and P is in line AB, (See Figure 2,2) Suppose

PC'C. A similar argument holds if PCC1. Note that AC s AC',

BC s BC* , and AB £ AB, so ZCAB s Z.CAB, and thus iCh? a ^C'AP.

Therefore, PC h PC", but since C' is in line PC, C1 is in

the C-side of line AB, and C ^ C*, then PC p PC'. This case

is thus not true.

Fig. 2.2

Case 2. Suppose there does not exist a point such as

P in Case 1, in other words, every point in line AB is in

the A-side of line CC'. (See Figure 2.3) There exists a point.

M such that CMC and CM s MC*, according to Theorem 2.2.

Let 0' be a point such that AMO'. There exists a point 0

such that AMO and 0 is in the interior of ZCBC' because one

of the following is true concerning 0': (1) If 0* is in the

interior of /CBC1 , let 0' = 0. (2) If 0' is in </CBC', let

23

0 be such that MOO*. (3) If 0' is In the exterior of

LCBC* , there exists a point Q in Z.CBC' such that HQO1 and

0 is such that MOQ.

Fig. 2.3

The implication of AMO is that 0 is in the non-A-side

of line CC . Since B is in the A-side of line CC , there

exists a point N such that N is in line CC' and ONB. Con-

gruence Axiom IV implies that since MC. = MC , CA = CA,

MA = MA, and MO = MO, then CO s C'O. Also, CB s CB, BO s BO,

and CO s C'O imply by definition that £CBO s LCBO. Since

CB 5 C'B, BN = BN, and Z.CB0 = Z.CBN £ LCBO = LC'BN, then

CN = C'N. Thus N is a midpoint of CC different from M.

(If N = M, then ONB = 0MB and OMA imply that line AB inter-

sects line CC.) Theorem 1,34 implies that a line segment

does not have two midpoints, so a contradiction has been

found, and the theorem is proved.

Theorem 2.4. If IABC = lA'B'C', ZACB s ZA'C'B', and

BC £ B'C , then AB s A'B». (Also, AC £ A'C and

ZBAC = ZB'A'C'.

24

Proof. Suppose /ABC s/A'B'C', ZACB £ ZA'C'B', and

BC = B'C'. (See Figure 2.4) The definition of

tABC s ZA'B'C implies that there exist points X in b3 and

Y in bI such that B'C' £ BX, B'A' e BY, and A'C« 3 XY. Since

X is in BC and BG = B'C' s BX, then X = C.

6*3 O X 6'

Fig. 2.4

Because ZACB s^A'C'B', there exist points W in CA and —^

Z in BC such that A'C' a XC, B'C' s ZC, and B'A» s ZW.

Since Z is in BC and BC = B'C1 = ZC, then Z = B.

The point W is in the A-side of line BC since W is in

ct. Thus Y and W are points such that Y is not in line BC,

W is in the Y-side = A-side of line BC, and BY = BW (since

BY a B'A1 = ZW = BW) and CY £ CW (since CY = XY = C'A» s CW).

Theorem 2.3 implies that W = Y.

Since Y and A are both in line BA and line AC, and

Y = W, then I = A = W. Thus AB £ A'B1, AC £ A'C', and

Z.BAC = Z.B'A'.e1.

Theorem 2.5. For noncollinear points A, B, and C,

BA a BC if and only if ZBAC s Z.BCA.

Proof. Suppose A, B, and C are noncollinear points.

First, suppose that ZBAC = Z.BCA. Theorem 2.4 implies that

since LBhC = iBCk, ZBCA s ZBAC, and AC £ AC, then BA = BC.

25

Second, suppose BA H BC. LBAC S ZBCA since there exist

points B and A, one lying in CB and the other in CA such

that BA = BC, BC = BA, and AC = AC.

Theorem 2.6. There exists a right angle.

Proof. There exist noncollinear points A, B., and C by-

Order Axiom II, Let Q be a point such that QAC,. (See

Figure 2.5) The first congruence axiom implies that there

exists one and only one point B* such that QAB' and AB & AB',

Let M be the point such that BMB' and BM = MB1. Since

AB a AB», Theorem 2.5 implies that ABB1 s ZAB'B. Also,

Theorem 1.33 implies that since ^ABB' 5 /AB'B, BM £ MB,

AB = AB1, then £BMA = £B'MA. ZBMA is a right angle since

BMB1 and ^BMA = ZlB'MA.

