Intersection of compact convexes 
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*Let $C_1,C_2,C_3,C_4$ be compact convexes of $\mathbb{R}^2$ such that $C_1\cap C_2\cap C_3\neq\emptyset,C_1\cap C_2\cap C_4\neq\emptyset,C_1\cap C_3\cap C_4\neq\emptyset,C_2\cap C_3\cap C_4\neq\emptyset$.
Show that $C_1\cap C_2\cap C_3\cap C_4\neq\emptyset$


*Let $C_1,\dots,C_n$ be compact convexes of $\mathbb{R}^m,(m,n)\in\mathbb{N}^2$. We suppose that $\displaystyle\forall 1\le j\le n,\bigcap_{1\le i\le n,i\neq j} C_i\neq\emptyset$.
Do we have $\displaystyle\bigcap_{1\le i\le n} C_i\neq\emptyset$ ?

The first one seems quite logical by drawing a graph, but I can't find a proper mathematical 'rigorous' proof. I don't know what to do for the second.
 A: For second question, let $m=2, n=3$. We can easily draw tree circles each pair of which intersects and all three of them have empty intersection. So the answer is negative.
As for the first one, a convex set $X$ has, by definition, a property that for any two points $u, v \in X$ point $au + (1-a)v$ belongs to $X$ for any $a \in [0, 1]$. In any words, whole segment between $u$ and $v$ lays in $X$.
Let us take four points $p_1 \in C_2 \cap C_3 \cap C_4$, $p_2 \in C_1 \cap C_3 \cap C_4$, $p_3 \in C_1\cap C_2 \cap C_4$ and $p_4 \in C_1 \cap C_2 \cap C_3$.
From the definition of convex set, $[p_1, p_2] \subset C_3 \cap C_4$, $[p_1, p_3] \subset C_2 \cap C_4$, $[p_1, p_4] \subset C_2 \cap C_3$, $[p_2, p_3] \subset C_1 \cap C_4$, $[p_2, p_4] \subset C_1 \cap C_3$ and $[p_3, p_4] \subset C_1 \cap C_2$.
Next we consider the convex hull of these points. It can be a single point, a segment, a triangle or a quadrangle.


*

*If it is only one point - it is in the intersection of all sets

*If it is a segment, then all its points belongs, we can assume w.l.o.g we can assume that its ends are $p_1$ and $p_2$. So $C_1\cap C_2 \supset [p_3, p_4] \subset [p_1, p_2] \subset C_3 \cap C_4$ which means that intersection is not empty.

*For a triangle (w.l.o.g.) on $p_1, p_2, p_3$ and $p_4$ is inside of the triangle. It is easy to see that triangle's vertexes are in $C_4$, so the whole triangle belongs to $C_4$ including $p_4$ which also belongs to three other sets.

*If the convex hull is a quadrangle, then again, w.l.o.g. we can assume that $[p_1, p_2]$ and $[p_3, p_4]$ are its diagonals. And since the quadrangle is convex, the diagonals intersects. Again it is easy to verify that intersection point belongs to all sets.


q.e.d.
A: I like Valeriy's answer, but let me also point out that the first question is a special case of Helly's theorem, which says that in $\mathbb R^d$, if you have at least $d+1$ sets such that any $d+1$ of them have a non-empty intersection, then the intersection of all the sets is non-empty.  
