COMPARISON OF GEOMETRIC FIGURES

Transcription

1 COMPARISON OF GEOMETRIC FIGURES Spyros Glenis M.Ed University of Athens, Department of Mathematis,

2 Introdution the figures: In Eulid, the geometri equality is based on the apability of superposition of Common notion 4 Things whih oinide with one another are equal to one another. ([]) The geometri equality, with respet to Klein s view, is based on the group theory as well as on the set theory: Definition Let a set X, G a subgroup of Aut(X) and the figures S, S X. We shall say that these figures are G geometrially equal if and only if there is an f G so that ( ) f S = S. The equality, indiretly defines the inequality of geometri figures. Eulid onsiders that a figure is smaller than another one if with an appropriate rigid motion the first oinides with part of the seond. Although for any two figures S, S easy to deide whether they are equal or not, however it is not that simple to deide if one of them is smaller than the other. Obviously a triangular region is never equal to a irular disk, but an we say that a triangular region is smaller than a irular dis if the radius of the dis is greater than or equal to the radius of the irumsribed irle of the triangle? In Eulid, the omparison involves only similar figures. On the ontrary, Klein s view of equality, leads us to define a geometri inequality using the notion of being subset and enables us to ompare even non-similar figures: it is We will say that is equal to or smaller than S whenever there is a eulidean rigid motion f so that ( ) S f S S. Then we will write S S.

3 This natural definition of inequality provides a paradox as we will immediately illustrate using the following example given by the professor V. Nestorides: Let us onsider a losed half plane A and let B be the half plane A with a line segment attahed vertially to the edge of the half plane. Sine A B we an say that A B. Moreover, there is a translation of B, so that it is fully overed by A and in this ase we may write B A. It seems logial to assume that A and B must be geometrially equal, in other words, that they an oinide if we apply a ertain rigid motion. But this is impossible to happen, beause every half plane remains half plane whenever we apply a rigid motion to it and obviously it an t oinide with a geometri figure that is not a half plane. Sine the geometri relation " " is not antisymmetri it is neessary to restrit the omparison to ertain lasses of geometri figures. We already know that in the lass of the line segments or in the lass of the ars of a irle, the relation " " is a total order. Therefore the question is, if there are other lasses of figures where the relation " " is a total or a partial order. We shall all good lasses (of geometri figures) those that among the figures they ontain we an t find a paradox like the one mentioned above. A good lass, but not the only one, is that of the ompat figures (sets). In fat, ompat figures have the property not to generate paradox with any other geometri figure whether ompat or not. Those figures will be alled good figures. Besides the ompats, good figures are also the open-and-bounded sets. On the ontrary, just bounded figures may not be good as we will prove later using a ounterexample, given again by professor V. Nestorides. The study, onerns not only the Eulidean Geometry, but it is also expanded into the Hyperboli and the Ellipti Geometry and some parts may be formulated in a pure algebrai language so that they over uniformly all three geometries. The onlusions we have reahed, are fully ompatible with our previous knowledge about the omparison of geometri figures. In the speial ase of the Eulidean Geometry we proved that there is a good lass, ontaining all the fundamental geometri figures, where we an ompare even non-similar ones. Therefore a omparison between a irular dis and a triangular region is meaningful in the new ontext. 3

4 Comparison of Figures in Eulidean Geometry. Basi definitions We adopt Klein s view for the Eulidean Geometry. Our spae is endowed with the eulidean metri and the group ating on is ISO( ), the group of eulidean isometries. The ouple ( ( ), where we will develop our study. ISO ) generates the eulidean geometri spae Definition. Two figures and S are geometrially equal when there is a eulidean isometry f : S so that ( ) f S S =. In that ase we will write S S. Remarks I. Figure is any subset of. From now on we will not distinguish the terms subset of and figure. II. We use the terms rigid motion and isometry synonymously. Definition. For any two figures and we shall say that is equal to or smaller than when there is a eulidean rigid motion S S. S S S S f so that ( ) f S S. Then we will write This natural definition does not satisfy in general the antisymmetri property, as we will prove later. Proposition. The relation " " is a pre-order of figures i.e. it is reflexive and transitive, and the reflex ion is meant in the sense of the geometri equality defined in. based on the eulidean metri ρ of where ρ (( x, y),( a, b) ) = ( x a) + ( y b) 4

