# Is $0$ a member of the following special kind of a convex compact set?

Let $$(V, \lVert \cdot \rVert)$$ be a normed space. Let us consider the set $$C = [-1,1]^{\dim V}$$. The boundary of this set consists of closed subsets $$B_i$$ (indexed by some set $$I$$) of affine hyperplanes (the family of whose we will call $$B$$). Now, consider the set $$K$$ with the following properties:

• $$K$$ is compact and convex,
• $$\forall i \in I\quad B_i \cap K \ne \emptyset$$.

## Question 1:

I would like to know whether from these properties of $$K$$ it follows that $$0 \in K$$.

It seems to be true if $$\dim V < \infty$$, with cases $$\dim V \in \{ 1,2\}$$ being rather easy to visualize. However, I wasn't able to find a simple argument which wouldn't result in checking many cases...

## Question 2:

I'm also interested in whether $$0 \in K$$ if $$B_i$$ are not necessarily subsets of a boundary of some set $$C$$, rather:

• they are closed subsets of a boundary of possibly different convex sets $$C_i$$ (in fact, we can assume that $$C_i$$ are closed balls in $$V$$),
• if $$B_i \in B$$, it follows that $$- B_i \in B$$.
• There exists a Hamel basis $$P$$ of $$V$$ such that $$p \in P \implies \exists i \in I \quad p \in B_i.$$
• $$\forall i \in I \quad 0 \notin B_i$$.

Are there any references that could help to answer these questions?

Unfortunately, these properties do not imply that $$0 \in K$$ even if $$\dim V = 3$$. The reason being that we can choose a vertex of such a cube, then take 3 vertices such that there exist an edge with that vertex and the chosen one. The convex hull of these 4 vertices do not contain $$0$$, but it does satisfy the necessary conditions.

Without loss of generality, $$K$$ is a polytope with its vertices in the $$B_i$$'s. By Farkas' lemma, it's equivalent for the vertices to reside on the same side of a subspace of codimension $$1$$. With this insight, it's easy to find counterexamples. For example, in three dimensions, we can take $$K$$ to be the convex hull of $$(1,1,1)$$, $$(-1,1,1)$$, $$(1,-1,1)$$, and $$(1,1,-1)$$: