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Prove that a segment in a convex closed set covered by closed subsets can be divided in a way that consecutive points lie in the same closed set.

Suppose tat $D \subseteq \mathbb{R}^d$ is a convex and closed set such that $D = \cup_{i = 1}^n C_i$ with $C_i$ closed. I need to show that for a segment $[x,y] \subseteq D$ I can give a partition with points $t_i$ such that $t_i,t_{i+1}$lie in the same $C_j$ for all $i$ (so $t_1,t_2$ lie in $C_1$,$t_2,t_3$ lie in $C_j'$...)

My problem

With different ideas I can start with $x \in C_1$ and get a point at the frontier $Fr(C_1)$. The problem is now how to prove that points in the frontier must be shared with another $C_i$.

Another way would be to take the maximum of $\alpha^{-1}$ ($\alpha(t) = tx+(1-t)y$) in $C_1$ and realize that $\alpha(t_{\max}+ \frac 1 n)$ lies in another $C_i$ and take limits but potential each $n$ cou.d lie in a different $C_{i_n}$

user1868607
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1 Answers1

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Clearly the proposition holds for $[x,x]$.

Assume that the proposition holds for $[x,z]\subset [x,y]$. This is, there are $x=t_1<t_2<...<t_n=z$ (the order being the linear order induced by $(1-t)x+ty$ from $[0,1]$) such that consecutive $t_i$ lie in a common $C_i$.

If $z<y$, then there is a sequence $z_1>z_2>z_3>...>z$ of points in $(z,y]$ such that $z_i\to z$. There is some $C_{i_0}$ such that infinitely many elements of this sequence lie in it. After all, the whole sequence is included in $\bigcup_{i=1}^n C_i$. So, at least one of them contains infinitely many elements of the sequence. Therefore $z\in C_{i_0}$. Take $t_{n+1}=sup_{u\in[x,y]\cap C_{i_0}}(u)$. Observe that $t_{n+1}\neq t_n$ because $C_{i_0}$ contains points of $(z,y]$. Also $t_{n+1}\in C_{i_0}$. Therefore, the proposition holds for $[x,z]$ for $z>y$.

By continuous induction the proposition holds for $[x,y]$. If uncomfortable with continuous induction, let $z$ in the paragraph above, the largest such that the proposition holds. If $z<y$ you get a contradiction by finding a larger point in which it also holds. Therefore, the maximum is $y$.