2

Let $f$ a continuous injection $\mathbf{R}^2 \to \mathbf{R}^2$, and $X$ a non-empty subset of $\mathbf{R}^2$ which is not contained in any straight line of the plane. Denoting with $\mathrm{cv}(Y)$ the convex hull of $Y\subseteq \mathbf{R}^2$, is it possible that $$ f[\mathrm{cv}(X)] \setminus \mathrm{cv}(f[X]) $$ is non-empty subset contained in the bourdary of convex hull of $f[X]$?

Paolo Leonetti
  • 15,554
  • 3
  • 26
  • 60

1 Answers1

3

Let $X$ be a closed square with one point removed from the interior of a boundary edge. Let $f$ take $X$ to a closed disk with one boundary point removed. Then $f(X)$ is convex, and $f(cv(X))$ is its closure (the closed disk).

Eric Wofsey
  • 342,377
  • 1
    For instance, $f$ could map the boundary of the square to the boundary of the disk, and then extend radially to the whole plane. – Eric Wofsey Mar 14 '15 at 22:07
  • 1
    I believe that Eric Wofsey's function is continuous. Where is it discontinuous? – user24142 Mar 19 '15 at 21:35
  • 2
    @jordan it is a well-known topological result that the disk and the square are homeomorphic, though people would rarely bother to write down the homeomorphism explicitly. Here is a link to a book by Gamelin, see example 6, p.209, a map contracting along radii: Send $(x,y)=(r\cos(t),r\sin(t))$ to $|\cos(t)|(x,y)$ if $|x|\ge|y|$, and to $|\sin(t)|(x,y)$ if $|y|\ge|x|$. A similar example would be to send $(x,y)$ to $(u,v)$ where $u=\dfrac{x\sqrt{\max(|x|,|y|)}}{\sqrt{x^2+y^2}}$ and $v=\dfrac{y\sqrt{\max(|x|,|y|)}}{\sqrt{x^2+y^2}}$ (and send the origin to itself). Your question has been answered. – Mirko Mar 19 '15 at 22:14
  • 1
    @jordan here is the link to the book by Gamelin (would not fit in my previous comment). Also, a closely related MSE problem: http://math.stackexchange.com/questions/365920/prove-a-square-is-homeomorphic-to-a-circle – Mirko Mar 19 '15 at 22:17
  • You are absolutely correct. My mistake was in the use of Invariance of Domain theorem, which implies that the image of the open unit square is exactly the open ball. Then the [cutted] bourdary of the square had to be mapped into the [cutted] bourdary of the circle. Here, I though no such continuous function could be constructed, and I was wrong. Thank you. – Paolo Leonetti Mar 19 '15 at 23:20