Context in source publication

Context 1

... n ∈ A , x n → x . If x ∈ B then as we know the contradiction is obtained. Hence x does belong to A and A is a closed set. Similarly B is closed. The curve Γ is connected and therefore either A or B is empty. It means that Γ = A or Γ = B which implies that the curve Γ either contains a circular disk ∆ x or is contained in a circular disk ∆ x and it contradicts the fact that Γ is the convex curve of constant width and has a length πd . Hence C = ∅ and there exists a point x ∈ C . Some neighborhood of x is also contained in the set C . We shall move along the curve Γ to the left from the point x to the first point y that does not belong to C . It is obviously that y cannot belong neither to A nor to B and therefore Γ = C . Then by applying the Heine-Borel lemma we conclude that Γ is a circle and theorem 1 is proved. The row of similar opened problems in the plane and in n -dimensional case appears in connection with Mizel’s problem. Problem 1. Let C be closed Jordan curve in R 2 and for arbitrary algebraic closed curve L of order n from property that intersection C ∩ L contains m points follows, that C ∩ L contains no less then m + 1 points. Does there exists a number m , that from property above follows that C be algebraic curve of order n ? Problem 2. Let in previous question L be a circle and m = 3, is it true, that C also be a circle? Problem 3. Let in problem 1 L be an ellipse and m = 4, is it true, that C also be an ellipse? Problem 4. Will be a compact C a sphere in R n , if C divides the space, and if from the belonging n + 1 tops of the arbitrary rectangular parallelepiped to a compact C , it follows that one more top lies in C too? Last question is interesting even if C be ( n − 1)-dimensional manifold or boundary of a convex set. Problem 5. Let C be a ( n − 1)-dimensional manifold (or boundary of a convex domain) in R n and not exist ( n − 1)-dimensional sphere S n − 1 that intersection C ∩ S n − 1 contains n + 1 points exactly. Is it true, that C be a ( n − 1)-dimensional sphere? Corollary. On two-dimensional plane R 2 there exist compact sets, which divide the plane and such that no circles has with them in accuracy three crosses point. In particular class of similar sets includes the Shottka sets and the Sierpiński’s carpet. Problem 6. Remain or not result of [14] true, if we will consider compact set C ⊂ R 2 , where the complement R 2 \ C is not connected? Problem 7. Are cited results and problems 1-6 true, if we will consider that one point (top) on C is fixed? Next examples will show that in problems 1-3 it is impossible instead of curve, in analogy with Tkachuk’s result, consider compact set dividing plane. Example. We shall consider the domain D on plane, bounded by circle S 1 . The Surge from it thick in the domain D infinite ensemble of opened balls D i , which are not intersect pair wise even on border and also not intersect the circle S 1 . Then we receive fractal compact set K = D ̄ \ ∪ D i without interior points, which divides plane on countable set of components. But it is easy to see that arbitrary circle can intersect K on only one point or on infinite number of points (see Figure 1). Other examples we will receive if domains D ( D i ) be domains bounded by ellipses or squares. In this case intersections with circle or ellipse can contain one, two, four or infinite many points but never three points (see Figures 2 and 3). Those examples give negative answer to problem 8 from [19]. Other close problems possible to find in the work of Grünbaum ...