Covering a sphere with congruent great-circle arcs

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Covering a sphere with congruent greatcircle arcs

J. H. Conway and H. T. Croft

Mathematical Proceedings of the Cambridge Philosophical Society / Volume 60 / Issue 04 / October 1964, pp 787 800DOI: 10.1017/S0305004100038263, Published online: 24 October 2008

Link to this article: http://journals.cambridge.org/abstract_S0305004100038263

How to cite this article:J. H. Conway and H. T. Croft (1964). Covering a sphere with congruent greatcircle arcs. Mathematical Proceedings of the Cambridge Philosophical Society, 60, pp 787800 doi:10.1017/S0305004100038263

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Proc. Oamb. Phil. Soc. (1964), 60, 787 7 8 7With 18 text-figures

Printed in Great Britain

Covering a sphere with congruent great-circle arcs

B Y J. H. CONWAY AND H. T. CROFT

Gonville and Caius College, and Peterhouse, Cambridge

(Received 2 July 1962; and in revised form 24 January 1964)

1. Mikusinski ((2), Problem 84) asks: can we express the surface of a sphere (inEuclidean 3-space) as the union of non-overlapping congruent great-circle arcs? Wegive a partial answer to this question.

We take the word ' congruent' in the above to mean both (i) equal in length, and(ii) topologically the same; thus there are three distinct cases to consider: when thearcs are all topologically open (i.e. containing neither end-point), when they are alltopologically closed (i.e. containing both), and when they are all half-open, half-closed.

We shall call the three problems S i (half-open), S2 (closed), and S3 (open). Asanalogues, we consider the related plane problems P i , P2, and P 3 , involving line-segments. I t is also of interest to ask the related question: can such a constructionbe performed if in the hypotheses the topological character is retained but the con-dition of equality of length dropped?

The six original problems may also be generalized to n dimensions: we ask whenn-dimensional Euclidean or spherical space can be filled with segments or arcs of theappropriate kind. We retain the P, S notation for these new problems. Our results forthe Euclidean problems may then be neatly summarized by 'Pm is possible in n dimen-sions if and only if n > m', and we can give effective constructions to show this.(Obviously the solubility of Pm for any given dimension implies it for all larger ones.)However, we give in the Appendix a non-effective construction which disposes of allthese higher problems.

Of the original six 2-dimensional problems we have solved all save S 2. In thefollowing, we shall devote a single section to each problem, and we shall considervarious generalizations in the appropriate sections. The sections other than that forS 3 are fairly trivial, and we regard the solution of S 3 as the main part of this paper.

2. P i . It is obviously possible to fill a line with half-open segments of any givenlength—we merely put them together head-to-tail. A plane is then filled by fillingeach horizontal line separately.

P 2. I t is impossible to fill a line with closed intervals even with no condition ontheir lengths, indeed any set of disjoint closed intervals (of positive lengths) leavesnon-denumerably many points uncovered.

We give an example to show that a plane can be filled with equal closed segments.We perform the construction in a denumerable number of stages, of which the first feware sketched in Figure 1. First, we construct an infinite 'pillar' of parallel closed seg-ments (1), then half-pillars (2), (3), open at the finite end, perpendicular to (1). Then

788 J . H . CONWAY AND H. T. CEOFT

(4), (5), (6), (7), again open at the finite end, but the locus of their end-points makingangles of £77-as shown. Then (8)-(15), one in each of the 8 angular gaps created, and bi-secting the angle there. And so on, each time putting one wedge in each angular gap,and bisecting the angle there. Each stage consists of inserting a finite number of suchwedges; and it is clear on consideration that each finite point is covered just once.

Fig. 1

Fig. 2 Fig. 3

P 3. The proof that a plane cannot be covered with equal open segments is verysimilar to the corresponding proof for a sphere. We therefore consider in detail onlythe latter, following which the reader will find a description of the modificationsrequired in the planar case.

A slight relaxation of the conditions removes the impossibility. Thus it followsimmediately from Figure 2 that the plane can be covered by open segments each of oneof any two distinct lengths. (A modification of this figure, Figure 3, proves thecorresponding result for closed segments.)