5 Let Α and Β two geometrially equal figures. Then, by definition, there is a eulidean isometry f so that f ( A) = B. Then f ( ) relation If is reflexive with respet to the geometri equality. A B and B C then there are isometries, Then for the isometry transitive ٱ A B also holds and we onlude that the f g so that ( ) ( ) f A B and g B C. g f holds g f ( A) C i.e. A C. Therefore the relation is In the following examples we shall prove that " " does not satisfy in general the antisymmetri property, with respet to the geometri equality of definition.. Example. Let the half lines ( ) losed subset of thus A / B. {,0 : 0} A= x x, ( ) {,0 : 0} B= x x>. Sine A is a while B is not, there is not an isometry f so that f ( A) B For the isometry f ( xy, ) = ( x+, y) ( ) B A, so we have both A B and B A while A / B. =, f A B holds. But it is also obvious that Example. Let the figure ( ) {(, ) : } {, : 0 } {(,) :0 } A= x y x x x and the half plane B= x y x. Obviously A B therefore A B A B O Sine every isometry maps half planes into half planes there is not an isometry f suh that f ( A) B =, so A / B. Every isometry maps losed sets into losed sets. 5

6 A B O But for the isometry f ( xy, ) = ( x 3, y) ( ), f B A holds. So we onlude that A B and B A while A / B. Example.3 Let the figure ( ) figure B produed by triangle Obviously B { xy, : x 0, y } Α= 0 whih is the right angle xoy and the A when we subtrat the inner part of the isoseles right {(, ) : 0, 0, } S = x y x y x+ y<. A so B A holds. By translating A parallel to the axis x x f xy, = x+, y) is an isometry. Thus A B. where ( ) ( But A suh that and g( A) by two units we also have that ( ) f A B, B are not geometrially equal, beause in ase there is an isometry right angle, whih is absurd. = B and sine isometries preserve the angles then B should also be a g B B f(a) O O 6

7 Example.4 Let an angle ω so that ω π is irrational, for instane we an hoose ω = π where may be replaed by any irrational number. We set a ( os kω,sin k ) k ω a sequene of points lying on the irumferene of the unit irle. By definition ak a m for ω λ k m otherwise we would have integers n, λ so that π = n. The set A { a : k 0,,,3,... } = = is a dense and equally distributed subset of the k irumferene. We also onsider the set \{ } B = A a A, therefore B A. If there is an isometry f of the plane suh that f ( A) = B then we will arrive at a ontradition. For every r a A = sin( ω ). Sine ( ) that the distane ontradition. Let T ( ) there is at least one a A so that the distane d a, a = r where f A = B then and for every b B there is at least one b B so ( ) be a rotation by and ( ) { :,3,... } k d b, b = r. But this does not hold for ao B and we arrived at a ω with enter the origin of the axe. Then T is an isometry T A = a k = B. So A B and B A hold, while A / B. relation: We improve the definition of the pre-order " " so that we arrive at an order Definition.3 In the set of figures we define a relation λ suh that: Aλ B { A B} or {A B and not B A} Proposition. λ is an order relation. simple Definition.4 A lass E of figures is said to be good when there are not any figures A, B in the lass E so that A B, B A and A / B. 7

