. Transactions of the American Mathematical Society . We have shown above that not every point of S2 is in D\and that no point of S% is on B. Let P0 denote a point of S* without is a simple continuous arc from P0 to Pi. Let BI denote the first pointof B on this arc in the order P0 PI . Suppose that MI is that one of the setsMI and MZ to which BI belongs. Since MI is a closed set not disconnectingS, PO can, by the lemma, be joined to PI by an arc not containing any pointof MI . Let B2 be the first point of B on this arc J in the order P0 Pi. Then t Hausdorff gives a proof for the case
. Transactions of the American Mathematical Society . We have shown above that not every point of S2 is in D\and that no point of S% is on B. Let P0 denote a point of S* without is a simple continuous arc from P0 to Pi. Let BI denote the first pointof B on this arc in the order P0 PI . Suppose that MI is that one of the setsMI and MZ to which BI belongs. Since MI is a closed set not disconnectingS, PO can, by the lemma, be joined to PI by an arc not containing any pointof MI . Let B2 be the first point of B on this arc J in the order P0 Pi. Then t Hausdorff gives a proof for the case when one of the sets is bounded. Cf. F. Hausdorff,Grundzuge der Mengenkhre, Leipzig, Veit, 1914, p. 342. J Hereafter in this paper, arc and simple continuous arc will be considered synonymousterms. 152 ANNA M. MULLIKIN [September B2 belongs to M2. The set P0 #1 + Po B2 contains as a subset an arc BI #1 be a simple closed curve enclosing #1 but neither containing norenclosing any point of M2 (Fig. 3) and containing only one point LI of BI B2,. and let H2 be a simple closed curve neither containing nor enclosing any pointof HI or MI , but enclosing B2 and containing only one point LI of BI a point of D\ within H\ be joined to a similar point within H% by an arclying within D\. There will be a subset of this arc, an arc RI R2, such thatRI lies on Hi, R2 on H2 and all other points of RI R2 lie without both HI andHz. Then there is a simple closed curve J\ composed of LI L2 + RI RZtogether with either arc LI RI on H\ and either arc L2 ^2 on H2. It isevident that the points of B on or within Ji that belong to Mi can be enclosedin a finite number of circles no one of which contains or encloses a point ofLI L2 + L2 R2 + RI #2 + M2. And similarly those points of B on or withinJi that belong to M2 can be enclosed in a finite number of circles no one ofwhich contains or encloses a point of LI L2 + L\ R\ + RI R2 + MI or a pointon or within a circle of the first set. Then, clea
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