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Let $\mathcal{Q}=\{Q_1,...,Q_n\}$ and $\mathcal{P}=\{P_1,...,P_m\}$ be partitions of $\mathbb{R}^d$ with $n<m$. Assume all $Q_k\in\mathcal{Q}$ and $P_i\in\mathcal{P}$ are close and convex sets. Then there are some $P_i,P_j\in\mathcal{P}$ and $Q_k\in\mathcal{Q}$ s.t. $\text{Int}(Q_k)\cap\text{Int}(P_i),\text{Int}(Q_k)\cap\text{Int}(P_j)\neq\emptyset$.

Note that I assume all elements in each partition have a nonempty interior and the intersection of the interior of each pair in each partition is empty.

I'm not sure about the correctness of this claim, however, it seems intuitive to me, and I believe it can be proven with relatively basic topological knowledge.

I've attempted to use the Pigeonhole principle and I think I can show that $Q_k\cap\text{Int}(P_i), Q_k\cap\text{Int}(P_j)\neq\emptyset$. However, I'm unsure how to move from here to the statement about the interior of $Q_k$.

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  • $\begingroup$ Hinr (I haven't worked out the details). If the point you found in one of the $Q$s is on the boundary then there is a nearby point in the interior. (You probably want to require as a part of your definition of a partition that the interiors of the pieces be nonempty,) $\endgroup$
    – Ethan Bolker
    Commented Jul 8 at 18:39
  • $\begingroup$ @EthanBolker Indeed I assume that the interiors are nonempty, thank you, I've edited the question. I will look into your hint. $\endgroup$
    – Staltus
    Commented Jul 8 at 18:55

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