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CONCERNING CONTINUOUS CURVES
CONCERNING CONTINUOUS CURVES
THESIS
Presented to the Faculty of the Graduate School of The University of Texas in Partial Fulfillment of the Requirements
For the Degree of
DOCTOR OF PHILOSOPHY
By
Raymond Louis Wilder, Ph.B.,M.S. Austin, . June,l923
214965
PREFACE
Continuous curves have been the subject of almost continuous research by mathematicians during the past halfcentury. Jordan first characterized them in his Cours d’Analyse. On the basis of Jordan’s parametric definition it is very difficult to determine whether a given point-set is a continuous curve, as demonstrated by the work of Peano, E.H.Moore, etc., on space-filling curves.
In 1907 Schoenfliess succeeded in characterizing continuous curves by means of properties of the complementary domains and accessibility of their boundaries. His work was good only for two dimensions, however.
In 1913 Hahn characterized continuous curves by means of the property of "connectivity im kleinen", which enables us to determine with remarkable ease whether a given point-set in any number of dimensions is a continuous curve.
In the past ten years the names of Hahn, R.L.Moore, Mazurkiewicz, and Sierpinski stand out most prominently in the work on this subject, although research on the topological structure has not been confined to them alone.
It is the aim of this thesis not only to give a new characterization of continuous curves, suitable for any number of dimensions, but also to characterize and analyze the boundaries of the domains complementary to a plane continuous (thus supplementing the work of Sohoenfliess in this regard), and to study the analogy between ordinary two-dimensional space and the space constituted by a plane continuous curve.
In regard to characterizing continuous curves in any number of dimensions, the following result has been obtained: In order that a bounded continuum M (in any number of dimensions) should be a continuous curve, it is necessary and sufficient that every two points that lie in a connected subset of an open subset of M lie in a sub-continuum of that open subset. The "necessary" part of this theorem was established by R. 1. Moore in 1921. The "sufficient" part is established in this thesis by the discovery of a new property of continua that are not connected im kleinen.
With reference to the boundaries of the complementary domains of a plane continuous curve the most important result obtained in this thesis is the theorem: In order that the boundary of a simply connected domain should be a continuous gairve, it is necessary and sufficient that every connected subset of it be arc-wise connected. It is also shown that the words "connected in the strong sense" may be substituted for "arc-wise connected" in this theorem. Other results in this connection are that : If B is the boundary of a complementary domain of a continuous curve, then (a) B can be divided into a countable set of arcs and simple closed curves together with a totally disconnected set of limit points, and (b) every sub-continuum of B is itself a continuous curve.
In regard to the analogy between ordinary twodimensional space and a plane continuous curve, it is shown that after suitable definitions of "region” and "domain” a plane continuous curve may be regarded as a space that possesses all the properties of ordinary two-dimensional space stated in the first 16 theorems of R. 1. Moore T s Foundations of Plane Analysis Situs. Regarding a continuous curve as a space has greatly facilitated the work in this thesis and enabled the atthor to establish new accessibility conditions which hold also for ordinary space.
Raymond louis Wilder
The University of Texas, May 22, 1923.
CONTENTS
Page Introductory 1 I. A continuous curve in the role of a space 3 11. An analysis of the point-set which constitutes the boundary of a comple- , mentary domain of a plane continuous curve 24 lll.Characterizations of continuous curves, and of the boundaries of the complementary domains of a plane continuous curve . 50
Introductory
CONCERNING CONTINUOUS CURVES
It has been shown by Hahn*, and independently
by Mazurkiewicz**, that in order that a closed, bounded and
connected! point-set should be a continuous curve (une ligne
de Jordan) it is necessary and sufficient that it he connected im kleinen at every point. A point-set Mis connected im kleinen at a point P if for every £>o there corresponds an H>o, such that if and Kg are circles with radii S and respectively, and centers at P, every point X of M that lies
* interior to Ko is joined, to Pby a connected, subset of M
that lies wholly interior to A point-set that is connected im kleinen at every point is said to be connected im kleinen.
The present paper has three main objects, viz.,
(1) To study the analogy between ordinary two-dimensional space and a plane continuous curve,
(2) To characterize and analyze the boundaries of the domains complementary to a plane continuous curve, and
(3) To give a new characterization of continuous curves, suitable for any number of dimensions.
I wish to thank Professor R. L. Moore for many valuable suggestions and criticisms, and to express my gratitude to him for first interesting me in a field of mathematics that is as fascinating as it is fruitful.
*Mengentheoretisohe Oharakterisierung der stetigen Kurve, Wiener Akademie Sitzungsberichte, OXXITI Band, Abt. Ila, pp. 2433-2489.
**Sur les lignes de Jordan, Fund. Math., Tom I, (1920), pp. 166-209.
Ta point-set is said to he connected if it is not the sum of two mutually exclusive point-sets neither of which contains a limit point of the other. A point-set is hounded if it lies wholly in a finite portion of the space under consideration.
>K According to Eahn T s definition X and, P must lie together in a closed and connected subset of that lies wholly interior to The word "closed” is unnecessary, however, when dealing with closed point-sets.
I. A continuous curve in the role of a space.*
For the present, I shall consider a space S, which consists of all the points of a plane continuous curve. I shall define a region in that space as follows: If Pis a point of £5, and k a circle with center at P, then the set of all points of S. which (1) lie interior to k, and (2) are Joined to P by a connected subset of S that lies wholly interior to k, constitutes a region R. A point P is a limit point of a point-set M in space S if and only if every region that contains P contains at least one point of M distinct from P. This definition of limit point is equivalent to the ordinary definition of limit point for twodimensional space, in that a point P which is a limit point of a point-set Min space is also a limit point of Min the ordinary sense, and vice versa. The set of all limit points of R that do not belong to R constitute the boundary of R. Every boundary point of R lies on k. A domain with respect to S**, or an S-domain, is a connected subset D of
S having the property that if P is a point of D, P lies in some region that is a subset of D. The set of points, that are limit points of D but that do not belong to D, constitute the boundary of I).
Using the above definitions, it is interesting to note that many of the properties of ordinary two-dimensional space are also properties of space S. In particular, Theorems 1-16, inclusive, (with the exception of the latter part of Theorem 16) of R. 1. Moore’s paper On the foundations of plane analysis situs* all hold true for space S with no change in the wording, except that the word ’’domain” should be replaced by ”S-domain”.
Two of these theorems which are of fundamental importance Are:
If A and B are distinct points of a domain M, there exists a simple continuous arc** from Ato B that lies wholly iM M.
Every two points of a regioa R caia be joined by ait arc lying entirely in R.
The close analogy between space S and ordinary twodimensional space can be further exhibited by a consideration of accessibility conditions*. For this purpose, I shall
establish the following theorem:
Theorem 1. In order that a boundary point xof an S-domain D should be accessible from any point y of D, it is sufficient either that (1) there exist a circle with center at x, such that the set of all points of the boundary, B, that lie interior to o]_ is a subset of a connected im kleinen subset of S-D, or that (2) x belong to no continuum of condensation** of B.
Proof. I shall consider these two conditions separately.
(1) There exists a circle Ci with center at x, such that if B x is the set of all points of B that lie within C lt is a subset of a connected im kleinen subset Kof S*P. The circle may be taken so small that it does not enclose y or contain £.
Since x is a boundary point of P, and therefore a distinct limit point of P, there exists in P a sequence y lt y 2, yg, •••• having xas a sequential limit point*. encloses
some point of this sequence. Call one suoh point C 2 be a circle concentric with C-p of radius < r/2, (where r is the radius of and such that exterior to Og. 0 2 encloses some point of the above sequence; call one such point Pg. let Cg be a circle concentric with C 2, or radius < r/S, and such that Pg is exterior to it. Og encloses a point Pg of the above sequence. .... Continuing this process indefinitely, there is obtained a sequence of circles, C 2, Cg, .... with centers at x and such that for every positive integer n>l, C n is of radius < r/n; and a sequence of points, 2 , Pg, ... belonging to P, such that for every positive integer n, lies interior to C n , but exterior to G n+ ]_. ♦
Since, as pointed out above, any two points of P are the extremities of an arc lying wholly in P, and Pj are the extremities of an arc P]_ and. Pg of an arc tg, Pg and. Pg of an arc tg, etc., such that t 2, tg, •••• all lie wholly in D. Let x 2 be the last point of on the arc from ?1 to P 2, in the order from Pq_ to Pg. . That portion of tg from x 2 to Pg constitutes an arc a 2. Since +a 2 is a closed set of points, there will be a last point of it on the arc tg in the order from Pg to Pg; call this point Xg. That portion of tg from Xg to Pg constitutes an arc ag. The set + a 2 * ag is closed, and there therefore exists a last point of it, X 4, on the arc t 4 in the order from Pg to That portion of t 4 from to constitutes an arc Continuing this process indefinitely, there is obtained a sequence of arcs, , .... such that for every positive integer n, has only one end point, x n , in common with the set a n , a„, a*, ••••a _. 1 2 5 n~l
There are two cases to consider: Either (1) for every value of n there exists a positive integer k, such that is the last arc of the sequence a 2, ag, • ••• having any points on C n ; or (2) there exists an n, such that an infinite number of arcs of the sequence a,, a 9, a*, • ••• have points on G n .