The solubility of P 3 in three dimensions also follows from Figure 2. To show this,we regard the figure as an orthogonal projection onto the horizontal plane. The

Covering a sphere with congruent arcs 789

'boundary' segments represent horizontal lines, but the 'internal' segments may bethought of as the projections of lines of just sufficient slope to ensure that their lengthsare right. (One thinks of wood in a timber-yard.)

S 1. As remarked above, the solubility of Pm in n dimensions implies its solubility inall higher dimensions. This is not obvious for Sm, although the existence of various' Hopf-like' fibrings enables us to deduce implications for certain pairs of dimensions(see (3), page 105 etseq.). However, we may reinstate this principle for S 1, provided thelengths of the arcs are less than \n. We give the proof for the step from one to two dimen-sions as an example. A circle may be filled in the obvious manner with arcs of length2,TIlie if k > 1, and so we can deduce that a 2-sphere may be covered with arcs of length2nlk if k Js 5.

We shall first cover one open ('Northern') hemisphere with equal half-open arcs.Let the length of each arc be 2n/k (k integral, and k > 5). We proceed downwards as

Fig. 4

shown in Figure 4, covering one annular region after another. It is clear that we maychoose the steepness of the arcs that fill each annulus so that, if dn be the latitude ofthe horizontal small-circles, we have dn\-0. For the moment we consider all the arcsto be closed at their lower end, open at the higher end. In the limit we have coveredthe whole hemisphere except for the North Pole N. Now pick a chain of arcsNA1A2A3,..., each of which has the lowest point of the previous one as its own(missing) highest point. We 'reverse' the arcs of this set, so that their top points areincluded in them, and not their lowest ones, e.g. NA1 includes N,A±A2includes Av....I t is clear that, with this modification, we have succeeded in covering one openhemisphere; we can deal similarly with the complementary hemisphere. Finally—using the induction hypothesis—we can 'put a girdle round the earth': we fill in theequator with k successive head-to-tail arcs of the same length. The sphere is nowcovered.

S 2. This problem in one dimension is very similar to P 2, and the same remarksapply. We have made little progress towards its solution in two dimensions; onedifference between the 'open' and 'closed' problems S 2, S 3 is that no result such asthe vital Lemma 1 of section S 3 is true for the closed case.

790 J. H. CONWAY AND H. T. CltOFT

We can, however, cover a sphere with closed arcs of only two lengths, namelyTT — e and n— 2e for any positive e < \n. Since e may be arbitrarily small, the lengthsmay be arbitrarily close. Figure 5 shows how equal closed arcs of any length cancover a closed lune of sufficient thinness: the figure obtained from this by removingthe two extreme arcs we shall call a partially open lune.

We now fit together one closed lune L, of angle a say, and denumerably manypartially open lunes of angles fii (— oo < i < oo) {fit > 0) so that all of them have thesame two vertices as L, and a + E/^ = 2TT (see Figure 6).

Fig. 5 Fig. 6

Fig. 7

On a little consideration we see that we have covered the surface of the sphere, withthe exception of a denumerable number of uncovered 'slots' each of length n — 2e.These we can just fill with closed arcs of this length.

S3. As has previously been stated, the (negative) solution of S3 is our mainresult. We shall give the proof in a moment, considering first some modifications.S 3, like S 2, is readily soluble if we relax the conditions.

Thus a sphere of unit radius may be covered with open arcs, all except six of length\TT, and these six of length v; see Figure 7. Let N,S be the North, South Poles,respectively; let A, B, C, D be 4 points in order equally spaced round the equator.The 6 open arcs noted in the statement above are the interlocking semicircular arcs


NAS, NCS, BND, BSD, ABC, ADC. There remain 8 spherical equilateral triangles,of side \n. And these may be filled in with open arcs of length \-n, e.g. in &BNC, wetake all the open arcs one of whose missing end-points is at N, and the other somepoint of arc BC, as in Figure 7.

We may remove the semicircular arcs by more complicated constructions at theexpense of the condition on the lengths: in particular, we can cover the sphere withopen arcs all of length less than any positive d, and greater than some positive edepending on 8.

3. The impossibility of S 3. We shall suppose throughout the proof that such aconfiguration does exist, and finally obtain a contradiction. The proof is dividedinto 7 lemmas. We use the following terminology: interior points of the open arcswill be said to 'belong' to them; we shall use the phrase 'end-point of an arc' torefer to the (missing) end-points, it being understood that these points then 'belong'to another arc of the system. We shall suppose the sphere to have radius 1, and thearcs to have length a.