Fig. 2.5

Definition 2.1. Line n is perpendicular to line m

means that nflm = [pj and if X and Y are points different

from P such that one is in n and the other is in m, then

Z.XPY is a right angle.

Definition 2.2. The statement that there exists a

perpendicular to line m at X means that there is a line q

perpendicular,to m such that m fl q = X .

26

Theorem 2.7. If p is perpendicular to m at X, then

there exists a line q perpendicular to m at Y £ X in m.

Proof. Suppose p is perpendicular to m at X and I ^ X

is in m. Let M be such that XM = MY and XMY. Let W ^ X be

in p. (See Figure 2.6)

Fig. 2.6

Congruence Axiom I states that since W and M are points,

and W and M are points, there exists one and only one point

Z such that WM = MZ. The following congruences imply that

ZZYM s /LMXW: XM s MY, ZXMW s LZKI, and WM s MZ.

Angle ZYM must, therefore, be a right angle since it

is congruent to a right angle. (Theorem 1.37) Let

line YZ = q. Since for A £ Y in m and B ^ Y in q, /AXB

is a right angle, and q A m = {Y}, then q is perpendicular

to i at Y and the theorem is proved.

Theorem 2.8. If T is a point in line m, then there

exists a perpendicular to m at T.

Proof. Suppose T is a point in m. Let P be a point

not in m, and Q and R be points in m. (See Eigure 2.7)

Because B and Q are points and P and Q are points, Congruence

Axiom I implies that there exists one and only one point Y

27

such that BQY and PQ = QY. Let M be in line PI such that

PM s MY. Since PQ 2 PI, MQ s MQ, and PM = MY, by definition,

dPMQ s Z.YMQ. Thus, since Z.PMQ and Z.YMQ are supplementary,

Z.YMQ is a right angle.

Eig. 2.7

Let S be such that PYS and U be such that QYU. Note

that ZQYM 5 LSYU since they are vertical angles. The points

M and Y are not the same and Q and Y are points, so there

exists one and only one point z such that MYZ and YZ = QY.

Likewise, Q and Y are points, and M and Y are points, so

there exists one and only one point X such that QYX and

MY a YX. Since ZQYM s ZZYX, YZ = QY, and MY = YX, then by

Theorem 1.33, ZYMQ = Z.YMZ, and is thus a right angle.

By the previous theorem, since there exists a perpen-

dicular to m at X, if T is in m, then there exists a line

perpendicular to m at T. The theorem is proved.

Theorem 2.9. Suppose A, B, and C are noncollinear points

and M is in line AC such that AM = MC. Line BM is perpen-

dicular to line AC if and only if AB £ BC.

Proof. Suppose A, B, and C are noncollinear points, M

is in line AC such that AM = MC, and line BM is perpendicular

28

to line AC. Because £BMA s LBMC, AM = MC, and BM.a BM,

AB s BC by Theorem 1.33*

Suppose A, B, and C are noncollinear points, M is in

line AC such that AM = MC, and AB = BC. Because AB = BC,

AM as MC, and BM = BM, then ZAMB = ZBMC. Thus, line BM is

perpendicular to line AC.

Theorem 2.10. There is one and only one perpendicular

to m at X.

Proof. Theorem 2.8 implies that there exists at least

one perpendicular to m at X.

Suppose there exist lines p and q such that p and q are

perpendiculars to m at X. (See Figure 2.8)

Fig. 2.8

Let Y / X be in m. There exists one and only one point

Z such that YXZ and IX = XZ. Let W ^ X be in p. Lines q and

m are different and X is in q and in m, so YXZ implies that

either Y is in the non-W-side of q or Z is in the non-W-side

of q. Let V be a point in m such that YYZ if Y is in the

non-W-side of q or YZV if Z is in the non-W-side of q. Let

U be a point in q such that WUV. Note that W ^ D since W

29

is in p and U is in q. By the previous theorem, UY s UZ.

and WY - WZ. But since Y and Z are such that UY = UZ,

YW = WZ, and Z and Y are in the Z-side of line VW, by

Theorem 2.3, ¥ = Z. This is a contradiction since YXZ.