Case 1* For every value of n there exists a positive integer k, such that is the last arc of the sequence a lf a 2, ag, • ••• having points on os. Then for any arc, a it there can be only a finite number of arcs of this sequence having points in common with a it since a i must lie wholly without some circle of the sequence Gg, Gg, • ••• Then,
of the finite number of arcs having points in common with a l# at least one, a forms, without a connected set with an infinite number of arcs of (where [a i J denotes the sequence a lt ag, a 3, ....). Likewise, of the finite number of arcs of subscript > that have points in common with a at least one, a n , forms, without a n , a connected set with an infinite number of arcs of Of the finite number of arcs of subscript >ng that have points in common with a n , ■ 2 at least one, a , forms, without a n , a connected set with an infinite number of arcs of And so on. In this manner there is obtained an infinite sequence of arcs a arm* a n -, •••• such that for every positive integer i, 2 has in common with a no being the ara only the end-point x ni , and with a onl y bbe end-point x The sequence of arcs yx n p has the property that for every value of i_, x ni x n^ has only one end-point in common with the arc directly preceding it in the sequence, and only one end-point in common with the arc directly succeeding it in the sequence, and with any other arc in the sequence no points in common. Furthermore, for every n, there exists but a finite number of arcs of this sequence having points without or on C n . It can be shown that the set x + * + + •••• constitutes a simple continuous arc from £to x lying, except for x, wholly in D. That is, x is accessible from
Case 2. Suppose there exists an n, such that an infinite number of arcs of have points on C n . It is certain that an infinite number of arcs of have points exterior to C n+l » and points interior to on+g.o n+ g. For every such arc, ai, let Ai be the first point of C n on ai in the order from P i to Xi • Then, in the order from Ai to P it let Bi be the first point of 0 n +g on that portion of ai from Ai to Pi, in the order from Ai to Pi. Then from Aj to Bi there exists an arc which is a proper subset of a it and such that if I n +1 is the set of all points of the plane between C n +i and C n+2 , Aißi is a subset of I n+l except for the points Ai and Bi. The set of all such arcs call [ai*J.
No two arcs of the set have any points in common. Por, suppose and. are two arcs of the set [ai*] having a point, x*, in common, and. that k < m. Then a m was taken subsequently to and. can therefore have at most one end-point, in common with Other than x m , and. a ffi can have no points in common. Hence x* must be identical with But A m is the first point of 0 on a m in the order from P m to x m , and is therefore a subset of that portion of a m from P m to A m , and unless is identical with x m , AjJß m can have no point in common with But if A m is identical with x m , a m can have no points exterior to C which is contradictory, since a m was taken as one of the arcs of having points exterior to o n+l . Hence A m ß m and have no point in common.
The infinite set of points of the type has at least one limit point on C Let Abe one such point. If z is any other point on then at least one of the arcs into which A and £ divide C n+l must contain an infinite set of points of the type having A as a limit point; call this arc Az. Then from the points of the type can be selected an infinite sequence A x *, Ag*, Ag*, .... in the order from £ to “A on Az, and having Aas a sequential limit point. The set of points Bj_*, Bg*, Bg*, .... where B n * is the other endpoint of the arc of to which A n * belongs can be arranged in an analogous manner on as a sequence Bg, Bg, .... with a sequential limit point K. For each n, let A n be the point of the sequence A^*, Ag*, Ag*, .... belonging to the arc of which B n is an end-point. There is obtained thus a sequence Agßg, .... of arcs of the set arranged in a definite order. Call this sequence the set [AiB i].
The set has a closed and connected limiting set, which has at least one point on 0 and at least one point on C n+2 . let wbe a point of within I n+1 » let be a circle with center w and lying wholly in I n+l , but not enclosing C n+2 . Because of the property of connectivity im kleinen of S, there exists, concentric with, and lying interior to a circle Kg, such that all points of 3. interior to Kg are joined to w by an arc* of S lying wholly interior to Let A k B k be the first arc of the set having points in Kg.
It is certain that no point on any arc A n ß n can be joined to w by an arc lying wholly interior to unless this arc contains points of all arcs of the set of subscript >£•
There are two cases:
(a) If w is a point of B, can be taken so small as to contain no points of S ~ B, in which event all arcs of the set of subscript can be joined by an arc of B lying wholly interior to K]_.
(b) If w is a point of *B:
For each n, A^B n and A^+i® n +i» together with those arcs A n A n+l and of G n+l arL<i °n+2* respectively, that contain no end-points of the set form a simple closed curve K n which cannot enclose w. Let tbe a point of interior to Kg. An arc of S, from t to w lying wholly interior to Kj, must contain, for every K n (where an arc b , such that b n lies wholly interior to K n except for its end-points; but for each K n there can be at most a finite number of arcs such as b n . If w is not the first point of on the arc from t to w, in the order from t to w, let m be that first point. Let [b n ] denote the set of all arcs of the type b n on the arc from £ to m.
If only a finite number of arcs of [b n ] contain points of S - B, then there exists an n such that if l>n, Ajßj and A^JS^ +l are joined by an arc of B lying wholly interior to
I shall show next that it is impossible that an infinite number of the arcs [b n ] contain points of
If any arc, b f contains points of S - D, it must, since its end-points are in B, contain points of B. So, if an infinite number of arcs of £b n ] contain points of S - D, the same arcs contain points of B; that is, every such contains at least one point, d n , of B. Call the set of such points
Clearly, m is the limit point in of the set of end-points of the set [b n 3. For the end-points belong to the set so that their limit point must lie in and if any other point, m*, of were this limit point, it would mean that a set of points in the arc from t to m has a limit point exterior to the arc (since m is the first point of on the arc), an impossibility, since an arc is a closed set. But if mis the limit point of the end-points of the set [b n ] it is the limit point of the end-points of those arcs to which the set [d n ] belongs, and hence also the limit point of the set £d n ]• Therefore m must belong to since B 1 is a closed set. But is a subset of K; thus m is a point of K and the set [d n ] belongs to K.
Let Kj 136 a circle with center -at m and. lying in n °t enclosing 0 n+ g. Since Kis connected, im kleinen, there exists, concentric with, and lying interior to a circle Kg, such that all points of K lying interior to Kg can be joined, to m by a connected subset of K lying interior to But Kg must contain points of [d n ], and a connected set containing a point of [d n ] and m must contain points of D; that is, points not belonging to K. Hence the contradiction.
In any case, then, there must exist a number such that for and 2>di» any arc Aißi can be joined to any other arc Ajß j by an arc of D lying entirely interior to K]_. This means, then, that those arcs of the set [ai] of which the arcs of [Aißi] of subscript are subsets, can be joined by arcs of Din the same manner. Gall the set of these arcs
For each arc ai of Sj, let Di be the last point of C n+ g on ai in the order from Xi to Pi, and $i first point of C n +g on that portion of a i from Di to P it in the order from to Pi# Then ai contains an arc I>i B i> such that DiEi lies wholly in I n+2 (where I n+2 is the set of all points of the plane between 0 n+ g and sn+g)$ n+ g) except for its end-points. The set of all such arcs call [DiEi]. It can be proved, by the method used above, that the arcs of the set [DiEi] can be ordered in a sequence •••• having the property that there exists a d* 2 , such that for i. and 2 an y arc Qan Joined to any other arc/DjTj by an arc of D lying wholly in I n+2 . Those arcs of which contain arcs of of call S 2.
having the
This process may be repeated. indefinitely, since S 2 contains an infinite number of arcs of the set and. thereafter any Si will have the same property# Furthermore, for any n, S n will be a subset of S
The first arc of the set belonging to call t^ 1 . The first arc of the set after t^ 1 that belongs
2149G5
to call And so on. In this manner, an infinite sequence of arcs, tg 1 , t^ 1 , .... is obtained having the property that (1) if i >l, has at most an end-point in common with t 3; and (2) there exists a positive number i 1 w x , such that for i> w 1 > and j >w lt t ± can be joined to t by an arc of D lying entirely interior to
In the same way, any set S nT (nJ =2,3,4,....) can n T n 1 n T be ordered in a sequence ,t 2 ,tg , .... having the . I property that (1) if i_>J., has at most an end-point n r ' in common with t J ; and (2) there exists a positive number n T w n r, such that for and j.>w n r, can be joined to t • by an ara of D lying interior to some circle which lies M. - . in the set of all points between C n+n t and O n +n T +l does not enclose °n + n T +l*
Now, let be any integer and j_ any integer ) w 2. ,In addition let J. be so chosen that in the original arrangement of arcs, tj 2 is of higher subscript than Then t^ 1 can be joined to tby an arc, £, of D, which lies entirely interior to It can be shown, as in Case 1 above, that the set +a 2 + .... +a n (where contains an arc £ from y to P n . Let be the last point of the arc £ on fk® arc ® in th® order from t^ 1 to t j 2, and the first point of tj 2 on s after That portion of the arc £ which has y and as end-points, together with that portion of £ which has as end-points, forms an arc
Let kbe any integer Wg, and. such that t^ 2 is of 9 ' higher subscript than the arc tj in the original arrangement of arcs, tj 2 and t k S can be joined by an arc x T y T of D (x T on X % and on lying entirely interior to a circle that lies in I n+2 but does not enclose C n+ g. let 2 be the last point on the arc x’y*, in the order from x T to jr’, of the set composed of t-j_ together with that portion of tj 2 to (tj 2 Sa k ); and y 2 the first point of after j9g on x T y*. The sum of the arcs an arc tg.
If this process be continued indefinitely, the arcs of are joined by other arcs of D in such a way that a set of simple continuous arcs satisfying the conditions for Case 1 is obtained, so that x is accessible from 2?
(2) x belongs to no continuum of condensation of B.
