This gives us a continuous transformation of G into itself. The implication being that, although I have formulated the Brouwer's Theorem for triangles only, it applies to a much wider class of planar sets. Loosely speaking, these are bounded closed sets with no holes . For simplicity, position D such that its base is horizontal and the third vertex is located above the base. With each such triangulation we associate a Sperner's labeling.
P is the fixed point of f.
Thus assume that this has not happened. Then the triangulation T n is assured to have a complete triangle D n . Now I am going to cut corners a little. Bounded sets in the plane have what's known as the Weierstrass-Bolzano property according to which every infinite sequence of points has at least one point near it. We have three sequences of points.