8 Remark The relation λ is defined so that it expels all the pathologial ases of " " where the antisymmetri property does not hold true. But the definition is not a natural one and gives no information to the question: Whih figures form a good lass? So the problem of the geometri order relation still remains unsolvable under the definition of λ. It is muh wiser to stik to the definition of " " and onentrate our study on those sets that satisfy the antisymmetri property. Definition.5 We will say that a figure A is good when for every figure B, if A B and B A hold, then A B also holds. Obviously a lass onsisting only of good sets is a good lass. The onverse does not hold true. A trivial ase is a lass onsisting of one figure (and all the geometri equals) that is not good. Sine there is no other figure in the lass to provide a ounterexample then the lass is good. Another non-trivial ase is the lass of the open or losed angles. We proved in example. that an angle is not a good set but it is quite easy to verify that using open or losed angles only, we an not provide a ounterexample.. Quest for good lasses of figures Proposition.3 If A is a good set and A B then Let B is also a good set. C B and B C. Sine A B then A C and C A hold. As A is a good set then there is an isometry isometry g of the plane suh that g( B) f of the plane suh that f ( A) = C. There is also an = A. Then g f ( B) = C i.e. B C ٱ Proposition.4 If A is a good set then its omplement is also a good set. 8

9 Let B f ( B) A Sine, ( ) f, g, of the plane, so that A and A B. Then there are isometries and g( A ) B. But then f ( B) A and g( A ) B hold. f g are - and onto, ( ) f B = f ( B ) and g( A ) g( A) ( ) f B A and g A B whih is equivalent to good there is an isometry as is - and onto h A = B i.e. h ( ) of the plane suh that A h A B and h ( ) A B ٱ = hold. Therefore B A. As A is ( ) = B. Then h A = B and Theorem. Let X, d be a ompat metri spae. If f : X X is an isometry then f ( X) = X. Well known Proposition.5 Every ompat subset of Let is a good set. A ompat and B an arbitrary set so that A B and B A isometries f, g so that f ( A) B and g( B) For the isometry ompat (theorem.). But then A.. Then there are g f : A A we have already seen that g f ( A) = A beause A is g( B) A g f ( A) = whih implies that B f ( A) and A B. Therefore A is a good set ٱ. So ( ) f A = B holds Now we shall introdue a new definition that will be partiularly useful. Definition.6 A figure A will be alled strongly good if for every isometry f : that satisfies f ( A) A, then the equality f ( A) = A holds true. Proposition.6 Every ompat subset of is strongly good. 9

10 Diret from definition.6 and theorem. ٱ Proposition.7 Every strongly good set is also a good set. Let A a strongly good set and B an arbitrary set suh that A B and B A. Then there are isometries f, g of the plane so that f ( A) B and g( B) ( ) A. For the isometry g f : A A, g f A = A holds sine A is strongly good. Then g( B) A= g f ( A) therefore B f ( A). Finally we onlude that f ( A) = B so A B and A is a good set ٱ Remark The definition of the good figure is diffiult to handle, as it depends on the interation with all the other figures. On the ontrary the definition of the strongly good figure is intrinsi, beause, in simple words, strongly good is any figure that does not fit (without deomposition) into part of itself. Proposition.8 Every open and bounded subset of Let is strongly good. A open and bounded and f an isometry of the plane suh that f ( A) A. As A is open then A A= and A= A A. Thus A= A\ A Also diam( A) diam( A) = < so A is losed and bounded subset of, hene ompat. Also A A and diam( A) diam( A) so the boundary is losed and A is bounded hene ompat subset of. f ( A) A f ( A) A f ( A) A and from Theorem 3. f ( ) ( ) ( ) ( ) ( ) ( ) f A = f A\ A = f A \ f A = A\ f A A\ A= A Hene f ( A) A and as f onlude that f ( A) A A f ( A) A = A holds. is an isometry then from Theorem 3. again, we = =. Therefore 0