Choose the sequence of circles Cg, os, ....» the sequence of points P-p Pg, Pg, .... and the sequence of arcs ag, as, •••• as in (1). As in (1), there are two cases to consider:
Case 1. Suppose that for every value of n there exists a positive integer k, suoh that is the last aro of the sequence ap a 2, • ••• having points on C • This situation is handled, exactly as in Case 1 of (1).
Case B. Suppose there exists an n, such that an infinite number of ares of have points on C n » Select the set as in Case 2 of (1), with the difference that to each has been added the arc forming the arc As above, it can be shown that no two arcs of [a i *] have points in common.
The points of the type have at least one limit point Aon 0 +1 • Let £be any point of 0 + - distinct from A. and 1 Then at least one of the arcs into which C n+l contains an infinite number of points of the type call this arc Az. Choose on Az, in the order from £to A, an infinite sequence of points of type A it namely, Ag*, Ag*, • ••♦ having Aas a sequential limit point. The set of points, P-.*, P ? *, Pg*, ...•, where for every positive integer n P r * is the othe r end-point of the arc of *] to which A n * belongs, has x as a sequential limit point, since x is a sequential limit point of the set P]_, Pg, Pg, .... The sequence of arcs A]-*?!*, Ag*Pg*, Ag*Pg*, .... has a continuum Mg as a sequential limiting set; Mg contains A and x.
Either there exists in Mg an infinite set of points of D having x as a sequential limit point, in which event a proof similar to that used in Case 2 (a) of (1) can be used to show the accessibility of x from jr, or there exists no such sequence. In the latter ®vent we can consider C n +]_ to be so I small that all points of Mg belong to 3- L. Then Mg is a subset of B, since each point of it is a limit point of D and does not belong to D.
Let be any point of Mg interior to C n+l and distinct from x. Let be a circle with e]_ as center, not enclosing x or containing x, and lying entirely interior to 0 determines a concentric circle Kg 1 such that all points of S interior to Kg 1 lie with e 1 in an arc off S that lies wholly interior to K^ 1 . There exists a positive number N such that for n>N, Kg r outs off at least one segment of
A *P„*. Let tbe a point of A n *P n * (k>N| interior to Kg, and let mbe the first point of M 2 on that arc of S from t to ei that lies entirely interior to Ki f (in the order from tto e-jJ. Let p n q n and P n +i<l n +i arQS of A n* P n* an 4 An+i* ? n+l*» respectively, cut off by (i.e., P n Q n lies entirely Interior to except for the points p n and q n , etc. ) and such that one of the arcs into which p n and P n divide K]/ does not contain q n or q n+l . Let P n <l n +i be that simple closed curve formed by the arcs p n q n , that arc p n Pn+l K l’ wil i Qll contains neither q n nor q n +l» an<i that arc q n Q n +i of Kj. 1 which contains neither p n nor p
Two possible situations have to be considered. I / shall say that possesses property E if the arc tm contains, no matter how is selected, subject to the conditions noted above, or how tm is selected, for an infinite number of distinct values of n, a point of S - D that lies interior to P n Q n +i« Either possesses property E or it does not; in the latter event, will be said to possess property G.
If possesses property G, let C n+2 * 136 a cirole concentric with C n+l , not enclosing ej_ and lying interior to There exists a sub continuum Mg of M g , which contains x and at least one point on C n + 2 *» but no points exterior to sn+2*« Either a point eg of Mg interior to C n+2 * and distinct from x can be found possessing property G or such a point cannot be found. If such a point can be found, let 0 be a circle concentric with C n+ 2* t not enclosing e g , and lying interior to C n+ s. There exists a subcontinuum M 4 of M- which contains x and at least one point on o n +g*, but no points exterior to o n +3*. A point eg of M 4 interior to on+g*0 n+ g* and
distinct from x can be found possessing property G, or no such point can be found. If such a point can be found, continue as before. In the event that this process can be kept up indefinitely, that is, if for every new o n+i * lying within °n+i a P° in ' t e i interior to C n+i * oan be found possessing property G, then there exists a circle with center at and lying withint C n+ such-that an infinite number of arcs of that have points interior to can be joined by an arc of D lying wholly interior to The method used in Case 2 (b) of (1) can now be used to show the accessibility of x.
In the event that this process can not he kept up indefinitely; that is, if finally a circle on+0 n+i * is found, such that every point of interior to sn+$ n+i * possesses property P, we may proceed, as follows: Consider C taken so that every point|of Mg other than x interior to possesses property P. But then every point of Mg other than x is a limit point of boundary points of B that do not belong to Mg, and since Mg is a continuum, x itself is a limit point of such points* That is, x belongs to a continuum of condensation, Mg, of B, contrary to hypothesis.
Hence, in every case, x is accessible from y if it belongs to no continuum of condensation of the boundary.
This completes the proof of Theorem 1.
It will be noticed that in the above proof no properties of space S are made use of, that are not also properties of ordinary two-dimensional space# The arguments used, apply,
therefore, in the latter space. Hence the following theorem:
Theorem 2: In order that a boundary point xof a connected domain P in ordinary two-dimensional space should be accessible from all points of P, it is sufficient either that (1) there exist a circle 0 with center at x such that the set of all points of the boundary, B, interior to 0 is a subset of a connected im kleinem subset of S - P (S being the set of all points in the space considered), or that (2) x belong to no continuum of condensation of B.*
It is clear that the conditions of Theorem 2 are satisfied if any one of the following simpler conditions are satisfied, viz., that (&) B be connected im kleinem, (b) B be a continuous curve, since in this case (a) is fulfilled**, (c) x belong to a connected im kleinen subset, K, mannigfaltigkeiten, Zweiter Teil, Jahresbericht der Deutschen Mathematiker-Vereinigung, vol. 2 (1908), p. 215.
of B, such that x is not a limit point of B - L
The accompanying figures are given to show that the two conditions of the above theorem are independent, and that they are only sufficient, not necessary.
Figure 1 is an illustration showing the independence of the conditions. ABCD is a square, E the center point of its interior, and P and F the mid-points of DC and AB, respectively. Pg, Pg, • ••• is a sequence of points such that if the distance BP—l, P n lies on the straight line EP and at a distance 1/n from P. Gj_ and Gg are the mid-points of AF and FB, respectively, and and GgHg are perpendiculars to AB of length 1/2; os, Gs, and G$ are mid-points of FGg, Ggß, respectively, and HgGg, H4G4, HSGS, HsGg, are perpendiculars to AB of length 1/4; and so on. Consider the hounded domain R whose boundary is the set of points consisting of the square ABCD, the sequence of points Pg, Pg, •••• and the straight line intervals EF, GgHg, GgHg, .... Every boundary point of R is accessible from any point of R; every point of the side AB of AB CD is on a continuum of condensation of the boundary of R, yet condition (1) of Theorem 2 is satisfied for all points of AB; condition (1) is not satisfied at P, yet P is accessible from any point of R by condition (2).
It can be shown that the above conditions are not necessary by taking as part of the boundary an isolated set of points of the domain having F as sequential limit point.
just as P is the sequential limit point of the sequence ? 1 , * ’ • • • ♦
A point of the boundary may be a limit point of continua of condensation of the boundary and yet be accessible from any point of the domain provided the point does not itself belong to a continuum of condensation of the boundary, as in figure 2. The points Pg, Ps, .... are situated on the straight line interval OPj_ in such a way that for every positive integer n, is at a distance 1/n from 0, and is a straight line interval perpendicular to oP lt of length 1/n and midpoint p n* The lines A n ß n are continua of condensation of curves of the type sin 1/x, and 0 is then a limit point of continua of condensation, but is accessible from any point of D.
An extension of space S can be made by considering S to consist of any connected im kleinen, closed and bounded set of points. For this purpose, the following theorem is introduced:
Theorem S• In order that a closed and. bounded pointset_should be connected im kleinem, it is necessary and sufficient that it should be the sum of a finite (or vacuous) set of mutually exclusive continuous curves t together with a finite Lor vacuous! set of isolated points.
Proof. 1. The condition is necessary. Let Kbe a closed, bounded and connected im kleinen set of points. By virtue of the property of connectivity im kleinen, every point Pofg is the center of two circles and Cg such that every point of K interior to Og lies with P in some continuuip of K
that lies entirely interior to If L denotes the set of all those points of K interior to C], which lie with P in some continuum of If that lies interior to o]_, together with the limit points of such points, then L is either the isolated point P, or a continuous curve which contains every point of K interior to Cg*. Since Kis closed and bounded, there exists,
by virtue of the Heine-Borel Theorem, a finite number of oiroles of the type Og, namely •••• $2 n » saoh that if x is any point of K, there exists a positive integer i. suoh that x lies interior to ea °h these oiroles corresponds a set, L, which is either a continuous curve or a single point P, and which contains all points of K interior to Cg. Hence, K consists of a finite (or vacuous) set of continuous curves together with a finite (or vacuous) set of isolated points. Since a finite number of continuous curves which have points in common form one continuous curve, the continuous curves in the theorem can be considered to be mutually exclusive**.
2. The condition is sufficient. Let Kbe a closed, I bounded set of points which is the sum of a finite set of mutually exclusive continuous curves Lg, lg, and a finite set of isolated points Pg, •••• Pj. If xis a
point of the continuous curve L n (n 2,. •• • i_) there exists a circle £ with center at x which contains no points of any other continuous curve of nor any point of the sequence p 2» •••• £ j* Since L n is connected im kleinen at x, it follows that Kis connected im kleinen at x. If x— P 1 t 2, • ••• j), there exists a circle 0 with center at x containing or enclosing no points of the set In + + .... + L n> and no point of the sequence P lt P 2> .... Pj. The conclusion is obvious*
*Presented. to the American Mathematical Society, in part, April 15, 1922, and. in more complete and. generalize ed. form, Feb# 24, 1923.