Therefore, there exists at most one perpendicular to m at X.

Theorem 2.11. If line p is perpendicular to m at X

and line q is perpendicular to ID at Y / X, then if P ^ X

is in p and Q / I is in q, ZPXY = ZQYX.

Proof. Suppose p is a perpendicular to m at X and q

is a perpendicular to IB at Y / X. (See Figure 2.9)

Fig. 2.9

Let M be such that YMX and YM s MX. Let Q / Y be in q.

Since Q and M are points, there exists P1 such that QMP',

and QM = MP'. Since QM s MP», (YMQ s £XMP', and YM s MX,

then £QYM = LMXP*, By Theorem 1.37, since Z.QYM is a right

angle, then Z.MXP' is a right angle. By the previous theorem,

P' must be in line p. Let P / X be in p, Note that

LPXM a £P«XM = ZP'XY 5 LQ.YM = iQYX. The theorem is proved.

Theorem 2.12. If /ABC is such that B and C are in m,

then if I ^ B is in m and Z / X is in ni, there exists a point

Y not in m such that IABC s £ZXY.

30

Proof. Suppose there exists /ABC such that B and C

are in m. Suppose X ^ B is in m and Z ^ X is in m. There

is one and only one perpendicular p to m at B, Let D ^ B.

be in p such that D is in the A-side of m. There exists

one and only one perpendicular q to ni at X. Let Q ^ X be

in q. (See Figure 2.10) Let C' be in BC such that C' ^ X.

E B rn

Fig. 2.10

Consider the following cases:

Case 1- Suppose A is in p. By the previous theorem

ZABC s ZABC' = ZDBC = ZQXC1 = ZQXZ = IXXZ for Y in q.

Case 2. Suppose A is not in p. Then A is in the interior

of ZDBC or the exterior of ZDBC. If A is in the interior

of ZDBC, by Theorem l.*K), there exists Y in the interior of

^QXZ such that ZABC = Z.XXZ. If A is in the exterior of Z.DBC,

then if E is a point such that CBE, A is in the interior of

Z.DBE, and the same argument follows.

Theorem 2.13* All right angles are congruent.

Proof. Suppose ZABC andZA'B'C' are right angles. Con-

sider these cases:

Case 1. Suppose line BC = line m does not intersect

line B'C' = line p, and suppose X is in p. Also for this

31

case suppose that every point of p is in the non-A-side

of m. (See Figure 2.11)

Fig. 2.11

There exists D in m such that ADX. By the previous

theorem there exist points Y in p and Z not in p such that

Z.YXZ a ZADB. For J such that JXY, since B and D are points,

and J and X are points, there exists a point Y1 such that

JXY' and BD = XY1. Also, A and D are points, and D and X

are points, so there exists a point Z1 such that DXZ' and

AD s XZ' . Since AD s XZ» , ZADB = ZlY'XZ' , and BD s XY» ,

then ZABD = Z.XY'Z'. Theorem 2.11 implies that £XY'Z' s^A'B'C',

so that /ABC = ZABD = ZA'B'C'.

Case 2. Suppose m does not intersect p and that every

point of p is in the A-side of m. Then there exists A'• in

the non-A-side of m such that BA = BA1' and AC = A'1C, and

a similar argument leads to the conclusion.

Case 3» Suppose m intersects p at H and line AB

intersects line B'C at G. (See Figure 2.12)

Fig. 2.12

32

Assuming G and H are points, and. since B and H are

points, there exists a point J such that GHJ and BH « HJ.

Also, since B and H are points, and G and H are points,

there is a point K such that BHK and GH = HK. Since BH = HJ,

Z-BHG s IJHK, and GH 5 HK, then £GBH £ £KJH. Thus,

LABC = /ABH s CQBU = /KJH a ZA'B'C .

Case 4. Suppose m intersects p at H and line AB does

not intersect line B'C'. A similar argument to Case 1 leads

to the conclusion.

Case 5. Suppose m intersects p at H and line AB

intersects line B'C' at H, Let n be any perpendicular to

m other than line AB. If n intersects line B'C', then the

same argument as Case 3 leads to the conclusion. If n does

not intersect B'C', then the same argument as Case ^ leads

to the conclusion.