11 ( ) ( ) ( ) ( ) f A = f A\ A = f A \ f A = A\ A= A, so A is strongly good ٱ Proposition.9 In, the union of a ompat with an open bounded set is a strongly good set. Let K ompat and A open and bounded subsets of, and the isometry plane suh that f ( A K) A K. f of the The set V = ( A K) o A K is also open and bounded. W = K \ V = K V is ompat as an intersetion of a ompat with a losed set. Obviously V W = Sine A is open subset of A K then A V and ( \ ) A K V K = V K V = V W A K. Hene V W = A K Also V W = f ( V W) = f ( V) f ( W) = Consequently f ( V W) V W f ( V) f ( W) V W f ( V ) is an open and bounded subset of ( o V = A K). From proposition.8 we onlude that f ( V) = V. A K therefore f ( V) V sine As f ( V) f ( W) V W and V W =, f ( V) f ( W) = then f ( W) But W is ompat and from Theorem 3. f ( W) = W. Therefore f ( V) f ( W) = V W f ( V W) = A K ( ) f A K = A K ٱ W Proposition.0 In, the intersetion of a ompat with an open bounded set is a strongly good set. Let K ompat, f ( A K) A K A open and bounded and f an isometry of the plane suh that We set X = A K whih is a ompat subset of K. Then f ( A K) A K f ( A K) X f ( X) X. Aording to Theorem. we onlude that f ( X) = X.

12 Sine A K is open in K, then it is also open in every losed subset of K. Therefore A K is open in X, so X \ ( A K) is ompat and it is obvious that ( ( )) ( f X A K X A K \ \ ). Aording to Theorem. we onlude that ( \( )) \( ) =. f X A K X A K But then f ( X \( A K) ) = X \( A K) and sine ( ) onlude that f ( A K) = A K i.e. A K is strongly good ٱ f X = X and A K X, we Remark The proposition holds even if A is not bounded. Proposition. The lasses X = { K A where Κ ompat and A open bounded subsets of the plane}, Y = { K A where K ompat and A open bounded subsets of the plane}, E= X Y and F= E { S : S E} are all good lasses. The lasses above onsist of strongly good sets. Then from definition.4 and the propositions.7,.8,.9,.0 the onlusion is obvious ٱ The lass F inludes almost all the fundamental figures of Eulidean Geometry: line segments, triangles, polygons, irles, ars et but not the open or losed angles whih, as we have already mentioned, form a good lass. Proposition. If W is the lass of open or losed angles then the lass F W is good. Sine we already know that the lasses FW, are good, then it is suffiient to examine whether a set from one lass provides a ounterexample to the other lass. Let A a set of the lass F. Then A E or A F\ E. If A E then it is bounded and it annot provide a ounter example with any angle of W. If A F\ E A E and provides a ounterexample with an angle then its omplement, will also provide a ounterexample with the omplement of the angle (whih

13 is also an angle). But this ontradits what we have already proved about the elements of E whih give no ounterexamples with the elements of W. So the lass F W is a good one and inludes the losed and open angles ٱ Remarks. The losed sets are neither good nor form a good lass aording to the example.. The onneted sets are neither good nor form a good lass aording to the example. 3. The bounded sets are neither good nor form a good lass aording to the example.4 4. The onneted and bounded sets are neither good nor form a good lass. This an be easily proved if in the set A of the example.4 we attah the inner points of the unit dis. 5. The onvex sets are neither good nor form a good lass aording to the example.3 6. The onvex and bounded sets are neither good nor form a good lass. This an be easily proved if in the set A of the example.4 we attah the inner points of the unit dis. If two figures A, B are good we have proved that their omplements are also good. However the union or intersetion of (strongly) good sets is not neessarily a good set as we will illustrate in the following ounterexamples: Claim The figure ( ) {,0 : 0} {( 0,) } L= x x is strongly good. We denote A= {( x,0 ): x 0} and {( 0,) } b = therefore L = A b. From now on we will regard the point { } (0,) and the set ( 0, ) as the same. Let an isometry f : suh that f ( L) L f ( A b) A b Sine all isometries are - and b A then f ( b) f ( A). 3