**Kura towski has used, an analogous definition for
what he calls ’’domaine connexe par rapport a 0.” Of. 0. Kuratowaki, Une definition topologique de la ligne de Jordan, Fund. Math. I, (1920), pp. 40-43.
transactions of the Amer. Math. Soo., Vol. XVII, No. 2, pp. 121-164, (1916).
**A simple continuous arc, or an arc, from A to B is a closed., bounded and connected set of points that is disconnected by the omission of any one point except A and B. A and B are called the end-points, or extremities, of this arc.
*lf x is a boundary point of a domain (in the ordinary sense, or an S-domain) D, and y a point of jD, x is said to be accessible from y provided there exists an arc from x to y which lies, except for x, wholly in D.
**A continuum is a closed and connected point-set consisting of more than one point. A point-set 0 is a continuum of condensation of a point-set M if 0 is a sub-continuum of M consisting of more than one point and such that every point of 0 is a limit point of M-£.
*A point P is a sequential limit point of a sequence of points Pj_, P 2, Pg, .... provided that if Ris a region containing P there exists a number N such that if n>N, P n lies in R.
*lt is permissible to say "arc" here instead of "connected set" since all points of S lying interior to K 2 belong to that region of £ determined by and since every two points of a region are the extremities of an arc lying wholly in that region.
*1 will remark here that, if P is a point of P, and arcs are taken from P to two points A and B’of B in such a way that P is divided into two domains P]_ and Pg, x being as well as on B, — — on the boundary of that x is accessible from provided that the conditions of the theorem are satisfied for P, at x. That is, the conditions of the theorem are sufficient for accessibility from all sides of the point x. I shall not, however, make use of the idea of accessibility from all sides in this paper.
**Schoenfliess showed, that all boundary points of an ordinary tw T o-dimensional connected domain are accessible from the domain if its boundary is a continuous curve. Cf. A. Schoenfliess, Pie Entwiekelung der Lehre von den Punkt-
*Hahn has shown that L is connected im kleinen. L is obviously what Hahn calls M*(P,r). Of. pg. 2448, loc.cit.
**Point-sets are said to be mutually exclusive when they have no points in common.
pi
II. An analysis of the point-set which constitutes the boundary of a complementary domain of a plane continuous curve*.
In the present section I shall analyze the boundary plane of a complementary domain of curve in terms of the elementary notions of point, arc and simple closed curve, and prove certain fundamental properties about such sets, including a necessary and sufficient condition that the boundary of a simply connected domain be a continuous curve*
If S is a continuous curve in ordinary two-dimensional space S r , and P a point of S 1 -S, that maximal** connected
subset of S' - S determined, by P is a complementary domain D of the continuous curve S. One of the complementary domains of S is unbounded, and the unbounded complementary domain always exists; S may or may not have bounded complementary domains* The boundary, of D, consists of all limit points of D that do not belong to D. Evidently is a subset of S.
Since S is bounded, there exists a circle C, which encloses all points of S but contains no points of S* If D is a bounded complementary domain of S, then the outer boundary of P is the boundary of the point-set composed of all points [x] such that x can be joined to some point of C by an arc which contains no point of P + R. L. Moore has shown* that B, the
outer boundary of D, is a simple closed curve. By a theorem due to Miss Torhorst** itself a continuous curve.
Theorem 4. If is the set of all simple closed curves (excluding the outer boundary in case the domain is bounded) contained in the t of a complementarydomain of a continuous curve S, then (1) Sq is countable, (2) if and C j are two distinct simple closed curves of the set have at most one point in common, and their interiors have no point in commoai;(3) in case I) is bounded and Cis any simple closed curve of S lt C lies interior to B, the outer boundary of D, or has at mofet one point P T in common with B, such that C - P 1 is interior to B.
Proof* In case Dis bounded, no points of can lie exterior to B, since all points ofmust be accessible from D by Theorem 2.
If £ is any closed curve of Sj, C cannot have more than one point in common with B. For, suppose £ has two points, A and E, in common with B. Then 0 will contain an arc, A*xE? which lies, except for A’ and E', wholly interior to B. A'xE' will divide the interior of B into two domains, and Rg, which are mutually exclusive*, and such that if I is the interior of B,
*Cf. R. I. Moore, Foundations of Plane Analysis Situs loc. cit. p. 141.
I =-Rl + R 2 + A’xE 1 ,- A' - B*.
D must lie wholly in either or Rg. Clearly points on one of the arcs into which A' and E' divide B are not accessible from P, a contradiction of Theorem 2. Hence 0 can have at most one point, P, in common with B, and the set 0 - P must lie interior to B.
In any case, D being bounded or unbounded, if C k and distinct — Cj are closed curves of the set the interiors of C k and 0j can have no point in common; for if their interiors have a point in points,of lie interior to oj, or vice versa, and a contradiction of Theorem 2 results.
Furthermore, C k and. 0j oan not have more than one point in common. For suppose they have two points, A and. E, in common. 0j must contain at least one point, x, which is exterior to C k , since C k and. 0 are not identical and. their interiors have no point in common, and. there will exist two points, A' and E l , common to G k and oj, such that the arc AJ^ 1 is exterior to C k . From Theorem 27, of R. L. Moore's Foundations of Plana Analysis Situs, it follows that one of the arcs into which A' and E’ divide C k is interior (except for end-points) to the simple closed curve formed by the other arc of C k and A'xE', a contradiction of Theorem 2 again.
Si, then, is a totality of simple closed curves whose interiors have no point in common, and such that any two have not more than one point in common. The closed curves of this set must therefore form the boundaries of a set of mutually exclusive domains. As every set of mutually exclusive domains is countable, must "be a countable set oq, Og, 0g» ••••
Theo rem 5• The boundary, p, of a complementary domain, D, of a continuous curve, S, cannot contain an uncountable infinity of simple continuous arcs no two of which have a point in common.
Proof. Suppose that does contain an uncountable set, T, of simple continuous arcs, such that no two arcs of T have a point in common. Then there exists some positive number C, such that there are uncountably many arcs of T of diameter* greater
than£. Gall the totality of these T l .
In each arc of T 1 there are two points x and £ such that
y& (i)
Assign xto a set [a] and to a set [Bj. The set {a] will have at least one limit point, A, since it is bounded. Then from £a] can be selected a sequence of points Xg, Xg, .... which has Aas a sequential limit point• Let yg, yg, • ••• be points of £b] such that for every positive integer n, x n and y n are points of the same arc of T*.
The set y-|_, yg, yg, .... must have at least one limit point, B*, and contains a subsequence Bjl, Bg, Bg, .... which has B_* as a sequential limit point, let A 2, Ag, .... be a subsequence of the sequence Xy, x 2, Xg, .... such that for every positive integer n, A n ,and. B n belong to the same arc of the set T r . Then A and. B* are sequential limit points of the sequences Ai, A 2, .... and. B lt 8 2 , Bs, ...., respectively, since if a sequence of points has a sequential limit point, every subsequence of it has the same sequential limit point. A and. B_* are distinct points, since, as a result of (1),
<f(A,B*) £ S. J(A,B*) =B>).
Let
Hahn has shown* that a connected, im kleinen continuum
is uniformly connected, im kleinen. That is, for every positive number E there exists a positive number such that if P 1 and. P’ ’ are two points of the continuum such that
(ftp’ ,p ”) <
then P’ and. P” are the extremities of an arc, 1, such that if x’ is any point of 1,
and. <J(x’ »£’ 1 £•
Since is a continuous curve, and. therefore a connected, im kleinen continuum, it is uniformly connected, im kleinen. Therefore, corresponding to the number there exists a number satisfying the conditions for uniform connectivity im kleinen.
There exists a positive number N- such that if n>N, is interior to a circle of diameter with A as center, and B n is interior to a- circle of diameter with B* as center. Hence, for any such value of n,
+ l) and cF(B n ,B n+ i) <
Then A n and A are the extremities of an arc of such that if P is a point of this arc,
and feintlK 'b
and, since
&P,A) < 2 0
Similarly, for n>N, B n and B n+l are the extremities of an aro of p satisfying similar conditions. These two arcs can have no point in common, and the arcs A n ß n and A n+ jB n+ j_ have no point in common by hypothesis. From the point-set composed of these four arcs can be selected a simple closed curve which consists of intervals of the four arcs. This simple closed curve bounds a domain which is bounded. The totality of such domains form a sequence >l. s2> Dg, . each of whose boundaries belongs to Si (except for one which may be the outer boundary, B, and which, for convenience, may be considered as omitted from the sequence). By Theorem 4 (2), these domains are mutually exclusive; furthermore, they constitute an infinite set of domains complementary to each of which is of diameter greater than 4F|. But this is a contradiction of a theorem due to Schoenfliess* to the effect that if £ is any given
positive number, there do not exist infinitely many domains complementary to a continuous curve, each of which is of diameter greater than £*
Hence cannot contain uncountably many arcs no two of which have a point in common*
Theorem 6. The boundary, p, of a complementary domain of a continuous curve, cannot contain an infinite set, countable or uncountable, of simple continuous arcs of diameter greater than any positive number 8, such that no two of these arcs have an interior point of both in common*
Proof* If such a set exists, select from it a sequence ti, tg, tg, •••• For every positive integer n, t n , being of diameter greater than £, must contain an arc t n ’ which does not contain the end-points of t n and is itself of diameter greater thang* Then the set t-^ 1 , , tg f , •••• is an infinite sequence of arcs no two of which have a point in common, and each of which is of diameter greater than 6. An argument similar to that used in the proof of Theorem 5 may be used to complete the proof of this Theorem.
