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August 21, 1998 Here is a hastily done translation of the parts of the Acta Mathematicapaper showing that the curve P is simple. In particular there is no attempt to typeset it nicely, but I hope it is not too bad to read. Please let me know of any errors. Dan Curtin Acta Mathematica, p. 147 Let C be such a curve, K(X) a point of C corresponding to the point X of AB; if there is a point X' on AB distinct from X (and different from B if X is A, different from B if X is A) such that the point K(X') coincides with K(X), that point is called a multiple point of the curve; otherwise K(X) is called a simple point. If all the points of the curve C are simple it is called a continuous simple curve, or else, in Hilbert's terminology, a Jordan curve. Last, such a curve is called closed or open according to whether the points K(A) and K(B) coincide, or not. Acta Mathematica, p. 157-161 8. Joining the points A, D and D, B (fig. 2) by segments one obtains the triangle ADB circumscribed about the curve P, in the sense that every point of P lies either in the interior or on the boundary of the triangle. In fact the construction shows that each vertex lies either in the interior or on the boundary of the triangle (thus, for example T, W, G, K, lie on the line AD) and so it is for the set S' of limit points of the vertices. Similarly, the triangle AGC is circumscribed about the part of the curve P lying between A and C (fig. 2), the triangle CKD is circumscribed about the part lying between C and D and so forth. As a result the part of P between C and E lies in the interior or on the boundary of the pentagon CKDNE (note that the sides CK and EN are perpendicular to AB.) Likewise one sees that if LM (fig. 3) is any side of the polygonal path Pa, every point of P between L and M lies in the interior or on the boundary of the triangle LRM where the angles at L and M each measure 30 degrees; and the points N and T divide LM in three equal segments, the part of P between N and T necessarily contained in the pentagon constructed with NT as base and similar to the one just discussed. To shorten our discussion we will call CE the lacunary segment of AB (fig. 2), FH the lacunary segment of AC, etc.; in general, where LM (fig. 3) is a side of Pa, the segment NT is called the lacunary segment of that side. Using this terminology, one sees that on AB there is 1 lacunary segment CE of length 1/3 (taking AB=1), two segments FH and PR of length (1/3)^2, 4 lacunary segments of length (1/3)^3 and so forth. The sum of all these segments is thus 1/3 + 2/ 3^2 +...+2^(v-1)/ 3^v+...=1 that is, equal to the length of AB. In the method used to define the curve P, one was led to use each lacunary segment as base of an equilateral triangle; and the operation ≥Omega≤ defined in no. 1 consisted exactly in replacing a lacunary segment (FH, for example) with a broken line (FGH) made up of the two other sides of the triangle in question. Designating by ≥Lambda≤ the operation which consists in applying the operation ≥Omega≤ simultaneously to all the lacunary segments of AB, one sees that , by this operation, the segment AB is replaced by a continuous curve C1 composed of straight segments meeting at angles equal to 60 degrees. This is the curve discussed in no. 6; its equation in rectangular coordinates (with A as the origin and AB the x-axis) can be written y=f1(x), where f1(x) is the function defined in no. 4. Rather than defining P using a sequence of operations ≥Omega≤, we can now define it using a sequence of operations ≥Lambda.≤ Having replaced the segment AB with C1, using ≥Lambda,≤ one can apply ≥Lambda≤ to each of the straight segments that make up C1 and so on indefinitely. One obtains an endless sequence of curves AB, C1,... and it is easy to see that one arrives at a limiting curve identical to P. Using this method of defining P one can show simply that each point of P is a simple point of the curve or, in other words, that distinct points X, X' of AB correspond to distinct points K(X), K(X') of the curve. Let (K) XX1, X1X2, ... and (K') X'X1', X1'X2', ... be the sequences of perpendiculars defining K(X) and K(X') respectively and assume K(X) and K(X') coincide; we will conclude that X and X' must coincide also. First let us consider the case when the two sequences (K) and (K') are infinite. The point X must lie on some lacunary segment of AB (excluding the endpoints of the segments); for otherwise X would be a vertex or a limit point of vertices and then K(X)=X contrary to hypothesis. Likewise X' must lie on some lacunary segment. Then clearly X and X' must lie on the same lacunary segment; otherwise one can state, from the preceding discussion, that K(X) and K(X') lie in two pentagons having no point in common, which contradicts the hypothesis K(X)=K(X'). By definition, X1 is the point where the perpendicular XX1 meets the polygon C1: thus X1 is a point of some straight segment of C1 and the same argument as above shows X1 must be on a lacunary segment; similarly X1' must lie on a lacunary segment and one concludes as above that X1 and X1' must be on the same lacunary segment. Continuing thus one shows that however large k may be Xk and Xk' lie on the same lacunary segment. Now the segment containing X1 and X1', by the construction used, makes an angle of 60 or 120 degrees with the segment containing X and X'. Letting XX' be the distance between the points X and X' one has the relation X1X1'=2XX'; the same argument applies to the points Xk, Xk' so in general XkXk'=2X[k-1]X[k-1]' (k=2,3,...) whence XkXk'=2^k XX' which, since Xk and Xk' tend by hypothesis to the same point K(X)=K(X'), forces XX'=0 that is the points X and X' coincide. There remains the case in which at least one of the two sequences (K), (K') consists of a finite number of terms. Suppose for example the sequence (K) is made up of the k terms XX1, X1X2, ... , X[k-1]Xk and that the sequence (K') contains a number of terms (finite or infinite) <=k. As in the previous case one sees that X and X' are on the same lacunary segment, X1 and X1' are on the same lacunary segment,..., that X[k-1] and X[k-1]' are on the same lacunary segment. Suppose X and X' are distinct points; then so are X[k-1] and X[k-1]' (since X[k-1]X[k-1]'=2^(k-1) XX') and, thus, so are Xk and Xk'; now Xk=K(X) is by hypothesis a point of the set S' (that is a vertex, or limit point of vertices), Xk' cannot be a point of S' for then Xk'=K(X') which contradicts K(X')=K(X). As a result Xk' must lie on a lacunary segment. Let NT (fig. 3) be the lacunary segment containing X[k-1] and X[k- 1]'; then Xk is a point of S' found on one or the other of the sides NR and TR and Xk' lies on a lacunary segment--let's call it N1T1-- belonging to one of these sides. From an earlier argument, the points of the curve P between N1 and T1 lie in a pentagon whose boundary has no other points in common with the sides NR and TR except the points of the segment N1T1; Xk is thus separated from this boundary and cannot coincide with the point K(X') (defined by the sequence (K')) which is a point on the curve P between N1 and T1. The theorem is thus proven. Thus P is a simple continuous curve.
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