14 The image of a half line through an isometry still remains a half line with point f ( a o ) as the origin, where a o = ( 0,0), thus ( ) f A A If f ( b) A then the points f ( b), f ( a ), f ( a ), where a ( ) o = 5, 0 A are ollinear. But then the points ba, o, a would also be ollinear whih is absurd! Therefore f ( b) = b. Also ( ( o), ) ( ( o), ( )) ( o, ) d f a b = d f a f b = d a b = As f ( ao ) A and the only point a A suh that ( ) d a, b = is, then o a f ( a ) o = a. o For every ( ) ( ( ) ) a A we have that d( a, ao) = d f ( a), f ( ao) = d f a, ao But there is only one point of the half line with this property, so f ( a) We proved that f = id so for L A b Claim The figure ( ) = we have ( ) {,0 : 0 } {(,) : 0} f L = L ٱ = a. M = x x x x< is strongly good. We denote B = {( x,0 ): x 0 }, B = {( x, ) : x< 0}. Then M = B B. Let an isometry f : where ( ) f M M. ( ) ( ) ( ) ( ) f M M f B B B B f B f B B B Sine B B f ( B B ) f ( B ) f ( B ) = = = The images of, B B are half lines therefore ( ) f B B and subsequently ( ) or f ( B ) B and subsequently ( ) f B B If ( ) and ( ) then f ( 0,0 ) ( x,0) f B B f B B ( ) ( f ( 0, ) = x,) for some x 0 f B B. = for some x and 0 4

15 (( ) ( )) ( ) ( ) ( ) (( ) ( )) ( ) = d 0,0, 0, = d f x,0, f x, = d x,0, x, = x x + Then x x = 0 and we onlude that x = x = 0. Therefore ( ) ( ( ) ( ) f ( 0,0) = 0,0) and f ( 0,) = 0, b = ) ( ) If o (0,0 and b =, 0 then ( o ) (( ) ( )) ( ) ( ) ( ) = d bo b = d f b f b = d f b,, 0,0, As f ( b ) f ( B ) B then f ( b ) b Sine f preserves the three ( 0,0 ), ( 0,) and (, 0 ) whih are non-ollinear then f = id If f ( B ) B and f ( B ) B then f ( 0,0 ) ( x,) x 0. (( ) ( )) (( ) ( )) ( ) =, x < 0 and f 0, = x,0), But = d 0,0, 0, = d x,, x,0 = x x + then x = x = 0 whih is ( ) ( absurd as x <. 0 So the only ase is f = id and then ( ) f M = M ٱ Example.5 Using the previous notations for the sets M and L, then the set {(,0) : 0} L M = x x is not good aording to the example., although LM, are strongly good sets. Example.6 We also use here the previously defined sets If g is a refletion with respet to the set ( ) ( ) L and M. yy axis then g( M ) is strongly good but the {,0 : } {(,) : 0} G = L g M = x x x x is not good beause for the set ( ) {,0 : } {(,) : 0} V = x x x x> we have that V G and 5

16 ( ) hg V, where h( x, y) = ( x+, y). Therefore V G and G V. But G is a losed set and V is not losed so there is no isometry f suh that V f ( G) =, hene V G. 6

17 Good figures and good lasses in non Eulidean Geometries. Hyperboli Geometry We will use the Poinare s dis as model. The non-eulidean plane is the interior of the unit dis D of the omplex plane i.e. { : } D= z z < The distane between the points A ( z ), B ( z ) of D is given by the formula d p z z, = artanh zz ( A B) () and we will all it the hyperboli metri. A A Ο B Β The isometries of the model are all funtions f : D D of the form: f ( z) f az + b = and aa bb = (without hange of orientation) or bz + a ( z) az b = και aa bb = (with hange of orientation) bz a 7

