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I have been trying to solve the following problem : Does there exist two disjoint connected subsets of the closed unit disc in $\mathbb R^2$ such that one contains the points $(1,0)$ and $(-1,0)$ and the other contains the points $(0,1)$ and $(0,-1)$.

It seems to me that path connected sets will not work here. I have tried to use sets which are connected but not path connected but could not solve it . Please help me.

5xum
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user137806
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  • You are right that path connected sets do not exist. This is a pretty application of Brouwer's fixed point theorem (although it looks more elementary). I am afraid that this method will not answer your question. – Jochen Mar 25 '14 at 07:51
  • @SpamIAm : No, this only holds for OPEN sets in $\mathbb R^d$. – Jochen Mar 25 '14 at 07:52
  • @Jochen : Can you kindly tell me how do I use Brouwer's fixed point theorem to answer this problem. – user137806 Mar 25 '14 at 08:16
  • You may consider the square $Q=[-1,1]^2$ instead of the disc. Let $f,g:[-1,1] \to Q$ be continuous connecting $(0,1),(0,-1)$ and $(1,0),(-1,0)$, respectively. Assume that the curves $f$ and $g$ "do not intersect" and apply Brouwer to $F:Q\to Q$, $(s,t)\mapsto (f(s)-g(t))/|f(s)-g(t)|\infty$ where $|(a,b)|\infty= \max\lbrace |a|,|b| \rbrace$. – Jochen Mar 25 '14 at 08:57

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