LEMMA 1. Let X be a point belonging to an arc si in the configuration. Then it isimpossible that X is the limit-point of a sequence {Bn} of end-points of arcs 38 n with theproperty that the angle fin between si and 38n (both produced, if necessary) satisfies

as n^-co.

Proof. It is clear from Figure 8 that, for sufficiently large n, we have the arcs 3Sn

passing arbitrarily close to one (at least) of Y, Z, the end-points of si', say to Z.Because si and each of the 38n are of the same length, the 38 n stick out beyond Z.That is, given an e > 0, we can take n so large that the angle 38 n subtends at Z isgreater than n — e.

Fig. 8

But now we know that the point Z, being an end-point of si, lies on an arc otherthan si itself, call it eS. The angle between si and ^ cannot be 0; nor can it be n, forotherwise, being open, si and *£ would overlap. Hence *& makes some angle a withsi (a #= 0,77).

There are then parts oftfin both directions a, n + a. from Z, for # is open. It is nowclear that this would imply the cutting of the arcs ̂ , 38n for some sufficiently large n.Hence the impossibility is established.

LEMMA 2. Suppose that the length a of the arcs is not \u. Then it is impossible thatpart of the sphere may be covered in the following way: that all arcs s/g = X Ye (6 an index)of the correct length lying in a certain angle shall have the same end-point Xasin Figure 9.

Proof. Since a # \n, the points YB fill up an arc of a small circle £? whose ' centre'(on the sphere) is X (if a < \n) or X's antipode (if a > \n). Consider points Yg that

792 J . H . CONWAY AND H. T. CROFT

are interior points of this arc of SP; no open great-circle arc of the covering system cancontain 2 or more such points; for, suppose it contained Ya, Yf. then (since ̂ i s a smallcircle) it will overlap some s/y, where, if a < \u, Yy is taken between Ya, Yfi, and, ifa > \n, Yy is taken beyond Ya or Yg.

X

Fig. 9

Now each covering arc containing such a point Ye is part of the great-circle theretangent to ̂ ( and hence we must now have a < \n). Take such a point Yx, and such anarc containing it, <$/x; then a point Y2 such that <3f1 subtends an angle very nearly n atY2; and let &2 be the arc containing Y2. Since Y2 is an interior point of the openarc ^2 , there is a portion of <&2

o n ea°h side of Y2. Take Ys (on Sf) between Yx and Y2,and so close to Y2 that C2J2 subtends nearly n at Y3. I t is clear from the figure that anypossible arc of the requisite length containing F3 overlaps either ^1 or <%/2 or one ofthe s4g. The result of Lemma 2 follows.

4. We need two new definitions.We shall say that a point X that belongs to an arc si is an ordinary point or O-point,

if all points sufficiently close to X on either side lie on arcs which make arbitrarilysmall angles with the arc si.

We shall say that a point X that belongs to an arc si is a limiting-point or L-point,if it is a limit-point of end-points of arcs that make angles with s# that are greater thansome non-zero a.

We now have

LEMMA 3. (i) The set of O-points and the set of L-points are exclusive and exhaustive.(ii) / / one interior point of an arc si is 'an L-point, so are all the others. (We then call

stfan L-arc.)(iii) The (missing) end-point of any arc is an L-point, and so in particular L-points

and hence L-arcs do exist in the configuration.(iv) There are only a finite number of L-arcs.

Proof, (i) First, we observe that we would obtain an identical set of i-points ifthe second definition above were reworded by replacing ' . . . limit-point of end-pointsof arcs that make angles with s/ that are greater than some non-zero a ' by ' . . . limit-point of interior points of arcs that make angles with si that are greater than somenon-zero a' . A little consideration of Figure 10 convinces one that if the point Xsatisfies either condition it necessarily satisfies the other.


Comparing now the amended definition of an L-point, and the original definitionof an O-point, we see that every point X must satisfy just one of them.