18 Definition. Two figures and S of the hyperboli plane D, are geometrially equal whenever S there is an isometry 3 f : D D so that ( ) f S = S. Then we will write S S Definition. For two figures and of the hyperboli plane D, we will say that S is equal to S S or smaller than S whenever there is an isometry f suh that f ( S ) S. Then we will write S S. The definitions of a good set, strongly good set and good lass in the Hyperboli Geometry are similar to those of the Eulidean Geometry. The relation is reflexive, transitive but not antisymmetri as we will prove using the following examples. Example. The example.4 of the irrational rotation ω in the eulidean plane, is also applied in the hyperboli plane provided that the enter of rotation is the origin and the points are k ( os ω, sin ω) a = b k b k with 0< b <. It is noted here that the angles of the hyperboli plane D are the same with the orresponding eulidean ones. It an be proved, exatly as in the eulidean plane, that the sets = { : = 0,,,3,... } and B A\ { a} A a k k It is also true that B The rotation ( ) i ase of f ( z) A beause B A. = are not geometrially equal. ω T z = e z, is an isometry of the hyperboli plane D beause it is the az + b = bz + a with b = 0, a i = e ω and aa bb Performing the operations we reah that T( ak) a k + ( ) { :,3,... } T A = a k = B i.e. A B. k =. = so Sine A B, B A while A / B then A is not a good set in Hyperboli Geometry. a k 3 with respet to the hyperboli metri d p 8

19 Example. Another model for the Hyperboli Geometry is given by the half plane In H the metri is given by d { :Im( ) 0} H = z z > H z, = artanh z ( z z ) z z We point out that the models of the dis and of the half plane are isometrially isomorphi and the orresponding mapping is: h: H We onsider ( ) and B { z H :Re( z) } z i D µε h( z) = z + i A { z H :Re z 0 } = z H : z = µε Re( z) 0 και Im( z) =. Obviously B A (see figure 4.). i i/ O It is easy to verify that the mapping g ( z) z 3 Then A is a subset of g ( B) = z H :Re( z). H H { } = is an isometry in H. i i/ - - O 9

20 Consequently we have that A B and B A. But there is no isometry f suh that f ( A) = B sine in B every point has infinitely many points within a given hyperboli i distane ρ > 0 ; however in A the point + for small ρ > 0 has only one point in A at hyperboli distane ρ. The Moebius transformation z i w z i = + with ( ) Re z > 0 maps onformally the hyperboli plane H to the hyperboli plane D and transforms the sets AB, and the isometry g H of H into the sets A, B and the isometry g D of A B, B A and A / B. D so that: Now we will searh for good sets and good lasses of the hyperboli plane D Lemma. If zw, Dthen zw > w Well known exerise. Proposition. The eulidean and the hyperboli metri on the unit dis D are equivalent. Well known Proposition. Let A D then A is bounded with respet to the eulidean metri. However D itself is bounded in eulidean metri but unbounded in hyperboli metri. Obvious. Corollary. If A is a losed and bounded subset of is ompat. D with respet to the hyperboli metri then it 0

21 Sine the metris are equivalent and A is losed with respet to the hyperboli metri then it is also losed (and bounded) with respet to the eulidean metri, hene ompat. As the metris are equivalent then it is also ompat with respet to the hyperboli metri ٱ Proposition.3 In D, every ompat and every open and bounded (in hyperboli distane) set is a strongly good set. In addition the union, the intersetion and the omplements ot them are also strongly good sets of the Hyperboli Geometry. The proofs are similar to those of proposition.6 Proposition.4 If we are studying only the figures in the interior of the unit dis, then the lass F W of proposition. is also a good lass of the hyperboli plane. It is obvious, by the definitions and the proposition.3.. Ellipti Geometry As model of the Ellipti Geometry we will use the surfae hemisphere of the Riemannian sphere i.e.: 3 {(,, ) :, 0} S + = x y z x + y + z = z S + of the upper with the assumption that two antipodal points of the equator oinide in other words the points a( x, y,0) S + and a( x, y,0) S + make up of the same point a. The lines of the model are the intersetions of S + with planes that pass through the enter of the sphere. Then we an define the distane between two points of the ellipti plane as: If ab, S + then d a, b = aros a b s ( ) where a b 3 is the ordinary dot produt of. We will all d s the ellipti metri.