Case 6. Suppose m = p. Theorem 2.11 implies that since

line BA is perpendicular to m at B and line B'A' is perpen-

dicular to m at B1, then IABC s ZA'B'C'.

Therefore, all right angles are congruent.

33

Theorem 2.1^. If AB a DE and C is a point, then there

exists a point F such that AC = DF and CB = FE.

Proof. If A, B, and C are collinear, then the theorem

follows from Congruence Axiom III or from Theorem 1.25*

Suppose A, B, and C are noncollinear. Let X be a point

such that AXB. By Theorem 2.8, there exists a line p perpen-

dicular to line AB at X. By Order Axiom IX, since A, B, and

C are noncollinear and X is such that AXB, there is a point

Y in p such that AYC, CYB, or.Y = C. Consider the case for

AYC. The other cases are similar.

Suppose ACY. (See Figure 2.13) Let G be a point such

that GDE. Since G and D are points and A and X are points,

there exists a point X' such that GDX' and DX' = AX. There

exists a perpendicular q to line DE at X*• Let Z ^ X* be

in q. Because Z and X' are points and X and Y are points,

there exists a point Y' such that ZX'Y" and X'Y' « XY.

Fig. 2.13

Theorems 2.5 and 2.13 imply that Z.YAX s Z.Y'DX' since

AX s DX" , /AXY = Z.DX1 Y' , and XY s X'Y1. Let H be a point

such that HDY1. There exists a point F such that AC = DF

since G and D are points and A and C are points. The point

3*

P is also such that CB s PE because AC = DP, ZCAB s ^FDE,

and AB s DE. The theorem is proved.

Theorem 2.15. If P is a point not in line m, there

exists at least one perpendicular to m containing P.

Proof. Suppose P is a point not in m. Let A and B- be

points in m. By Theorem 2.14, there exists a point P' in

the non-P-side of m such that AP = AP' and Pfi = P'B. (See

Figure 2.14)

p

Tna

Fig, 2.14

Since AP b AP' , PB s P'B, and AB s AB, then PAB s Lp'AB,

Consider the case when B7 is in the interior of ZPAP' and the

case when B is not in the interior of £PAP',

Case 1. Suppose B is in the interior of £PAP*. By

Theorem 1.38> there is a point C in AB such that PCP'. Since

AP b AP', ZPAB = Z.PAC = Z.P'AB = ZP'AC, and AC s AC, Theorem

1.33 implies that £PCA = Z.P'CA. Also, since supplements of

congruent angles are congruent, /.PCA = ZpCB, and there exists

at least one perpendicular (line PC) to m at P.

Case 2. Suppose B is in the exterior of £PAP*. Let

B' be such that BAB". The point B' is in the interior of

Z.PAP' and ZPAB1 s Z.P'AB' , since they are supplements of

35

congruent angles. The rest of the argument is similar to

Case 1. The theorem is proved.

Definition 2.3. Angle A'B'C' < /ABC means that there

exists a point X in the interior of /ABC such that

ZABX a /A'B'C' or /XBC £ ZA'B'C.

Theorem 2.16. If /ABC is an angle and /.DEF is an angle,

then one and only one of the following holds:

(1) £ ABC a /DEF, (2) /ABC < IDEE, or (3) /.DEF < /ABC.

Proof. Suppose /ABC is an angle and IDEE is an angle.

Let G be such that GEP. There exists a point C* such that

GEC1 and BC a EC', because B and C are points, and G and E

are points. Theorem 2.14 implies that there exists A' such

that CA s C'A', AB = A'E, and A' is in the D-side of line EE.

One and only one of the following holds:

Case 1. Suppose A' is in EF. This is not true or else

A is in line BC.

Case 2. Suppose A' is in ED. This implies by def-

inition that /ABC = /.DEP.

Case 3. Suppose A' is in the interior of /.DEP. The

Definition 2.3 implies that in this case /ABC < /DEP. (See

Figure 2.15)

& E

Fig. 2.15

36

Case Suppose A1 is in the exterior of ^DEF. This

implies that £DEF < Z.ABC.