(ii) Suppose, if possible, the contrary: that one point X of s/ is an Z-point, andanother point Y of s/ is an O-point. Then arcs 88 near to s? (and on either side of s#)at Y make arbitrarily small angles with s/. Bearing in mind that sf, 88 have thesame length, and that, after Lemma 1, no sequence of end-points of 88's can have aslimit-point any interior point of sJ', it follows that the end-points of the 88 have theend-points of si as their (only) limit-points.

X

Fig. 10

But again, some arcs ̂ have points arbitrarily close to X, and make angles greaterthan a with si (for X is an Z--point). But then some # intersects one of the 88.Hence it is impossible in the configuration for an arc si to contain both 0- and Z-points.

(iii) See Figure 11.

Fig. 11

Let X, the end-point of si, belong to 88. As we have seen previously, si', 88 cut atan angle a, where a =t= 0, n. Again, as in (ii) above, if X were an O-point, there wouldbe arcs close to 88 and on each side of 88, their end-points having end-points of 88 aslimit-points. Then some of these would cut si'. So X is an i-point.

(The mere existence of Z-points can also be seen to follow from the 'furry-ball'theorem. For if all points were O-points, there would be a well-defined direction ateach point X of the surface (i.e. the direction of the arc containing X) and this directionwould vary continuously. We know that this is impossible.)

(iv) We shall suppose, if possible, that there are an infinite number of Z-arcs, anddeduce a contradiction.

First, by giving the end-points (or if a = n, one end-point and the mid-point)suitable spherical coordinates, and applying the Bolzano-Weierstrass theorem in the

794 J. H. CONWAY AND H. T. CROFT

appropriate 4-dimensional space, we easily obtain: there are points A, B, and arcsAnBn of the system such that An->A, Bn^B, and the arc ^.B^-^-the arc AB in theobvious sense. See Figure 12.

It is not prima facie obvious that AB is an arc of the covering. This is the first thingwe must prove. Suppose first, if possible, that some point X of the arc AB belonged toan arc %> of the covering that cut AB at an angle a (a + 0, TT). Using the fact that %> isopen, we see that there are sections of ̂ sticking out each side of AB. It follows that^ intersects some AnBn, which is not possible; hence every point of AB belongs tosome arc lying along AB; because two arcs (being open) cannot abut at angle TT, oroverlap, it follows that AB (being of length = Mm (AnBn) = a) just coincides with anarc 3> of the system. Thus, on the hypothesis of an infinite number of L-arcs, we havesuch a figure as Figure 12, with AB an arc of the system.

Kg. 12

Again A and B, being merely end-points of 2&, each belong to other arcs s&, SB,respectively, which make angles neither 0 nor TT with 2).

Consider now a point Y, the mid-point of AnBn for large n. This is, by hypothesis,an 2/-point, and hence has, arbitrarily close to it, an end-point of an arc $'. It is clearfrom the figure that € lies between AB and An-Bn. where n' < n and n' ->co with TO.Because, then, S has also length a, $ cuts either j / or 38, if n be chosen sufficientlylarge. This contradiction to the 'non-overlapping' condition of the configurationestablishes (iv).

5. We now investigate the configuration which the L-axcs can make.

LEMMA 4. The L-arcs form a closed network covering the sphere, each separate part ofwhich is a convex polygon, with a finite number of sides, and there are but a finite numberof such polygons.

Proof. First, we observe that the set of i-points is a closed set, since by Lemma 3 (iii)the (missing) end-points of any L-axc are Z-points, and so belong to other L-axcs.Next we see that this set intersects every great-circle. For suppose the equator con-


tains no i-point, and consider an i-point Q with the numerically least (positive ornegative) latitude. Q is an interior point of a great-circle arc of L-points (byLemma 3 (ii)); and so other i-points have smaller latitude, for circles of latitude aresmall circles.

We now take any O-point P, and consider the set of i-points which are 'visible'from P, that is to say, have no i-point between them and P. It follows from thepreceding paragraph that there is exactly one such point on each ray through P. Weshall now show that a visible i-point Q which is not the end-point of any i-arc isthe centre of an open interval of visible L-points on its own L-axc s4. For suppose not.Then there will exist rays PQn tending to PQ for which all the Qn are visible points on-L-arcs other than jtf. Since there are but a finite number of L-arcs, we may supposethat all the Qn are on the same i-arc 38, and they will therefore have a limit point Rin the closure of 8S. Since the set of Z-points is closed, R is an i-point. Moreover,R is on the ray PQ; but it cannot be between P and Q (for Q is visible), nor can it beon PQ produced, for then the Qn could not all be visible, as some must be behind J/.Hence R — Q. But Q cannot be an interior point of both s& and 3$, and so must be anend-point of 38.