22 Definition.3 Two figures and S of the ellipti plane S +, are geometrially equal whenever S there is an isometry 4 f : S S + + so that ( ) f S = S.Then we will write S S Definition.4 For two figures and S of the ellipti plane S +, we will say that is equal to or smaller than write S S. S S whenever there is an isometry f suh that f ( S ) S. Then we S The definitions of a good set, strongly good set and good lass of the Ellipti Geometry are similar to those of Eulidean and Hyperboli Geometry. The relation is reflexive and transitive but not antisymmetri, as we will prove next. Before proving that is not antisymmetri we must note that: α) The lines of the ellipti plane have finite length π β) The ellipti plane is not oriented, in fat it is a one-sided losed surfae (suh a surfae is the Moebius bundle). As suh it is meaningless to talk about half planes or half lines and anything else defined by them. Therefore, the examples.,., and.3 annot be realized in the Ellipti Geometry. Claim The mapping os ω sin ω 0 x T : S + S + where T( x, y, z) = sinω osω 0 y 0 0 z is an isometry of the ellipti plane S +. a( x y z ) ( ) Let,, and b x, y, z then T a x y x y z ( ) = ( osω sin ω, sinω+ os ω, ) T b x y x y z ( ) = ( os ω sin ω, sin os, ω + ω ), 4 with respet to the ellipti metri d s

23 T a T b = x x + y y + z z = a b and the dot produt ( ) ( ) d T a T b T a T b a b d a b (, ) = aros ( ) ( ) = aros = s (, ) So ( ) ( ) s This isometry orresponds to a rotation of the ellipti plane by angle ω with enter the north pole N(0,0,) ٱ Example.3 We first onsider some ω suh that ω is irrational. π 3 Let the points ak = os kω, sin kω, of the surfae S +. It is easy to verify that aa = k k+ os ω + 3 is independent of k. 4 4 s k k+ = is onstant for every k = 0,,,... Then (, ) d a a r We onsider the figures A= { a : k = 0,,,3,... } and B A\ { a} Sine B A then B A k =. T ( ) Using the above isometry it is easy to see that T ak = a k + so ( ) { :,3,... } T A = a k = B i.e. A B. k If there was an isometry of the ellipti plane f : S S result in absurd as in the example.4. so that ( ) + + So the relation " " is not an order in the Ellipti Geometry. f A = B then we Proposition.5 S + is ompat (with respet to the ellipti metri) Let the sequene { u n } S +. { } n 3 u is bounded in with respet to the eulidean metri, so it has a onverging subsequene { Then ( ) k n } u where ( ) ( u u x, y, z x, y, z), and u S + sine it is a kn kn kn kn 3 losed subset of. d u, u = aros u u aros u u = aros = 0 s kn kn 3

24 We proved that every (bounded) sequene of S + has a onverging subsequene therefore S + is ompat ٱ Corollary. Every losed subset of S + is ompat. Proposition.6 In S +, every losed and every open set is a strongly good set of the Ellipti Geometry. Every losed set is ompat in S +, so it is also strongly good. Then its omplement it is also strongly good, therefore and every open set is strongly good ٱ Proposition.7 If A is open set and Κ is losed set of the ellipti plane then the set good. The same as in proposition.9 A K is strongly Corollary.3 If Α is open set and K is losed set of the ellipti plane then the set good. The set A K is strongly good aording to the proposition.7 Then its omplement ( A K ) = A K is also strongly good ٱ A K is strongly 4

25 3 Open issues A fundamental question is whether the definitions of the good set and the strongly good set are equivalent or there is a ounterexample of a good set that is not strongly good. In the appendix it is proved that in all good sets are also strongly good. So it is our belief that the definitions are also equivalent on the plane. Another question is whether the algebra produed by ompat or open and bounded} onsists only of strongly good sets. Finally, as in X { A : A = Ω= {set of good lasses} the assumptions of Zorn s 5 lemma are satisfied it would be quite interesting to find maximal lasses within the set Ω. 5 Every totally ordered subset of Ω is defined to be a set of good lasses Y every i, j Y. I it is true that i j F F or Fj Fi. Obviously the good lass i = { F i, i I } so that for F is an upper bound of i I