Definition: If M is a continuous curve, I shall define an ( end-point of M to be any point P of M, such that if a is an arc of M whose end-points are P and any other point P T of M, the set M - (a - P) contains no connected subset consisting of more than one point which contains P. If a point P’ can be found such that this condition is not satisfied, then P is not an end-point of M*
Definition: If M is a continuous curve, and P a point of M, then if M - P is not connected, P is called a of M* If M - P is connected, I shall call P a non-cfl.t-point of M.
Theorem 7. In order that a point of a continuous curve that contains no simple closed curves should be an end-point, it is necessary and sufficient that it be a non-cut-point.
Proof, (a) The condition is necessary. For, let Pbe an end-point of a continuous curve M which contains no simple closed curve, and let P’ be any other point of M. P and P’ are the extremities of an arc aof M. Since P is an end-point, M - (a -P) contains no connected subset consisting of more than one point that contains P. Suppose that Pis a cut-point of M. Then
M - P = + Mg,
where and Mg are two non-vacuous mutually separated** sets.
Now P, "being an end-point of a, does not disconnect a. And a -P, being connected, must lie wholly in M-|_ or Mg. Suppose it lies in Since Mi has no limit points in Mg, Mg contains a domain d with respect to M, whose boundary with respect to M is the point P. (This follows at once from the connectivity im kleinen of M and the fact that Pis a limit point of M.) If xis a point of Mg, x and P are the extremities of an arc a’ which lies wholly in d and hence in Mg, except for the uoint P, by Theorem 1. But since a P lies wholly in
M-(a-P)=> * Mg + P a'
That is, M - (a - P) contains a connected subset which contains P. Hence under the supposition that P is a cut-point of M a contradiction results. Therefore P must be a non-cut-point of M.
(b_) The condition is sufficient. Let Pbe a non-cutpoint of a continuous curve M that contains no simple closed curve. Suppose Pis not an end-point of M. Then there exists a point P’ of M, such that if a be an arc of M whose extremities are P and P T , M-(a - P) contains a connected subset Mg, consisting of more than one point, which contains P.
Sigce P is a non-cut-point, M- P is connected. Divide M- P into two sets, a- P and Mg’. Then
M 2 ' => Mg -£•
Let dbe that maximal connected domain with respect to M such that
Vo a
and
d Z>M 2 - P.
Then d must have some limit point x distinct from P in a.
Let £be a point of d. There exists an arc a 1 whose extremities are x and £ and which lies wholly in d except for x (see Theorem 1); also an arc a ’ T whose extremities are y and P, and which lies wholly in d except for P. Then a. a’ and a’’ contain a simple closed curve • But M contains no simple closed curve by hypothesis. Hence the supposition that P is not an end-point of M leads toa contradiction, and P must therefore be an end-point of M.
Theorem 8• If M is a continuous curve that contains no simple closed curve, then M cannot contain, for any given positive number &, an infinite number of arcs of diameter greater than and such that no two of these arcs have an interior point of both in common.
Proof: The proof of this theorem is nearly identical with the proofs of Theorems 5 and 6, except that a contradiction is obtained as soon as it is demonstrated that M contains a single simple closed curve. Or, it may be considered a special case of Theorem 6, since such a continuous curve is the boundary of only one domain, namely, the unbounded domain.
Theorem 9. If Mis a continuous curve, and Nis a closed proper* subset of M, then M - K is a countable set of
domains with respect to M whose boundary points with respect to M are contained in N, and such that if
r-M - (N + d)
where d is any one of these domains, R and d are mutually separated.
Proof: LetJE be a point of M- N. Then that maximal connected subset, d, of _M - N determined by P is a domain with respect to M. To show this, let
R s M - (N + d)
(1)
Then R and d. are mutually separated. For, R can contdn no limit point, x, of d, since d + x would be connected and x therefore a point belonging to d, contrary to (1). On the other hand, if d contains a limit point, y, of R, there exist two circles, and kg, with centers at y, neither of which encloses a point of N, and such that all points of M interior to kg lie with x in a connected subset of M that lies wholly interior to kjL, and furthermore such that at least one point, x, of R, lies within k 2. Then x and y lie in a connected subset L of M that lies wholly interior to But no points of N lie interior to and therefore
M - H 1.
As d is a maximal connected subset of M - N, y must belong to d. This is impossible, since R and d are mutually exclusive according to (1). Hence no points of R can lie interior to kg and y cannot be a limit point of R. As R contains no limit points of d, and d contains no limit points of R, and as, moreover, the two sets are mutually exclusive by (1), R and d are mutually separated.
Hence, if P’ is any point of d, there exists a circle which encloses P’, but no points of Ror N. It follows at once that dis a domain with respect to M. The boundary of d with respect to M must be a subset of N, since d can have no limit points in JR.
Si nee every point P of M - N determines a maximal connected subset of M - I, i.e., a domain with respect to M, M_ - n is a totality of domains, [d] , with respect to M, whose boundaries with respect to M belong to N.
To show that these domains form a countable sef, it is only necessary to make use of the fact that every uncountable point-set contains at least one of its limit points. For, if Q be a point set which consists of points of £d] such that one and only one point of each domain of [d] belongs to Q, Q contains at least one limit point of itself. Call such a point A. let dbe that domain of [d] of which A is a point. If R be defined as in (1), then
R Cd] - d.
(2)
But
[d] - d =>£ - A
(2)
Prom (2) and (j) it follows that A is a limit point of R, which is impossible, as shown above. Hence the set of domains [d] is countable.
Theorem 10: If J? is a continuous curve that contains no simple closed curve, and N a closed proper stabset of M, then for any positive number & M - JT contains at most a finite number of maximal domains with respect to M of diameter ■ r _.. •>—. _— . II W ■* ■ I l—a W — 'W—« greater7than £•’ i
The proof of this theorem is similar to that indicated for Theorem 12.
Theorem 16: If Mis a continuous curve that con\ \ A ~~\ tains no simple closed curve, IT a closed and connected proper subset of M, and d a maximil connected domain which is a sub-
Theorem 11: Every closed and connected subset of the boundary of a complementary domain of a continuous curve is itself a continuous curve.
Proof: Let be the boundary of a complementary domain of a continuous curve S, and N a closed and connected subset of To show that N is a continuous curve, it is necessary to prove it connected im kleinen.
let P be a point of N at which it is not connected im kleinen. Then there will exist* two concentric circles k and k 2,
and a countable infinity of closed and connected point-sets* M, Mj, such that (l)each of these point-sets is a - —— — subset of N and contains at least one point on and at least one point on kg, but contains no point exterior to or interior to k 9, (2) no two of these point-sets have a point in common, and indeed, no one of them is a proper subset of N which contains no point without or within kg, ($) the set M is the sequential limiting set of the sequence of sets M l# Mg, Mg, circles kn and ko can be determined in such a way that
Let p be the lower limit of the distance xy, where x * -ur is any point of Mg and yis a point of Mj + Mg* Sinoe is a continuous curve and therefore uniformly connected im kleinen, there exists a number such that if a and b are any two points Of such that
<T(a,b) <
then a and b are the extremities of an arc ab ofsuch that every point c of the arc ab satisfies the relations
/(a,o) < (f ) dVb.c) < f
Since Mg is a continuum, there exists in Mg a chain of points Xg, such that
»x^ + -^) » (1 = 1,2, ..... k-l) 7
is a point of and. a point of kg. Then there exists an arc x^x of such that if x’ is any point of this arc,
(/’(XpX' ) f 3 J (x 1+1 ,x»)
The set X»Zx^x^ is a continuous curve, since any connected set consisting of a finite number of arcs is a continuous curve, and therefore contains an arc whose end-points are and Let be the last point of this ®rc on in the order from ±1 to and the first point of this arc on k g in the erder from to Then = tg is an ®ro of flwhich lies, except for its end-points, wholly interior to and wholly exterior to k 2.
Now let p' be the lower limit of the distance xy, where x is any point of Mg, and y any point of M 4 + tg. There exists a number such that if a and _b are any two points ofsuch that
</(a,b) 7 J
then a_ancl b are the extremities of an arc ab of such thtt every point o of the arc ab satisfies the relations — -* ——
c/(a,o) < j cAb.o) <
As is a continuum, there exists in a chain of points x^ t
Xg, ...... x n , such that
< » (i - 1,2,.....
xq is a point of and x n a point of kg. By a discussion similar to that used above in showing the existence of the arc tg, it can be proved that there exists an arc having properties similar to those of tg.
Continuing in this manner indefinitely, it can be shown that there exists an infinite sequence of arcs, tg, ts, such that for every positive integer n>l,
/9—t au 8 j az u and such that t has no point in common with any other arc of the sequence tg, t 4, •••••• Clearly this is a contradiction 9 of Theorem Hence N must be connected im kleinen at P, and A as P is any point of N, il must be connected im kleinen.
It might be noted in passing that Theorem 15 is a generalization of a result obtained by Mazurkiewicz* to the
effect that every closed and connected subset of a continuous curve that contains no simple closed curve is itself a continuous curve.
Theorem 12: If M is a continuous curve that contains no simple closed curve, N a closed and connected proper subset of M, end d a maximal domain which is a subset ofM, 4 then d has only one boundary point with respect to M, and that point belongs to N.