Theorem 2.17. If UBC < Z.DEF < Z-GHI, then

LABC < ^GHI.

Proof. Suppose ZABC < ^DEF < £GHI. Since /ABC < ZDEF,

there exists a point X in the interior of ZDEF such that

Z.DEX s MBC or Z.FEX s ^ ABC. Since Z.DEF < Z.GHI, there exists

Y in the interior of LGHI such that £GHY £ LDEF or

IIHY £ Z.DEF.

Case 1. Suppose £DEX a ZABC, and £GHY s LDEF. (See

Figure 2.16)

Y_ t

/v 1

Fig. 2.16

Let G» be in HG such that ED = HG1. Let I1 be in HI

such that EF = HY'. Since Z.GHY = £G'HY = ZDEF, HG' 2 ED,

and HY' = EF, then G'Y1 = DF. Theorem 1.38 implies that

EX contains X1 such that DX'F. Theorem 1.25 implies that

since G'Y' = DF and DX'F, then there exists one and only

one point Z such that G'ZY* and G'Z s DX'. Now it must be

shown that Z is in the interior of £GHI and that

ZG'HZ = £GHZ 5 £ABC.

37

The point Z is in the interior of ZG'HY and. every

element of ZG'HY' is in the interior of /.G'HI since every

element is in the I-side of line G'H and in the G'-side of

line IH.

Because G'H - ED, G'Z = DX', and

ZHG'Z = ZHG'Y » ZEDX' = ZEDF, then HZ s EX'. Since H£ s EX',

HG' s ED, and X'D s G'Z, then

ZDEX' = ZDEX h ZABC s ZG'HZ = /GHZ. Therefore, /.GHZ £ /ABC.

The other three cases are listed below. Each leads

to the conclusion by a similar argument to Case 1.

Case 2. Suppose ZDEX « ZABC and ZIHY = ZDEF.

Case 3. Suppose ZFEX = ZABC and ZGHY = /DEP.

Case Suppose /.FEX s /ABC and /IHY = ZDEF.

For each case ZABC < ZGHI and the theorem is proved.

Definition 2.4. M is a bisector of ZABC means that M

is in the interior of /ABC and ZABM « ZCBM.

Theorem 2.18. Every angle has exactly one bisector.

Proof. Suppose ZABC is an angle. Let D be such that

DBC. There is a point E such that DBE and BA s BE since

D and B are points, and A and B are points. (See Figure 2.17)

Fig. 2.17

38

Let M. be such that AME and AM = ME. Since BA s BE,

BM = BM, and AM s ME, then IBM. = Z.BME, and there is a

bisector of tABC, namely, BM.

Suppose BN is also a bisector of 4ABC. By definition,

N is in the interior of /-ABC and ZABN = £EBN. Let H be. a

point such that HBN. There exists a point M* in BN (by

Congruence Axiom I) such that BM* = BM. Since

ZABN = £ABM» s ZEBN = Z.EBM1 , AB 5 BE, and BM' s BM' , then

AM' a M'E. Thus, M* is a midpoint of AE. No interval has

two midpoints, however, by Theorem 1.13. Thus, M « M1 and MI

BM = BM' « BN, and it is shown that every angle has exactly

one bisector.

CHAPTEB III

TOPOLOGICAL PROPERTIES

Chapter III deals with further implications of the

fourteen axioms, particularly the last. This sequence of

theorems concerns the topological implications, rather than

the geometrical, as in Chapter II.

Theorem 3.1* The set of points in an interval is at

least countably infinite.

Proof. Suppose AB is an interval. By Theorem 1.1,

there exists a point P- such that AP^B. Suppose there

exist n points such that APnPn__^.. .B. Then by Theorem 1.1,

there exists P n + 1 such that Ap

n+ipnpn_i

pn,_2* * *

B» a n d

Theorem 1.5» all these points are elements of the interval.

Thus, the set of points in an interval is at least countably

infinite.

Theorem 3.2. If pA^B^ and i s a sequence of

intervals such that A B C A^ ,8^ , C ... C.A-B,, then n n n-1 n-1 ^ 11*

there exists a point G such that [gJ ^ •

Proof. Suppose P, A^, and B^ are three points such

that PA^B^ and is a sequence of intervals as described

in the statement of the theorem.