We now consider the rays joining P to all the end-points of Z-arcs. Since there areonly a finite number of them, we may consider these rays as being arranged cyclicallyaround P. Now the visible points on the rays between two adjacent rays of this setmust all be on the same i-arc, and so the set of visible points forms a closed polygonwith finitely many sides. Consider now the points visible from P', another O-pointwithin this polygon. All such points must be within or on the polygon, and so thepolygon for P ' is contained in that for P. Reversing the roles of P and P ' we see thatthe two polygons must be identical. Hence if P" is a further point inside the polygon,P" is visible from P', that is to say, one great-circle arc P'P" is contained within thepolygon. Since P' and P" are arbitrary interior points, the polygon is convex.

Since there are only a finite number of possible vertices there can be only a finitenumber of such polygons, and since P was an arbitary O-point, these polygons coverthe surface of the sphere.

This concludes the proof of Lemma 4.

The configuration of _L-arcs envisaged in the statement of Lemma 4 is by no meansimpossible per se, e.g. the configuration of Figure 7 illustrates this possibility. Whatis impossible is the filling in of each of the faces thus created with equal arcs, equalalso to each of the bounding arcs.

We require some further definitions.In a polygon, if a side AB of length a is the limit of arcs AnBn of the system that

lie inside the polygon, then we call AB a 'finishing-side' or P-side of the polygon.In a polygon, an interior angle, neither of whose arms is an P-side of the polygon,

we call a i£-angle.LEMMA 5. Except perhaps when a = \n, we have:(i) each polygon has just 2 F-sides;(ii) some polygon has a K-angle.


Proof, (i) We observe first that the boundary of the polygon consists just ofjL-points, the interior just of 0-points, and so all arcs of the system must have theirends (which are Z-points, by Lemma 3 (iii)) on the boundary. Further, as no two arcscut,their end-points travel in opposite directions around the boundary. See Figure 13.

There must then be two limit-points Ax, A2 to one set of end-points, and similarlyB-y, B2 for the other set; moreover, we cannot have Ax = A2 nor Bx = B2, or else weobtain the type of configuration proved impossible in Lemma 2 (this is where we usethe hypothesis of that Lemma, that a =f= \TI). On a little consideration it is furtherseen that AXBX, A2B2 must necessarily be Z-arcs forming part of the boundary of thepolygon. And AXBX = A2B2 = a. Therefore, AXBV A2B2 are (non-adjacent) .F-sides,and no other sides of the polygon can be. This proves (i).

Fig. 14

The fact that a side can be an .F-side of a polygon and yet be an i-arc is not contra-dictory—it merely means that the configuration is 'well-behaved' on one side of it,and not on the other; but we cannot have an L-axc being an .F-side for two adjacentpolygons—we shall use this point below.

(ii) See Figure 14. This is similar to Figure 13. We have joined AXA2, BXB2 by great-circle arcs, which, by convexity, lie within the polygon. Now, B2 belongs to an arci r s ^ o f the system (labelled as in the figure, say). Part of ^ may form a side ofthe polygon, but f̂ cannot cut the interior of the polygon, nor can it be A2B2, for B2

is an end-point of that arc. Now, XB2A2 is cut into a finite number of parts by L-arcsB2P1,B2P2, ...,B2Pn_1 (n ^ 1) that have B2 as end-point. We shall show that at leastone of these angles is a isT-angle. Suppose not. Then B2PX must be an .F-side for thepolygon containing A2S2Plt for otherwise A2S2PX would be a isT-angle, since A2B2

cannot be an .F-side on both sides. Considering Px B2P2 similarly, we deduce that -62-̂ 2is an .F-side for the polygon containing PXS2P2. Repeating this process, we come


finally to the conclusion that B2X is an F-side for the polygon containing Pn_x£2X.Since B2X is not of length a this is a contradiction.