Proof: The boundary points of d all lie ini as a consequence of Theorem 14. Suppose there exist two boundary A points of d with respect to M. By Theorem 16 Nis itself a continuous curve and therefore contains an arc from to Pg. application of But 1 there exists another arc from to Pg which lies, except for and Pg, wholly in d.. As d and N have no points in common, these two arcs have in common only the points and Pg, and their sum is therefore a simple closed curve. But this is a contradiction of the hypothesis that M contains no simple closed curve. Hence d has only one boundary point with respect to M.
For convenience, the major results of Theorems 14, 15 and 17 are embodied in one Theorem as follows:
Theorem 13: If Mis a continuous curve that contains no simple closed, curve, and. N is a closed and. connected proper subset of H, then M - N consists of a countable set of domains with respect to M, viz., d_ , d o , d,, such that (1) for — 1 .9 every positive integer n, d n has one and only one boundary point with respect to_M, and this point is a point of N, (2) no two of these domains have a point in common, -dnmnin?, q Hmtt po-int of *7 and ($} if £. is any positive number, at most a finite number of these domains are of diameter greater than £.
Theorem 14: If x and. y are two distinct points of a continuous curve M that contains no simple closed, curve, then no point of the are xy of M, excepting the points x and. y, is an end.-point of M.
Proof: As shown by Mazurkiewioz*, there exists inM
only one arc from jx to y. Let Pbe a point of xy, distinct from x and y. Then
xy - P = + Mg,
where and Mg are mutually separated sets. As a consequence ..,n—■» — of Theorem 18. M - xy consists of a set of domains with respect to M each of which has one and only one boundary point with respect to M, that point being a point of xy. let £d’3 be the collection of all those domains of M - N whose boundary points with respect to M lie in and Ld T ’J the collection of all those domains of M - N whose boundary points lie in Mg.
If $ is the boundary with respect to M of a domain di of [d 1 ] + Cd 11 ], then and M - +P) are two mutually separated sets (see Theorem 18), whence by Theorem 11, P is not an end-point.
If P is not the boundary with respect to M of any domain of id’] + Cd’ I ], then the sets (d’] + and £d IT ] + are mutually separated, andP is not an end-point of M.
In any case then, P cannot be an end-point.
Theorem 15: A continuous curve M that contains no simple closed curve consists of (1) a sequence of arcs Cg, c^ t ...... no two of which have in common an interior point of both, and such that (i) if n is any positive integer, + + + + c R is a continuous curve and a proper subset of M, (ii) for £>o, there exists a number such that if n
< £
and the diameter of any one of the countable set of maximal domains with respect to M lying in M - M is less than £; l’ — and (2) a totally disconnected set of end- (or non-cut-) points, P w , each of which is a limit point of the sequence Mp Mg, Mg, ....
Proof: The continuous curve Mis of diameter greater than some positive number, say 1. Than there exist two points, x and y, belonging to M, that are not end-points of M, and whose distance apart is greater than I*. Let be that arc of M
which has x and yas end-points. As a result of Theorem 18, M - is a set of domains with respect to M, whose boundary points with respect to M lie on and such that only a finite number of these domains, dg, ds, .... d k , are of diameter greater than 1. let be the boundary point of with respect to _M (i -k). k). Since the diameter of d. is greater than 4J/U itcrV 1 “* 1, there exist in two points and such that
J'(x 1 ,y i ) > 1.
It is clear, than, that either
</’(Pi,y : £) > t
or
Suppose that
/(P i ,x I )>i
Then the set composed of d i and contains an arc from x i to whose diameter is greater than
Now the set .
+ Cg + •••••• +
where
c. = arc x P . .... k) i i
is a continuum. Then M - consists of a set of domains* ' ■ ■■ ll—*
only a finite number of which can be of diameter greater than 1$ let these be d k+l , d k+2 , •••• d k+j* Sinc ©
</U i+l )> 1, (1 = 1,2,5 j) **
there exist in d k+i two points x i and which are not end-points, such that
> 1.
Then, if be the boundary point of with respect to M,
or y ± ) >i .
Suppose that
/CP i ,x 1 ) > f .
Then by virtue of Theorem 1, d, .. + P. contains an arc cr. whose - x+i i k+i extremities are x. and P.. Thus a continuum
V= V + c k + i + + c k + 3
is obtained.
Continuing this process, there is obtained, eventually, a continuum
M m * -il + 0 2 + + °h
such that M - consists of a set of domains only a finite number of which are of diameter >4-, and none of which is of diameter > 1. For if domains of diameter >1 could be obtained indefinitely, plainly an infinite set of arcs of diameter would also be obtained, a situation contradictory to Theorem 9. M- V being non-vacuous, let its domains of diameter > if any exist, be denoted by d h+2 , ••••• d h*q* the boundary point of d h+i (i * 1,2,....q) be Since
( i
it can be shown that there exists in d. ~ a point x. such that h+ i * i
< /(r i ,x i )>i ,
and hence d^ + contains an arc c^ +i from to
If the process indicated above be continued indefinitely, there is obtained a connected set of arcs ...... such that (1) none of these arcs contains any end-point of M by virtue of Theorem 19, and no two of them have in common a point which is an interior point of both, (2) if &is any positive number, there exists a number p such that if ii. is any positive integer
£ (On) £ j
and the diameter of every one of the countable set of maximal domains with respect to M lying in M - where
= °1 + °2 + + °n
(a continuous curve since it consists of a finite and connected set of arcs) is less than £.
If the set of points consisting of the totality of arcs in the seauence c_, c o , c,, be denoted by M* than 1 2’ — M* is non-vacuous. For J!* contains no end-points of Mby (1) above, and Mmust at least two end-points, as already shown.
Every point of M- M* is a limit point of M*. For, suppose P is a point of M - M* that is not a limit point of M*. Thon there exists a circle 1c with center P that contains no point of M*. But P must belong to some proper connected subset s of M - M* that lies interior to k, by virtue of the property of connectivity im kleinen of M. Let denote a positive number such that
Now there exists a such that if n is any positive maximal number M - consists of a set ofyydomains with respect to Mno one of which is of diameter greater than/p. However,
s_ must belong to some domain of M - since
M - M - M* s,
and this domain cannot be of than /y. Therefore the supposition that P is not a limit point of leads to a contradiction; hence all points of M - _M* must be limit points of M*.
The set M - M* is a totally disconnected set, since if it contains a connected subset s a contradiction will result as in the preceding paragraph.
Since M* contains no end-points of M, and since M must have at least two end-points, all end-points of M must lie in M - M*. It remains to show that every point of If -M* is an end-point of M.
Suppose there exists in M - M* a point P that is not an end-point of M. Then Pis a cut-point of M, by virtue of Theorem 11. Hence
M - P + H 2,
where and Eg are mutually separated sets. M* is a connected set that does not contain P, and therefore P cannot disconnect M*. Hence M* must lie wholly in H, or H 9, say in H_ .As H o — anl as > cannot contain any limit points of points of M - M* are of M*, the set M - (M* + P) must be a subset of H n . —mil, . X -I— But
M- P = M* + M - (M* + P).
Hence, if
I H x J=> M* + M ~(M* + P)
Hg must be a vacuous set. That is, M- P does not allow of division into two mutually separated sets, and is therefore connected. Pis therefore a non-cut point of M, and by Theorem 11 an end-point of M. This completes the proof of the Theorem.
The following theorem is a generalization of Theorem 18:
Theorem 16: If Mis the boundary of a complementary domain of a continuous curve, and N a closed and connected proper subset of M, then M - N consists of a countable set of domains with respect to M, viz., d , d o , d„, ..... satisfying the don— d £ ditions of Theorem 18, except that (1) should be made to read, "for every positive integer n, d n has at most two boundary points —IM with respect to M, and these points belong to N."
The proof is very similar to the proof of Theorem 18.
Theorem 17. The boundary of a complementary domain of a continuous curve is the sum of three mutually exclusive point-sets S 2 and [P], where (1) is a countable set of simple closed curves no two of which have more than one point in common and whose Interiors have no point in common (unless one of these simple closed curves be the outer boundary in the case of a bounded complementary domain), (2) S 2 is a countable set of simple continuous arcs no two of which have in common an interior point of both, and (3) [P] is a totally disconnected* set of limit points of the set ’ + S 2.
Proof. (1) has been established in Theorem 4.
let the boundary of a complementary domain of a continuous curve, and P a point of S]_’. That maximal connected subset of determined by P I shall call a set of type Q, provided it consists of more than one point. A set of type Q, together with its limit points, I shall call a set of type N.
By Theorem 11, a set of type N is a continuous curve. Furthermore, it is a continuous curve that contains no simple closed curve. For, suppose N is a set of type N that contains a simple closed curve 0. Then
V O C,
and as
4
where Q represents that set of type Q which determines N, then must
N - DC.
That is, every point of 0 is a limit point of Q that does not belong to
Let d be that maximal domain with respect to which is a subset of ft- 0 and of which is a subset. Then every point of 0 is a boundary point of d with respect to Clearly this is a contradiction of Theorem 16. Hence no set of type N can contain a simple closed curve.
Applying Theorem 15, every set of type N is the sum of a countable set of arcs, no two of which have in common an interior point of both, together with a totally disconnected set of limit points of these arcs. All arcs of ft so determined assign to a set Sg.
