There are $(d+1)f$ points (denote the set of all points as $S$) in $\mathbb{R}^d$, that can be divide into $d+1$ disjoint sets $F\_1,...,F\_{d+1}$, each set of size $f$. If we have
$$
\mathcal{H}(F\_i)\bigcap \mathcal{H}(S-F\_i)=\emptyset
$$
for all $1\leq i\leq d+1$. How can we show the following:
$$
\mathcal{H}(S)-\bigcup\_{i=1}^{d+1}\mathcal{H}(S-F\_i)\subseteq \operatorname{conv}(v\_1,...,v\_{d+1})
$$
where $v\_i\in F\_i$, $1\leq i\leq d+1$



How to show it is contained in a convex hull?
$\def\conv{\mathop{\rm conv}}\let\emp\varnothing$This is the answer if @Douglas Zare is correct, so that $\conv(\cdot)$ means the same as $\mathcal H(\{\cdot\})$. 


We omit the condition that $|F\_i|$ are the same, thus considering any disjoint finite sets $F\_i$ with $S=\bigcup\_i F\_i$. Notice that we may keep only those points of $F\_i$ which are the vertices of $T=\conv S$; this just makes the claim sharper. Due to the compactness reason, we may assume that the vertices of the dual polyhedron of $\conv $ are in general position; thus every $d$ vectors connecting a vertex of $\conv S$ with other vertices are independent.


If $|F\_i|=1$ for every $i$, the claim is trivial. So assume that $|F\_1|>1$. In this case, since $F\_1$ is separated from $F\_j$ with $j>1$, there exists an edge $uv$ of $T$ with $u,v\in F\_1$. This edge belongs to some $(d-2)$-dimensional face $P$ of $T$ which is a part of a $(d-1)$-dimensional face $Q$. 


Now rotate the hyperplane spanned by $Q$ around the affine subspace defined by $P$. Let $w$ be the first vertex we meet. Then the hyperplane $\alpha$ spanned by $P$ and $w$ separates some vertex of $Q$ from the rest vertices of $T$. Then $\alpha$ contains the vertices of at most $d-1$ of the $F\_i$ (since it contains two vertices from $F\_1$), so the whole simplex it cuts out of $T$ is contained in some $\conv (S-F\_i)$. Now we may remove this vertex and proceed by induction.