We note in passing that the angle at B2 interior to the polygon is at most n — a,where a is the least iT-angle in the whole configuration. We now restrict our attentionto II, a polygon containing such a minimal if-angle.

6. LEMMA 6. It is impossible to cover II with equal arcs of length a.

Proof. See Figure 15. Let the minimal iiT-angle occur at O\ and let AXBX, A2B%

be the _F-sides of the polygon; join up, by great-circle arcs, the sides of the pentagonCAXBXB2A2C. By convexity, this pentagon lies within the original polygon. Asremarked above, the interior angle of the pentagon at B2 can be at most n — a, andthe same is true, for exactly the same reasons, at Ax, A2, and Bx. And of courseAX6A2 < a.

Fig. 15

We shall show that the shortest distance from C to the arc B1B2 is strictly greaterthan a. The impossibility of the covering will then be immediate from Figures 13, 15.Since the great-circle distance from a fixed point to a variable point on a fixed greatcircle is minimal when the line joining them is a perpendicular to that great-circle, theshortest distance from C to BXB2 is attained either (i) by CBX or GB2, or (ii), by agreat-circle arc, CD say, that cuts BXB2 perpendicularly at D, a point between Bx

and B2. We consider cases (i) and (ii) separately, and show that in either case thisshortest distance is greater than a.

We use two familiar facts—first, that a spherical polygon has positive sphericalexcess, and second, that in any spherical triangle 'larger side opposes larger angle'.

Case (i). We suppose that GB1 is the distance in question. See Figure 16. Now ifCBX ^ a = AXBX, then would CAXBX ^ AXCBX < a. This is absurd, as it implies thatthe angle sum S of the pentagon satisfies

377- < 2 < <x + a + (7T-a) + (n-a) + (n — a) = Zn-a.

Case (ii). Since A16D + DQA2 < a, one of these angles, say A1CD, is less than orequal to \a. We now restrict ourselves to one half of the pentagon. See Figure 17.We shall prove the inequalities CD > AXD > AXBX = a. If CD ^ AXD we shouldhave CAXD ^ AXCD ^ \CL, and then the angle sum £' of the triangle AXCD wouldsatisfy

n < £' < \a. + \a + \-n = a -f \TT50 Camb. Philos. 60, 4


and so a > \v. But this in turn implies

3TJ- < 2 «S a + (77 - a) + (n - a) + (n - a) + (n - a) < f n,

again a contradiction.If instead AXD «S A1B1 we should have A1B1D ^ ^-f iBj < \n, and in this case

the angle sum 2" of the quadrilateral AXBXDC would satisfy

2?7- < 2" s? (77 - a) + \u + \n + \a = 2n - \<x.

This last contradiction completes the proof of Lemma 6.

Fig. 16

7. We need finally to deal with the case a — \n.

LEMMA 7. It is impossible to cover the sphere with great-circle arcs, if a = \TJ.

Proof. If we do not use the characteristic property of the exceptional case a = \n,the proof runs as in the general case. The characteristic property is the non-satisfactionof Lemma 2. Lemmas 3 and 4 remain true. Then, for a = \n, we may suppose that wehave such a configuration as that envisaged in Lemma 2. That is, arcs of the systemhaving a common end-point at X say, with their other end-points Yg lying on anotherarc of the system, CS. Let P, Q be the furthest limit-points of the set Ye on (&; P, Q maybe at end-points of & or not. In the figure (Figure 18) P is and Q is not. We shall workwith this figure, which exemplifies the general case. We shall need that arc of thesystem 2 that contains P. Now the arcs XP, XQ must be arcs of the system, andwe call them s&', 88, respectively. Let i be the arc containing X. I t lies exterior to thetriangle XPQ, and makes an acute angle with either XP or XQ. In either of these caseswe shall get a contradiction: in the first by considering the convex polygon (or anyone, if there are several) that abuts AXPQ along XP; in the second by consideringthe similar one abutting along XQ. In the first case it lies (by convexity) within thespherical triangle whose sides are s#, S>, S; similarly in the second case, within thetriangle whose sides are 88, <&, S. And, on consideration we see that either of these is


too small to be covered with arcs of length \u (unless they all have one end at X, in contra-diction to the definition of P and Q). The result of Lemma 7 follows; and hence alsothe theorem.