Define S 1 and S 2 as follows:

S-j = [ x in A ^ : PX s PB^ for all n} U

[W in line A^B^ : B^A^wJ and

39

40

S 2 = {y In k 1B 1 : PAn < PY for all n} U

{Z in line A^B^ : PB^z}.

If s 2 ^ 0, then the theorem follows. Thus, suppose

that S 1 A S 2 s 0.

It must be shown that si u S2 = l i n e AlBl* SuPPOs© x i s

a point in line A^B^. If X is such that XA^B^ or A^B^X,

then X is in S-j or S2. If X is A x or Bx, then X is in the

nonempty set or the nonempty set S 2. Finally, suppose

X is in Since S ] [A S 2 = 0, either there is a B n such

that PB < PX or there is an Am such that PX < PA . If X H lu ia

is such that PB^ < PX, then PA, < PX for all i, since XI A

PAA < PBn for all i. If X is such that PX < PAffl, then

PX i for all i, since PAffl ! PBi for all i. Thus, in

each case, X is in S. U S2. Since every point of S x and

every point of S 2 is in line A^B^, then U = line A^B^.

Suppose A and B are points in S^. Consider the

following cases:

Case 1. Suppose A and B are in A^B^. If PA^AB and X

is such that PXB, then PX < PB i PBn for all n toy Definition

1.10 and the definition of S^. Thus, X is in S^ and

BA = [x : PXB} U [w in line A ^ : B^pw] C S ^ The same

argument leads to the conclusion if A is not in A^B^,- and

A is such that PAA^B, APA^B, or A = A^,

Case 2. Suppose A and B are not in A^B^. Then, either y

B-jA-jAB or B^A^BA. In the former, BA C S^, and in the latter,

AB C sl. The argument also holds if A or B is A^.

Suppose A and B are points in S9. Similar arguments to

those atoove show that either AB C Sg or BA C Sg,

%1

The Completeness Axiom implies that there exists G--4

such that for CT in S. and D in S^, either GC = S. or GD = Sg,

Suppose n is a counting number. Then A n is in and

BL is in S„, Since A G B . it is implied that G is in n l n n* 00 -F\ FA B Y . Therefore, there exists a point satisfying the / n n;

theorem. fl —>

Theorem 3.3. If ^ is a set of points C PAf such that

there is a point C in PA such that PX s PC for all X in 0,

then there is a point L in PA such that:

—y and (2) if I in PA is such that PY < PL, then there is a

(1) PX i PL for all X inl,

if

point S in -I such that PYS.

Proof. Suppose I is a set of points in PA such that

there is a point C in PA such that PX s PC for all X in h.

Define S^ and Sg as follows:

S ± = (Y : PI S PX for some X in SJ U

[Y : APY} U [?}, and

S 2 = [Y : PX < PY for all X in $].

It must be shown that S^ U S^ = line PA. Suppose X is — >

in line PA. If X is not in PA, then X is in S ^ Suppose X

is in PA and Y is a point in A Consider each of the following

cases:

Case 1. If X - Y, then X is in S^.

Case 2. If PXY, then X is such that PX < PY for some Y

in b, so X is in S^.

Case 3* If PYX, then, if there exists Z in J such that

PYXZ, then PX < PZ for some Z in <?, and X is in S^. But if

kz

there does not exist a Z in ^ such that PYXZ then PY < PX

for all Y in S t and X is in S2, or else, X is in in which

case PX = PQ for some Q in J , and thus X is in S^,

Case 4. If X = P, then X is in S^.

Prom the above arguments and the fact that all the

points in S 1 and all the points in S 2 are in line PA, it is

proved that U = line PA.

Also note that by Theorem 1.29, S ^ A S 2 = 0. Since

P is in S 1 and C is in S 2, the sets are nonempty.

Suppose G and H are in S^, Consider these cases: <

Case 1. If G = P and H is such that PH s PX for some

X in 4, then HG = [h] U [ j : Hjp) U £y : APYj C S r

Case 2. If G = P and H is such that APH, then

GII = £P} U [Y : APYJ C s 1 .

Case 3. Suppose PGH. Let J be such that PJH. Then

PJ < PH = PX for some X in 4, and J is thus in S^.