The sets Sj’ and Sg are mutually exclusive* For, suppose they have in common a point x. Then x belongs to some arc, a, of some set N of type N, and, as determined, no points of a are end-points of N (see proof of Theorem 15)* It follows, that x is a cut-point of N. Hence N- x is the sum of two mutually separated point-sets, N]_ and Ng. Both and Ng must contain points of Q, the set of type Q which determines N. But this is impossible, since Q is a connected subset of B- Sj/, and hence of a. It follows that the sets and Sg are mutually exclusive, and it also follows that every arc of Sg is a subset of some set of type Q.
If a and b are two arcs of Sg, then (1) if a and b belong to the same set of type Q, they have in common no point which is an interior point of both, and (2) if a and b_ do not belong to the same set of type £. they have no points in common. It follows that the arcs of Sg are countable (by Theorem 5).
The set of points
[P] — (81* * Sg)
is a totally disconnected set of points. For, if there exists a connected subset, t, of [P], then t is a subset of some set of type Q, as
Si'rs t.
If N is the set of type N determined by this set of type Q, and [a] the set of arcs common to N and Sg, then N - [a] is totally disconnected by Theorem 15. But
£ - La]
It follows that t cannot be a connected set, and. that [p] is totally disconnected..
This completes the proof of the theorem.
*Presented to the American Math* Soc*, Dec. 29, 1922.
**lf M is a point-set and P a point of M, the maximal connected subset of M determined by P is the set of all points of M that lie, with P, in a connected subset of M.
*R. 1. Moore, Concerning continuous curves in the plane Math. Zeitschr., Band 15, (1922) pp. 254-260.
**Uber den Rand der einfach zusammen hangenden ebenen Gebiete. Math. Zeitschr. 9 (1921), S. 64 (73).
*The diameter of a point-set M is the upper limit of (T(x,y), where x and £ two points of M and cT means distance; i.e., stands for the distance from x to y.
*Loc. cit.
*loc. oit., p. 221, IX.
*This definition was first given by R. L. Moore in
Concerning the cut-points of continuous curves and of other closed and connected sets. Proc. Nat’l Acad. Sci. vol. (1922).
**Two point-sets are said to be mutually separated when they are mutually exclusive and neither contains a limit point of the other.
*The symbol should be read "contains or is identical with”.
*lf N is a subset of a point-set M, than K is a proper subset of M if M - N is not vacuous.
' " ' ” ■ * . " ' —— n-.- ■ — Of. R. L. Moore, Concerning the cut-points of continuous curves and, of other closed and connected point sets, loc. cit.
S. Mazurkiewicz, Un theorerne sur les lignes de Jordan, Fund. Math., Tom 11, pp. 123-125.
*See Un'theorerne sur les lignes de Jordan, Loo* cit.
*That two such points can be found is a direct consequence of a theorem proved by R. 1. Moore, viz., in order that a closed, connected and bounded point-set M should be a continuous curve which contains no simple closed curve, it is necessary and sufficient that every closed and connected subset of M should contain uncountably many cut-points; (See R. 1. Mpore, Concerning the cut-points of continuous curves and of other closed and connected'point-sets, loc.cit. )~ and of Theorem 11.
*M - is not vacuous, since contains no end-points by Theorem 19 (the points and P i being also non-cut points) and it has been shown by R. L. Moore that every bounded continuum contains at least two non-cut points. Of. R. L. Moore, Ooncerning simple continuous curves. Transactions of the American Mathematical Society, vol. 21 (1920), pp. MO-Ml, Theorem 2; Concerning the cut-points of continuous curves and of other closed and connected point-sets, Proo* of—the -Nat' 1 Acad, of Sciences, vol. Also see S. Mazurkiewicz,
**Where the notation for a point-set is placed in the parenthesis, the J 1 should be read ”the diameter of”.
*A point-set is said to be totally disconnected provided that it contains no connected subset consisting of more than one point.
III. Characterizations of continuous curves, and of the boundaries of the complementary domains of a plane continuous curve*.
In this section I shall establish a condition which continue that are not connected im kleinen must satisfy* By means of this condition I shall characterize continuous curves for any number of dimensions, and the boundaries of the complementary doipains of a plane continuous curve.
Lemmaf/e If a bounded continuum Mis not a continuous curve, then there exist two concentric circles and Kg, and a sequence of sub-continua of M,
» ......
such that (1) each of these sub-continua contains at least one point on and Kg, respectively, but no points exterior to or interior to K 9, (2) no two Of these sub-continua have a point in common, and no two of them points of a connected subset of M which lies wholly in the set + Kg+ I, (where lis the annular domain bounded by and Kg), is the sequential limiting set* of the sequence Mg, Mg, , (4) if K is
is that maximal sub-continuum of M containing IQ and. lying wholly in the set K, + K o + I, then all of the continua M,, M o , Mg, ...... lie in a connected, subset of M - K. Proof: Re L. Moore has established, conditions (1), (2), and of this Lemma**# It remains to establish condition
being the radius of the circle Ki, let Kg and K 4 be two circles concentric with Ki, and such that
R KI > R K g > k k 4 > r k 2
For every value of n, (n~ ,1,2,3, ) there exists a continuum such that
and. such that if in the statement of this Lemma each is replaced, by Mj.*, and. Kg replaced, by Kg and. K 4, respectively, and. K replaced. by T, statements (1), (2), and. (3) hold, true, but (4) may or may not hold. true. If statement (4) does hold, true, the proof is complete. If statement (4) does not hold true, let Kg and K$ be circles concentric with Ki and such that
5 K4 > ®Kb > ®E6 •>
For every value of n, (n ,1,2,3,.....), there exists a continuum M& such that
and such that if in the statement of this Lenina each is replaced, by , Ki and Kg replaced by K§ and Ks, respectively, and K replaced by W» statements (1), (2), and (3) hold true, but (4) may or may not hold true. I shall prove that (4) must hold true.
There exists an infinite sequence of distinct sets, Lg, ..... such that (1) for each n, L n is a maximal connected subset of M - and contains at least one set, but at most a finite number of sets of the sequence M lt Mg, Ms, , (2) every set of the sequence Mg, Ms, ••••• belongs to some set of the sequence Li, Lg, Ls,
For every value of n, (n ±1,2,3,...), has a limit point in T. For, suppose is a set of this sequence that has no limit point in T. Then, since is a maximal connected subset of M - T, is closed. and T are then two mutually separated continua.
There exists a connected domain D 1 containing but containing no point of _T nor having a point of T on its boundary.
If A is a point of and B a point of T, then M is a continuum containing A and B. That is, M contains a point A interior to and a noint B exterior to Then there exists a sub-continuum Q of M, which contains A and at least one point x on the boundary of but no point exterior to
x cannot belong to T and hence belongs to M - I; furthermore x is joined to A by a subset Q of M - T, and must therefore belong to Ip But no points of lie on the boundary of Therefore the supposition that has no limit point in T leads to a contradiction. Hence every set L n (n • 1,2,3,....) has a limit point in T.
Not more than one set of the sequence !•$,•••• has a limit point in W. For if one of these sets, say Lj, has a limit point in W, then must
w,
and. no two of these sets have a point in common.We oan consider as "being omitted from the sequence Ls, ••••• We have, then, and infinite sequence of connected sets, Lg, Is, •... such that for every n (n * 1,2,3,...),
M - W l n
and l n has a limit point in T. Since Tis connected, the set
U * + 1g + + ••• • • + W
is a connected subset of M- W. Since every set of the sequence Mi , Mg t , .... belongs to some set of the sequence lg, and therefore to U, condition (4) of the theorem is satisfied by the sequence W, Mg , M$ ...... as also are conditions (1), (2), and (3), if, as already pointed out, is replaced by W, each by , and and respective! y.
Although the above proof is given for two dimensions, it should be observed that a similar proof can be given to show that the lemma holds for any number of dimensions.
Lemma 2: If x and y are two points of the boundary, of a complementary domain of a continuous curve, there exist at most two distinct arcs from x to y; i.e., arcs that have in common at most their end-points, x and y. U /t Mi J * (This lemma is a corollary of Theorem 4)
Theorem 18: In order that the boundary of a simply connected domain should be a continuous curve, it is necessary and sufficient that every connected subset of it be connected in the strong sense*.
A. The condition stated in the theorem is necessary. Let Ml be a continuous curve which is the boundary of a simply —- connected domain D, and let Nbe any connected subset cf A] • If x and y are any two points of N, then x and y are the extremities of a simple continuous arc that lies wholly in N. For, suppose that no such arc exists. There are three cases to consider:
(1) x and y the extremities of two distinct arcs, and a 2 of M.
Since N contains no arc from x to y, there exists at least one point, P lt on and at least one point, P 2, on a 2, such that neither nor Pg belongs to N.
Add to N its limit points, and call the resulting set NJ. By virtue of Theorem 16, N* is a continuous curve, and hence contains an arc, a, from x to y. As a result of lemma 2 there are at most two distinct arcs in M from x to y: hence either & = or Suppose that a = Then belongs to N T , but not to N, as stated above. But the set NJ - must then contain a connected subset, N, containing x and y, and therefore a domain d with respect to N' containing x and y. But d must contain an arc from x to y, as in the - — of properties of space at the beginning of this paper, analogous to the properties of ordinary space stated in R. 1. Moore’s Foundations of Plane Analysis Situs.
