Please Email comments or suggestions to:curtin@nku.edu

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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