H?" = : PJH} U £Y : APYj C S ^

Case 4. Suppose HPG. The £j : P J H ^ ^ S ^ by Case 3»

and [Y : APYj C S 1, thus HP C S^.

Suppose G and H are in S 2 # Consider these cases: .

Case 1. Suppose PGH, Let J be such that PGJ.

Since PX < PG < PJ for all X in J, J is in Sg.

G? = GH C s 2.

Case 2. The case where GHP cannot hold since

PX < PH and PX < PG implies PGH or PHG.

Case 3. The case where G or H is P cannot hold, or

else that point would be in S^.

^3

The Completeness Axiom implies that there exists a

point L such that if K is in and J is in S2, then either

LK ~ 01* LJ == -4 <

Suppose LK s S^t Since L is in S1# PL = PZ for some Z

in •$. Suppose PL < PQ, where Q is in <&. But Q is in S1, and

LQ 4- S » so PX = PL for all X in Since PL 2 PY for some

Y in I and PX = PL for all X in d, then L is in h. Suppose

Y is in PA such that PI < PL. Then PYL. Thus:: (1) PX i PL for all X in *5, and

•4

(2) if Y is in PA such that PY < PL, then L is such

that PYL. Suppose LJ = S9. Then PX < PL for all X in i, and thus

c 0

PX « PL for all X in 6. Suppose Y is in PA such that PY, < PL»

The point Y is in S^ because LY 4- Suppose there does

not exist S in % such that PYS. That implies that PX = PY

for all X in J and thus Y is in Let M be such that YML.

The point M is such that PX < PM for all X in J , Thus, M

is in S2, but this is a contradiction since LM S^.

The theorem is proved.

Theorem If 4 is a set C PA, then there is a point

G in PA such that:

(1) PG £ PX for all X in &, and

(2) if Y in PA is such that PG < PY, then there is a

point T in I such that PTY.

Proof. The argument is similar to Theorem 3»3> con-

sidering the following sets and then applying the Completeness

Axiom:

S x = {y : PI < PX for all X in 4} U

[y : APYj U [pj.

S2 ~ fo i PX s PY for some X inij.

Theorem 3.5. If AB is an interval and £ is a collection

of segments that cover AB, then there is a finite sub-

collection of segments which cover AB.

Proof. Suppose AB is an interval and <3 is a collection

of segments that cover AB. Let P be such that PAB, Let

f[ m £x in AB : AX can be covered by a finite number of the

sets of 6 - J . Suppose B is not in H. Since B is such that

PX i PB for all X in /?, by Theorem 3.3, there is a point L

such that:

(1) PX = PL for all X in tf, and

(2) if I is in PA such that PY < PL, there exists a

point S in ^ such that PYS.

The point L is in AB or else (2) is not satisfied.

Since every point in AB is covered by some element of L is

covered by segment EF in 6. The point E is not such that

PX = PE for all X in Jf, so there exists G in H such that AEG.

Since G is in ft, there is a finite subcollection

JV . s S } of sets in t which cover AG. Thus, the

I 19 2* 9 riJ finite collection (si, ..., Sn, segment Ef} covers [a} U

segment AF. Therefore, each point of [L] U segment LF is

in t . Let H be such that LHF. The point H is in 7? but

AL < AH, which is a contradiction. Thus B is in and the theorem is proved.

Theorem 3.6. The set of points in an interval is

uncountably infinite.

^5

Proof. Suppose that the set of points in AB is at most

countably infinite. Consider the following sequence of

intervals contained in AB; Pn^Png, Pn^Pn^, Pn^Pn^, ...

determined as follows:

= {n : APnB}. Let n^ be the least element of

^2 = : Pn-jPnB}. Let n^ "be the least element of ig*

^ = £n : Pn^PnPn^]. Let n^ be the least element of

= £n : Pn^PnPn^. Let n^ be the least element of

• • • m m m

l2m = ' Pn2m-lPnPn2ra-2}" n2m ^ *eas^ element

$2nH-l = [ n : pn2m-ipnpn2m}- L e t n2n*l b e t h e l e a s t element

o f snrt-l'

• - • .

