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I have this version of the theorem: let $x_1,\ldots,x_n$ $n$ points from $\mathbb{R}^N$ and $F_1,\ldots,F_n$ closed set such that $$ \operatorname{cov}\{x_{1_i},\ldots, x_{i_k}\}\subset F_{i_1}\cup\ldots\cup F_{i_k}$$ then $\bigcap_{i=1}^{i=n}\neq\emptyset$.

The proof begin as follows: suppose that $C= \operatorname{cov}\{x_{1_i},\ldots, x_{i_n}\}$ and $\bigcap_{i=1}^{i=n} F_i=\emptyset$ as $C\subset F_1\cup\ldots \cup F_n$ then $$\forall x\in C, \exists i_0\in\{1,\ldots,n\}, x\not\in F_{i_0}$$

consider continuous function defined on $C$ by $$\Phi_i(x)=d(x,F_i)$$ for $i\in\{1,\ldots,N\}$ so $$g(x)=\sum_{i=1}^{i=n} \Phi_i(x)>0$$

the function defined for all $x\in C$ by $$\Phi(x)=\dfrac{\sum_{i=1}^{i=n}\Phi_i(x) x_i}{g(x)}$$ has a fixed point $x_0\in C$

as $C\subset\bigcup_{i=1}^{i=n} F_i$ there is $j\in\{1,\ldots n\}, x_0\in F_j$ so by definition $\Phi_j(x_0)=0$

we have $$\Phi(x_0)=x_0=\dfrac{\sum_{i=1}^{i=n} \Phi_i(x)x_i}{g(x)}$$

then $x_0\in \operatorname{cov}\{x_i, i\neq j\}\subset \bigcup F_i, i\neq j$

i don't understand this last step, and how to find the contradiction ?

Thank you

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