8. We describe briefly the modifications required for the proof of P 3. All goeswell until we reach part (iv) of Lemma 3, to which we must add the words 'in anybounded region'. A similar change is required in Lemma 4, but here there is also themore serious possibility that the polygons might themselves be infinite. However,they can at worst be 'semi-infinite strips', as all other possibilities are precluded bythe convexity conditions, and so they will still have at least one .F-side. But therestill exist some finite polygons, as any i-arc abuts some finite polygon. This is provedby applying the convexity restriction to the figure formed by the given Z-arc and the.L-arcs containing its end-points. This is all we need to prove part (ii) of Lemma 5.

We must insert epsilons into all the inequalities of Lemma 6, and redefine a as thelower bound of all i£-angles rather than the least such angle. It is easily shown thatthere exist acute if-angles, and further that the arc of the system whose end-point isthe vertex of an acute isT-angle must be perpendicular to the i-arc at its other end.This shows both that any polygon containing an acute if-angle is finite and also thatonly case (ii) of Lemma 6 can apply to it. Even when so modified, one of the inequalitiesrequired is strict, and so Lemma 6 still holds. P 3 has no exceptional case such as thatof Lemma 7.

APPENDIX

We prove here a theorem which disposes of all the P and S problems in three andmore dimensions. To this end, we suppose we have a space 8 and a collection C ofobjects with which we should like to fill 8. Then by & partial filling of S by C we shallmean a filling of a part of S by a part of C. If this part of C has a smaller cardinalthan G itself we speak of a small partial filling. The partial fillings are partially orderedby inclusion, and we shall say of two distinct comparable partial fillings that one is anextension of the other.

THEOREM. / / S and C have the same cardinal, a sufficient condition for the existenceof a complete filling (using all of C and all of S) is that any small partial filling may beextended so as to cover any unused point of 8 or so as to use any unused object of C.

50-2


Proof. We consider the least ordinal a whose cardinal is the same as that of S and C.Then we may enumerate the points of S as Pp (ft < a) and the objects of C as Op (ft < cc).We now perform the construction in an infinite number of stages, one for each ordinalft < a. At any stage we shall have a small partial filling of S by C, which we extend toa larger one. At the ftth stage, if/? is even, say ft = 2y, we extend the filling so as tocover the point Py of S, and if ft is odd, say ft = 25+ 1, so as to use the object Os of C.(For the notions of odd and even ordinals, see (l), page 54.) We can always arrangethat the new partial filling will be small, by using just one new object per stage. Ifwe do this, it is easy to see that the partial filling obtained prior to stage ft (ft < a) hasthe same cardinal as ft, and so is small even if ft is a limit ordinal. Since a is an initialnumber, 2a = a, and so the filling obtained after all stages ft (ft < a) have been per-formed will be a complete filling. This completes the proof.

As an example of the use of this theorem we show that the P problems are solublein three dimensions, without any use of ad hoc devices. I t suffices to show that anysmall partial filling of three-dimensional space by line-segments of the given typemay be extended so as to cover any unused point P, since here we need not distinguishbetween distinct objects of C. In the rest of this proof' small' will mean ' smaller thanthe cardinal of the real numbers'. A given line-segment belongs to at most one planethrough P, and so there are only a small number of planes through P which containused line-segments. Hence there is a plane II through P containing no used segment.Ft intersects the used segments in a small number of points, and so only a small numberof lines I through P in II will intersect used segments. Hence there is some linethrough P meeting no used segment. We can now extend the filling so as to cover P byintroducing a new segment lying along this line. This completes the proof.

This argument generalizes immediately to show that all the P and S problems aresoluble in higher dimensions, and indeed that •w.-dimensional space may be completelyfilled with m-dimensional objects of any reasonably well-behaved type, providedm ^ n — 2. As an amusing example, three-dimensional space may be expressed as theunion of (the perimeters of) unit circles. The continuum hypothesis is occasionallyuseful for particularly restive sets of objects.

REFERENCES(1) BACHMAN, H. Transfinile Zahlen (Berlin, 1955).(2) FAST, H. and SWIEHCZKOWSKI, S. (editors). The new Scottish book (Wroclaw, 1958).(3) STEENBOD, N. E. The topology of fibre bundles (Princeton, 1951).

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