Since • • •Pn2m+iPrl2ra

Pn2m-lPn2ra-2 C """ C AB> by

Theorem 3,2t there is a point C such that C is in

/~l Pn2nrflPn2m' Pn2m-lPn2m-2' ***' AB^*

Because the number of points, in AB is at most countable,

C b p for some positive integer x. By the Archimedian A

f

Principle of Counting Numbers, there is some n^ such that

ny> x.

Consider these cases:

Case 1. Suppose y is odd. Then = £n : P^_2P;nPny.-il*

The least element of & is n , so x is not in J . Thus,

Px is not in which is a contradiction.

Case 2. Suppose y is even. Then y+1 is odd and

x < so the same argument as Case 1 follows.

Thus, the set of points in AB is uncountably infinite.

Definition 3,1. The point I is a limit point of set k

in line k means that if segment XZ in k contains Y, then

segment XZ also contains a point different from Y in i.

Definition 3.2. Set 1 is bounded means that there

exist points A and B such that for all P in I , APB. Set i

is said to be bounded by A and B.

Theorem 3*7. Every bounded infinite set of points in.

a line has a limit point.

Proof. Suppose & is an infinite set of points in

line k such that 1 is bounded by A and B. Suppose 1 does

not have a limit point. That implies that there exists &

such that <3= £segmentx segmentx is a segment which contains

X and contains at most one point in % for all X in AbJ.

Theorem 3*5 implies that because AB is an interval and

C is a collection of segments that cover AB, then there is

a finite subcollection of segments which cover AB. But,

that implies that there is a finite number of points in J,

which is a contradiction of the previous theorem.

Therefore, & has a limit point.

Definition 3.3. Set 0 is a topology on a set X means

that (9 is a collection of sets such that:

(1) 0 is in <9 and X is in 6),

(2) if OJI and 0 2 are in <9, then 0^/10 2 is in $, and

(3) if 1 is an index set, and if 0^ is in (9 for all.. i

in I, then ,U 0, is in <9. •ceX i

Theorem 3*8. Suppose k is a line. Let (9 be the collection

to which set S belongs if and only if S is empty or S is a

47

subset of k such that if P is a point in S, then there is

a segment XI such that P is in segment XI and segment XI

is a subset of S. <5> is a topology on k.

Proof. Suppose 0 is the collection described in the

statement of the theorem.

By definition, 0 is in 0-.

The line k is in (9 since k is a subset of k such that

if P is in k, then by Order Axiom I and IV, there exist

points J and K such that JPK, and segment JK is a subset

of k. Property (1) of Definition 3*3 is satisfied.

Suppose 0^ and 0 2 a^e in 0, This implies that 0^ is

in k and 0 2 is in k, thus 0^-A. o 2 is in k.

If m then 0 is in by definition. Suppose

°1"^" °2 ^ ^ ^1"^"^2" Since Q is in 0^, there

is a segment AB such that Q is in segment AB and segment AB

is a subset of 0^. Also, Q is in 0^, so there is a segment CD

such that Q is in segment CD and segment CD is a subset of Og.

Segment AB = (H in k : ABB}.

Segment CD = [s in k : CSD}.

Segment A B / L Segment CD = [T in k : A T B and C T D J . Consider

these cases:

Case 1. Suppose A C B D . This implies that

°l-fi. ° 2 = IT in k : CAB} = segment CB C O ^ O g .

Case 2. Suppose ADBC. This implies that

^1 ^2 = I? * n ^ : DTb} = segment DB C 0^ A- Og.

Case 3» Suppose BCAD. This implies that

A o 2 = £T in k : CTA^ = segment CA C 0^-H. q

^8

Case Suppose BDAC. This implies that

C^A 02 = [T in k : DTa} = segment DA C 0^ «fl 0^.

Thus, is in 0,

Suppose I is an index set and 0^ is in <9 for all i in I.

.U 0. is a subset of k since each 0. is a subset of k. <e.i i i

Let P be in .U 0.. Then for some i, P is in 0.. By j.a i i

definition, there is a segment EP such that P is in segment

EP and segment EP is a subset of 0^. Since

segment EF C 0. C U 0., then segment EP is a subset of U 0.. 1 1 iej 1

Thus, U 0. is in (9. The theorem is proved. AMI 1

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