Hence NJ • contains an are from x to y. As a result of £ — lemma/we must conclude that this arc is the arc However, the set NJ - + Pg) also contains a connected subset, N, containing x and and hence an arc a# from x to y. We have, then, two distinct arcs, and ag, from x to y, and a third arc, ag, which is identical with neither nor ag. There mu6t point P of ag that does not lie on the simple closed curve formed "by + ag. Then P determines an arc ag 1 of ag which lies, except for its end-points, wholly within or without the simple closed curve * ag. Let the end-points of a$J be x' and y’. There exist' thpee distinct arcs from x 1 to z y’. But this is a contradiction of Lemma 2. In this case, then, N must contain an arc from x to y.
(2) x and y the extremities of only one arc, a, of A| •
In this case, if P is a point of a not belonging to N, the set - P contains an arc from jx to y which cannot be identical with a, thus giving an immediate contradiction of the hypothesis that M contains only one arc from xto y. In this case also, then, must N contain an arc from x to y.
(3) x andy neither the extremities of two distinct arc of #| , nor of only one arc of ft, but the extremities of a totality of arcs, T, such that if and a 2 are any two arcs of T, and a 2 have interior points in common, and each has interior points that do not belong to the other.
let P denote a set of points such that (1) if P is any point of P, P lies on every arc of T, and (2) if P is a point that lies on every arc of T, P belongs to J?. That such points exist, distinct from x and y, can be easily shown as follows: Let be any arc of T. Suppose a 2 is any other arc of _T, and let P be a point of a 2 not on J? is an a 2 ’ — — interior point of an arc^ which is a subset of a 2, and which has in common with only its end-points; cill these end-
points A and B. Either_A is distinct from x and yorß is distinct from x and y. let Abe distinct from x and y. Then Ais a point of F. For, suppose Ais not a point of F. Then — a some arc, ag, of T, does not contain A. Let A’ be the last point of ag on the arc xA of in the order from x to A; on the arc A’y of in the order from A’ to y, let B T be the first point that has in common with the point-set consisting of the arc Ay of and the arc ag’• That arc of from A’ to B’ call a 2 f .. The set + ag* + as' contains two points that are the end-points of three distinct arcs, a contradiction of Lemma 2. Therefore A must be a point of F distinct from x and y. Similarly, all points common to a 1 and a g are common to all arcs of T and hence belong to F.
Either N contains all points of F or it does not.
Suppose N contains all points of F. If Pis any point of a, that is not a point of F, let a 9 be an arc of T that does not contain it. In the order from xto y let x f be the last point a 2 has in common with the arc xP of and y’ the first point after x’ that ag has in common with the arc Py of Since, as shown above, x’ and y’ are points of F, they belong to N, and as they are the end-points of two distinct arcs of M, the proof of case (1) of this theorem shows that N contains one of these arcs. Similarly, every point P of that does not belong to F determines two points x 1 and y’ of F which are th e end-points of an arc that belongs to N. Denote the totality of such arcs by L.
The point-set 1 + F contains an arc t from x to y. This fact can be established by showing that L + F is a continuous curve. L ♦ F is a closed set. For, (i) F is closed. If P is a limit point of F, then, since _F is a subset of and is closed, P must belong to If ag is any other arc of T, P lies on for the same reason. That is, P is common to all arcs of T, and hence belongs to F» Also, (ii) all limit points of L, if they do not belong to L, belong to F. For, let Pbe a limit point of L that does not belong to L. Let kbe a circle of arbitrary radius r and center P, and k* a circle of radius r/2, concentric with k. k T must contain points of an infinite set of arcs of L. But since L is a subset of M, and M cannot contain an infinite set of arcs of diameter greater than r/2 that have only endpoints in common (Theorem 9) then must an infinite set of arcs of 1 lie wholly interior to k. But the end-points of each arc of L must then lie interior to k, and as the radius of k is arbitrary it.follows that P is a limit point of such endpoints; i.e., of F. Therefore if Pis not a point of L it is a point of F. Since F is closed, and I is either closed or those limit points that it does not contain belong to F, it follows that F + L is closed. As L ♦ F is connected and a subset of L + F is a bounded continuum, and by virtue of Theorem 16 a continuous curve, therefore containing an arc from x to y. As 1 + I is a subset of N, it follows that N contains an arc from x to y.
Suppose N does not contain all points of f; that is, that there exists some point P of P that N does not contain.
If, as in case (1) of this theorem, N 1 denotes the set_N together with its limit points, then N’ is a containuous curve and therefore contains an arc from x to y; this arc must belong to T, and therefore contains P. That is, P belongs to N T , but not to N. Then NJ - P contains a connected subset, N, containing x and y, and as shown in case (1) of this theorem must therefore contain an arc from x to y that does not contain P. But this is impossible, since P is a point of F and therefore common to all arcs from x to y. Hence N must contain all points of F, and therefore, as shown above, an arc from x to y. B. The condition is sufficient. For, suppose it is not sufficient; then there exists a simply connected domain, D, whose boundary, M, satisfies the condition that every connected subset of it is connected in the strong sense, but which is not a continuous curve.
M is a bounded continuum*, and it therefore follows
that if it is not a continuous curve it is not connected im kleinen. Then the conditions of Lenina 1 must hold true of M.
Let P. be a point of that lies in X* Then K maybe re-defined as the set of all points of M that lie, with P, in a connected subset of M that lies wholly in the set + Kg + I. Since M* K contains a connected subset N such that
Nrs +M x +Mg + +
then must M • (K * P) contain the connected subset N + P, since a set can not be disconnected by the addition of limit points. By hypothesis, N_+ P is connected in the strong sense. That is, if x is any point of H, there exists a continuum L which is a subset of H + P and contains x and P. But this is impossible. For, suppose such a set exists. Let obe a circle with center Ka. at P and lying wholly in I. Then there exists a sub-continuum — L x of L which contains P and a point y on 0, but no point which is exterior to o**.
Hence y is a point of K, since lies wholly in I. But
M - (K - P) D N + P 010 J,
and y cannot be a point olff both K and M- K. The supposition that M is not connected im kleinen leads, therefore, to a contradiction. Hence Mis connected im kleinen and is a continuous curve.
It will be noted that the above proof also establishes the following theorem:
Theorem 19: In order that the boundary of a simply connected domain should be a continuous curve, it is necessary and sufficient that every connected subset of it be arc-wise**
connected..
It has recently been shown by R. I. Moore that every two points that lie together in a connected subset of an open*
subset of a continuous curve, lie in a sub-continuum of that open subset.** Using Lemma 1, it is easy to show that Any
bounded continuum which has this property is a continuous curve.
Theorem 20: In order that a bounded continuum M should be a continuous curve, it is necessary and sufficient that every two points that lie in a connected subset of an open subset of M lie in a sub-continuum of that open subset.
[The proof of this theorem for the two-dimensional case is given below. An aiialogous proof may be given to show that the theorem holds in m dimensions.]
(1) The condition stated in the theorem is necessary, as shown by R. L. Moore.
(2) The condition stated in the theorem is sufficient. For, suppose M is a bounded continuum satisfying the condition stated in the theorem, but which is not a continuous curve. Then the conditions of Lemmal are satisfied.
Let A and B be two points of M that lie in I, and let and Cg be two circles with centers A and B, respectively, such that (i) and Cg have no point in common nor have their interiors any point in common, and (ii) and C g lie wholly in I j Ka .
If Q-l and Q 2 denote the sets of all points of K interior to and C 2 * respectively, let
T - K - (Qi ♦ Q 2 >«
Tis closed, and therefore M-Tis an open subset of M. Since
KOT,
all of the continua M 2, Mg, •••• lie in a connected subset, U, of M - T.
Since A and B belong to M , they are limit points of the sequence M 2, Mg, .... and therefore limit points of U. The set U + A + B is therefore a connected subset of M * T.
By hypothesis every two points that lie in a connected subset of an open subset of M lie in a sub-continuum of that open subset. Then A and B must lie in a continuum N, such that
M - T N.
Since H is a continuum containing a point A interior to and a point B exterior to there exists a sub-continuum of N which contains A and at least one point P on but no point exterior to 0-p Since lies wholly in I, Pis joined to A by a continuum of M that lies in I and must therefore be a point of the set K, but not of the set Qi or the set Q 2. That is, P must be a point of the set T. But P also is a point of the set N, which is a subset of the set M* T. Hence the supposition that M is not a continuous curve has led to a contradiction, and M must therefore be a continuous curve.
*Th© theorems and lemmas in this section were presented to the American Mathematical Society, Dec. 29, 1922.
*A point-set t is said to be the sequential limiting set of a sequence of point-sets tg, tg, provided that (a) each point of t is the sequential limit point of an infinite sequence of points, Pg, Pg, .... such that, for every n, P n belongs to t n , and (b) if Pg, Pg, .... is a sequence of points such that, for every n, P n belongs to t , then contains the sequential limit point of every subsequence of P_, P , which has a sequential limit point.
**Concerning the cut-points of continuous curves and of other closed and connected point-sets, loc t cit.
*of. Anna M. Mulliken, Certain theorems relating to plane connected point sets, Trans. Amer. Math. Soc., XXIV (1923), (as yet not printed).
*A point-set M is said, to be connected, in the strong sense if every two points that lie in a connected, subset^ of M lie also in a sub-continuum of N. Of. Theorem 24.
Of. R. L. Moore, Concerning continuous curves in the plane, Math. Zeitschr., Band 15 (1922), Theorem 2.
** Cf. Anna M. Mulliken, loc. cit.
**A point-set M is said to be arc-wise connected when every two points of it are the extremities of a simple continuous arc that lies wholly in
*lf L is a closed proper subset of a continuum, v. M, then M - L is an open subset of M.
**R. L. Moore, Concerning continuous curves in the plane, loc. cit., Theorem 1.
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