26 Appendix In the appendix we prove that good sets oinide with strongly good sets in do not know what happens in. Lemma Let a set A, a A and an isometry f : suh that A A\{ a}. Then A is not a good set. If f : is an isometry then f ( x) x ( ) f x = x+, 0. If f ( x ) = x then f ( A) A\ { a} A A\ { a} f ( A) A\ { a} = or f ( x) = x or ( ) = = i.e. a A, absurd. Both f ( x ) = x and f ( x) = x+ have the property From the assumption we have that f ( A) A\ { a} ( ) \{, ( )} \{, ( )} f A A a f a A A a f a If ( ) = therefore f. We = i.e. f x = x+, 0 or = id. = = i.e. a A, absurd. f x = x+, 0. Then f ( a ) = a, so Also a A ( ) ( ) \{ } f a = a+ f A = A a. Let the set B A\ { a } B A. = +. Then a+ B and a B sine a A and Obviously B A and f ( A) = A\ { a, a+ } B Let us assume that there is an isometry g : so that g( A) = B For every x A we have that f ( x) = x+ A\ { a} therefore there is at least one = ( ) so that d( x, x ) x x x A = (for instane x = x+ ). But then we will also have that ( ( ), ( )) Sine g( A) = B there is some xo d g x g x =. A suh that g( x ) = a B. Then there is also some x = x ( xo ) A so that (, ) (, o o ( o) ) o d x x = d x x x = 6

27 It follows that d( xo, x ) d( g( xo), g( x )) d( a, g( x )) = = =. We may say that there is some b B= g( A), where b g( x ) d( a, b) =. = so that But then b= a or b a, whih in either ase are not points of = + B g( A) =. This is a ontradition and we onlude that A is not a good set. Proposition Let A be a good set, then A is strongly good. We assume that A is not strongly good. Then there will be an isometry T : so that T( A) A. Therefore there is a A\ T( A) and T( A) A\ { a} A ( ( )) i.e. T( A) Sine A T ( A) A T T T( A) then A T( A) then ( ) A T A. We an also prove that A A\ { a} beause: A\{ a} A and A T( A) A\ { a} A \{ a} A. A and as A is a good set sine A is a good set we onlude that But from the previous lemma suh a set A is never a good set, absurd! ٱ 7

28 Bibliography In Greek th. D. Anapolitanos: Introdution in the Philosophy of Mathematis 4 edition, Nefeli publiations.. I. Argyropoulos, P. Vlamos, G. Katsoulis, S. Markatis, P. Sideris: Eulidean Geometry, shool textbooks publiations D. Lappas: History of Modern Mathematis, spring semester V. Nestorides: About the inequality of geometri figures, leture at 0//004 in the Department of Mathematis Athens University. 5. Ε. Stamatis: Eulid s Geometry, vol. Ι shool textbooks publiations H. Strantzalos: The Evolution of Eulidean and non-eulidean Geometries (part I), Kardamitsas publiations. 7. H. Strantzalos: Introdution of geometri transformations in the seondary eduation, Notes for the partiipants of the seminar on History and Mathematis Eduation held in Dep. of Mathematis Athens University, spring C. Boyer U. Merzbah: History of Mathematis, nd edition, G. Α. Pneumatikos publiations. In English 9. C. Caratheodory: Theory of Funtions of a omplex variable, Chelsea Publishing Company New York. 0. L. Carothers: Real Analysis, Cambridge University Press.. J. Dugunji: Topology, Allyn and Baon In. Boston.. Sir Thomas L. Heath: Eulid the thirteen books of the elements, vol. nd edition unabridged, Dover. 3. D. Hilbert: Foundations of Geometry, 4. E. Hlawka, J. Shoibengeier, R. Tashner: Geometri and Analyti Number Theory, Springer Verlag. 5. L. Kuipers, H. Niederreiter: Uniform distribution of sequenes, Wiley N.Y. 6. E. Moise: Elementary Geometry from an advaned standpoint, Addison Wesley. 7. E. G. Rees: Notes on Geometry, Springer Verlag. 8