955 
finite number of steps we have reached a representation «”) no more 
allowing a suchlike modification. 
We now construct on g all those polygons formed by base sides 
belonging to «®) which are represented by «w) vn a single point. 
These polygons divide u into a finite number of domains g,, J, … … Je 
Each domain g,, which by «@ is not represented nowhere dense. 
admits the property that there is no polygon lying entirely within it 
or partly within it and partly on its boundary, which is represented 
by « in a single point '). Any two base triangles belonging to the 
same domain y, can be connected within y, by a path transversing 
only base sides not represented in a single point, so that of the base 
triangles of g, either no one is represented negatively or no one 
positively. 
As each coherent part of the boundary of y, is represented on u’ 
by a single point, u’ is covered by the image of y, with a certain 
degree which we will suppose to be positive. Then there are no 
negative image triangles, but there are in general singular image 
triangles with two coinciding vertices. 
By considering each coherent part y,- of the boundary of yg, as a 
single point /,-, yg, is transformed into a sphere sp,, and we can 
deduce a simplicial division of sp, from the simplicial division of g, 
belonging to aw), by bisecting all those base sides of g, which touch 
the boundary but do not lie in the boundary, dividing by means of 
these bisecting points each base triangle one side of which lies in the 
boundary, into a triangle and a trapezium to be considered as a 
base triangle of sp,, and dividing those of the remaining base tri- 
angles of which sides have been bisected, into new base triangles 
corresponding to those bisecting points. The simplicial representation 
‘is then at the same time a simplicial representation 
of sp, on uw’, whilst by suitable subdivisions of the simplicial divisions 
aw) of g, on w 
1) For, as this property holds for polygons formed by base sides, any base 
triangle of g» possesses at most one base side represented in a single point. 
Therefore each broken line, lying in a single base triangle and not in a single 
base side, which is represented in a single point, must necessarily lie entirely in 
a straight line segment connecting two points of the circumference not coinciding 
with vertices. So a polygon represented in a single point must either consist ex- 
clusively of base sides, or it can transverse only such base sides as are represented 
in one and the same base side of x’. In the latter case however the series of the 
base triangles of . crossed in this way would have to be represented in that 
selfsame base side of yw’, so that each of the two limiting polygons of this series 
(of which at most one can be illusory) would be a polygon formed by base sides 
and represented in a single point of u’. 
